Bundle
MATH 225
10 Items
MATH 225N Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard Deviation
MATH 225N Week 7 Assignment Conducting a Hypothesis Test for Mean – Population Standard Deviation Known P-Value Approach Question What is the p-val... [Show More] ue of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.) z−1.8−1.7−1.6−1.5−1.4 Great work! That's correct. ________________________________________ 0.084 Answer Explanation Correct answers: • 0.084 The p-value is the probability of an observed value of z=1.73 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.73, or greater than z=1.73. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.73, or to the right of z=1.73. A normal curve is over a horizontal axis and is centered on 0. Two points are labeled negative 1.73 and 1.72 The area to the right of 1.73 and to the left of negative 1.73 is shaded. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.73 is 0.042, which is just the area to the left of z=−1.73. Since the Standard Normal curve is symmetric, the area to the right of z=1.73 is 0.042 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.042)=0.084 Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance level, does the data provide sufficient evidence to conclude that Mary's mean throw is less than 61 meters? Accept or reject the hypothesis given the sample data below. • H0:μ=61 meters; Ha:μ<61 meters • α=0.01 (significance level) • z0=−1.99 • p=0.0233 ________________________________________ Great work! That's correct. ________________________________________ Reject the null hypothesis because |−1.99|>0.01. Do not reject the null hypothesis because |−1.99|>0.01. Reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Do not reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Answer Explanation Correct answer: Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.01 is less than or equal to p=0.0233, so the decision is to not reject the null hypothesis. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Marty, a typist, claims that his average typing speed is 72 words per minute. During a practice session, Marty has a sample typing speed mean of 84 words per minute based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is greater than 72 words per minute? Accept or reject the hypothesis given the sample data below. • H0:μ≤72 words per minute; Ha:μ>72 words per minute • α=0.05 (significance level) • z0=2.1 • p=0.018 Select the correct answer below: ________________________________________ Do not reject the null hypothesis because 2.1>0.05. Do not reject the null hypothesis because the value of z is positive. Reject the null hypothesis because 2.75>0.05. Reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05. Perfect. Your hard work is paying off 😀 ________________________________________ Determine the p-value for a hypothesis test for the mean (population standard deviation known) Question What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27? (Do not round your answer; compute your answer using a value from the table below.) z0.10.20.30.40.5............ Well done! You got it right. ________________________________________ 0.788 The p-value is the probability of an observed value of z=0.27 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−0.27, or greater than z=0.27. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−0.27, or to the right of z=0.27. A standard normal curve with three points labeled on the horizontal axis labeled z. The mean is labeled at 0 and observed values of negative 0.27 and 0.27 are labeled. The areas under the curve and to the left of negative 0.27 and to the right of 0.27 are shaded. The shaded areas are both labeled p-value. Using the Standard Normal Table, we can see that the p-value that corresponds with z=0.27 is 0.606, which is the area to the left of z=0.27. However, we want the area to the right of 0.27, which is 1−0.606=0.394. Because the Standard Normal curve is symmetric, the area to the left of z=−0.27 is 0.394 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.394)=0.788. Determine the p-value for a hypothesis test for the mean (population standard deviation known) Question Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ>55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places. Right-Tailed T-Table probability 0.0004 0.0014 0.0024 0.0034 0.0044 0.0054 0.0064 Degrees of Freedom 54 3.562 3.135 2.943 2.816 2.719 2.641 2.576 55 3.558 3.132 2.941 2.814 2.717 2.640 2.574 56 3.554 3.130 2.939 2.812 2.716 2.638 2.572 57 3.550 3.127 2.937 2.810 2.714 2.636 2.571 58 3.547 3.125 2.935 2.808 2.712 2.635 2.569 59 3.544 3.122 2.933 2.806 2.711 2.633 2.568 60 3.540 3.120 2.931 2.805 2.709 2.632 2.567 ________________________________________ Yes that's right. Keep it up! ________________________________________ t=3.12, p-value=0.001 Answer Explanation Correct answers: • t=3.12, p-value=0.001 Since σ is unknown and the sample size is at least 30, the hypothesis test for the mean can be performed using the t-distribution. Here, the sample mean x⎯⎯⎯ is 56,500, the hypothesized mean μ is 55,000, the sample standard deviation s is 3,750, and the sample size n is 61. Substitute these values into the formula to calculate the t test statistic. tt=56,500−55,0003,750/61‾‾‾√≈3.12 Now find the p-value. Notice that the test statistic has 61−1=60 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>55,000. Find the p-value for a right-tailed test of a t-distribution with 60 degrees of freedom, where t≈3.12. That is, to find the p-value, find the area under the t-distribution curve with 60 degrees of freedom to the right of t≈3.12. The p-value that corresponds to these conditions is approximately 0.001. Determine the p-value for a hypothesis test for the mean (population standard deviation known) Question What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.59? (Do not round your answer; compute your answer using a value from the table below.) z−1.8−1.7−1.6−1.5−1.40...... ________________________________________ Correct! You nailed it. ________________________________________ 0.112 Answer Explanation Correct answers: • 0.112 The p-value is the probability of an observed value of z=1.59 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.59, or greater than z=1.59. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.59, or to the right of z=1.59. A normal curve is over a horizontal axis and is centered on 0.00 with ticks at negative 1.59 and 1.59. The area under the curve and to the left of negative 1.59 and to the right of 1.59 is shaded. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.59 is 0.056, which is just the area to the left of z=−1.59. Since the Standard Normal curve is symmetric, the area to the right of z=1.59 is 0.056 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.056)=0.112. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session, Nancy has a sample driving distance mean of 229.6 yards based on 18 drives. At the 2% significance level, does the data provide sufficient evidence to conclude that Nancy's mean driving distance is less than 253 yards? Accept or reject the hypothesis given the sample data below. • H0:μ=253 yards; Ha:μ<253 yards • α=0.02 (significance level) • z0=−0.75 • p=0.2266 ________________________________________ Perfect. Your hard work is paying off 😀 ________________________________________ Correct answer: Do not reject the null hypothesis because the p-value 0.2266 is greater than the significance level α=0.02. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.02 is less than or equal to p=0.2266, so the decision is to not reject the null hypothesis. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives. Kathryn still claims that her average driving distance is 207 yards, and the low average can be attributed to chance. At the 1%significance level, does the data provide sufficient evidence to conclude that Kathryn's mean driving distance is less than 207 yards? Given the sample data below, accept or reject the hypothesis. • H0:μ=207 yards; Ha:μ<207 yards • α=0.01 (significance level) • z0=−1.46 • p=0.0721 ________________________________________ Keep trying - mistakes can help us grow. Correct answer: Do not reject the null hypothesis because the p-value 0.0721 is greater than the significance level α=0.01. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.01 is less than or equal to p=0.0721, so the decision is to not reject the null hypothesis. Make a conclusion and interpret the results of a one-mean hypothesis test (population standard deviation known) using the P-Value Approach Question Ruby, a bowler, has a sample game score mean of 125.8 from 25 games. Ruby still claims that her average game score is 140, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Ruby's mean game score is less than 140? Given the sample data below, accept or reject the hypothesis. • H0:μ=140; Ha:μ<140 • α=0.05 (significance level) • z0=−0.52 • p=0.3015 ________________________________________ Select the correct answer below: ________________________________________ Reject the null hypothesis because the value of z is negative. Do not reject the null hypothesis because |−0.52|>0.05. Reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Reject the null hypothesis because |−0.52|>0.05. Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05 Explanation: Decision rule using the P−value and significance level.If P−value<α: Reject H0.If P−value>α: Fail to reject H0. Decision using the P−value and significance level.We compare the P−value with the significance level (α).P−value (0.3015000000)>α (0.05)Decision: Fail to reject H0. Great work! That's correct. Correct answer: Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.05 is less than or equal to p=0.3015, so the decision is to not reject the null hypothesis. Marie, a bowler, has a sample game score mean of 129.2 from 24 games. Marie still claims that her average game score is 143, and the low average can be attributed to chance. At the 5% significance level, does the data provide sufficient evidence to conclude that Marie's mean game score is less than 143? Given the sample data below, accept or reject the hypothesis. • H0:μ=143; Ha:μ<143 • α=0.05 (significance level) • z0=−1.07 • p=0.1423 • Accept the null hypothesis • No,at 0.05 significance level, the data does not provide sufficient evidence to conclude that Marie's mean game score is less than 143 . • Explanation: • Ho:μ=143Ha:μ<143 • values are given as , • • α=0.05 (significance level) • z-statistic =−1.07 • p-value=0.1423 • • [0.1423 > 0.05] p value is greater than alpha ,so accept the null hypothesis • No,at 0.05 significance level, the data does not provide sufficient evidence to conclude that Marie's mean game score is less than 143 . Perfect. Your hard work is paying off 😀. Correct answer: Do not reject the null hypothesis because the p-value 0.1423 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, if α>p-value, reject H0 because the results of the sample data are significant. There is sufficient evidence to conclude that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct. If α≤p-value, do not reject H0. The results of the sample data are not significant, so there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be correct. In this case, α=0.05 is less than or equal to p=0.1423, so the decision is to not reject the null hypothesis. • [Show Less]
- Exam
- $9.45
- 1
- 523
MATH 225N Week 8 Assignment Coefficient of Determination
MATH 225N Week 8 Assignment Coefficient of Determination Compute and interpret the coefficient of determination Question A medical experiment on tum... [Show More] or growth gives the following data table. x y 57 38 61 50 63 76 68 97 72 113 The least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 3922.8 and the sum of squares of regression (SSR) was 3789.0. Calculate R2, rounded to three decimal places. CORRECT ANSWER: 0.966 Correct! You nailed it. R2=SSRSST R2=3789.03922.8 R2=0.966 Compute and interpret the coefficient of determination Question A scientific study on mesothelioma caused by asbestos gives the following data table. Micrograms of asbestos inhaled Area of scar tissue (cm2) 58 162 62 189 63 188 67 215 70 184 Using technology, it was determined that the total sum of squares (SST) was 1421.2 and the sum of squares due to error (SSE) was 903.51. Calculate R2 and determine its meaning. Round your answer to four decimal places. Correct answer: R2=0.3643. Great work! That's correct. Therefore, 36.43% of the variation in the observed y-values can be explained by the estimated regression equation. R2=1−SSESST R2=1−903.511421.2 R2=1−0.6357 R2=0.3643 R2=36.43%. Compute and interpret the coefficient of determination Question A new mine opened and the number of dump truck loads of material removed was recorded. The table below shows the number of dump truck loads of material removed and the number of days since the mine opened. Days (since opening) # of dump truck loads 2 45 5 53 8 60 9 60 12 67 A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 278.0 and the sum of squares of regression (SSR) was 274.3. Use these values to calculate the coefficient of determination. Round your answer to three decimal places. 0.987. Yes that's right. Keep it up! R2=SSRSST R2=274.3278.0 R2=0.987 Compute and interpret the coefficient of determination Question A new mine opened and the number of dump truck loads of material removed was recorded. The table below shows the number of dump truck loads of material removed and the number of days since the mine opened Days (since opening) # of dump truck loads 6 54 9 78 14 92 17 86 21 121 A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 2349 and the sum of squares of error (SSE) was 329. Use these values to calculate the coefficient of determination. Round your answer to three decimal places. CORRECT ANSWER: 0.860. Yes that's right. Keep it up! R2=1−SSESST R2=1−3292349 R2=1−0.140 R2=0.860 [Show Less]
- Exam
- $5.45
- 0
- 402
MATH 225N Week 8 Assignment Predictions Using Linear Regression
MATH 225N Week 8 Assignment Predictions Using Linear Regression Make predictions using a line of best fit Question The table shows data collected on... [Show More] the relationship between the time spent studying per day and the time spent reading per day. The line of best fit for the data is yˆ=0.16x+36.2. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Studying (Minutes) 50,70,90,110 Reading (Minutes) 44,48,50,54 (a) According to the line of best fit, what would be the predicted number of minutes spent reading for someone who spent 67 minutes studying? Round your answer to two decimal places. CORRECT ANSWER: 46.92. Great work! That's correct. Substitute 67 for x into the line of best fit to estimate the number of minutes spent reading for someone who spent 67 minutes studying: yˆ=0.16(67)+36.2=46.92. (b) Is it reasonable to use this line of best fit to make the above prediction? CORRECT ANSWER: The estimate, a predicted time of 46.92 minutes, is both reliable and reasonable. The data in the table only includes studying times between 50 and 110 minutes, so the line of best fit gives reliable and reasonable predictions for values of x between 50 and 110. Since 67 is between these values, the estimate is both reliable and reasonable. Yes that's right. Keep it up! Make predictions using a line of best fit Question Michelle is studying the relationship between the hours worked (per week) and time spent reading (per day) and has collected the data shown in the table. The line of best fit for the data is yˆ=−0.79x+98.8. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Hours Worked (per week) 30,40,50,60 Minutes Reading (per day) 75,68,58,52 (a) According to the line of best fit, what would be the predicted number of minutes spent reading for a person who works 27 hours (per week)? Round your answer to two decimal places, as needed. CORRECT ANSWER: 77.47 Correct! You nailed it. Substitute 27 for x into the line of best fit to estimate the number of minutes spent reading for a person who works 27 hours (per week): yˆ=−0.79(27)+98.8=77.47. (b) Is it reasonable to use this line of best fit to make the above prediction? CORRECT ANSWER: The estimate, a predicted time of 77.47 minutes, is unreliable but reasonable. Perfect. Your hard work is paying off 😀. The data in the table only includes the time worked between 30 and 60 hours, so the line of best fit gives reliable and reasonable predictions for values of x between 30 and 60. Since 27 is not between these values, the estimate is not reliable. However, 77.47 minutes is a reasonable time. [Show Less]
- Exam
- $5.45
- 0
- 335
MATH 225N Week 2 Assignment: Frequency Tables Questions and Answers
MATH 225N Week 2 Assignment: Frequency Tables Questions and Answers 1. A data set is summarized in the frequency table below. Using the table, determine t... [Show More] he number of values less than or equal to 5. Give your answer as a single number. For example if you found the number of values was 19, you would enter 19. Value Frequency 3 9 4 4 5 3 6 7 7 6 8 3 9 3 10 5 11 4 ________________________________________ Provide your answer below: 16 1. Given the frequency table, how many times does the data value 3 show up in the data set? Data Frequency 1 3 3 5 4 2 6 3 8 11 10 3 Answer : 5 2. A group of students were surveyed about the number of siblings they have. The frequencies and relative frequencies of their responses are shown in the below. Complete the cumulative relative frequency table. Number of Siblings Relative Frequency 0 0.18 1 0.33 2 0.16 3 0.14 4 or more 0.19 ________________________________________ Provide your answer below:.18, .51, .67, .81, 1.0 3. Given the relative frequency table below, which of the following is the corresponding cumulative relative frequency table? Value Frequency 4 0.35 5 0.2 6 0.05 7 0.4 HelpCopy to ClipboardDownload CSV ________________________________________ Select the correct answer below: ________________________________________ Value Frequency 4 0.3 5 0.6 6 0.65 7 1 Value Frequency 4 0.35 5 0.6 6 0.65 7 1 Value Frequency 4 0.3 5 0.55 6 0.6 7 1 Value Frequency 4 0.35 5 0.55 6 0.6 7 1 5. A group of students were surveyed about the number of books they read last summer. Their responses are summarized in the frequency table below. How many students responded to the survey? Number of Books Frequency 0−1 2 2−3 9 4−5 7 6−7 3 8−9 1 10 or more 2 ________________________________________ Provide your answer below: 24 ________________________________________ 6.The ages of the students in an art class at the community center are listed below. 9,11,14,14,16,21,24,26,32,33,37,38,38,52,53,55 Complete the frequency table. 1,4,3,5,0,3 ________________________________________ Yes that's right. Keep it up! ________________________________________ 7. A data set is summarized in the frequency table below. Using the table, determine the number of values less than or equal to 7 in the data set. Give your answer as a single number. For example if you found the number of values was 16, you would enter 16. Value Frequency 3 8 4 4 5 3 6 2 7 2 8 7 9 3 10 6 11 3 12 7 ________________________________________ Provide your answer below: 19 8. A data set is summarized in the frequency table below. Using the table, determine the number of values less than or equal to 6. Give your answer as a single number. For example if you found the number of values was 14, you would enter 14. Value Frequency 1 5 2 3 3 2 4 3 5 4 6 3 7 8 8 3 9 7 10 8 11 3 ________________________________________ Provide your answer below: 20 9. As the manager of a store, you wish to determine the amount of money that people who visit this store are willing to spend on impulse buys on products placed near the checkout register. You sample twenty individuals and records their responses. Construct a frequency table for grouped data using five classes. 8,18,15,10,29,4,15,2,4,9,16,14,13,8,25,25,27,1,15,24 ________________________________________ Provide your answer below: lower Class Limit Upper Class Limit Frequency 1 6 4 7 12 4 13 18 7 19 24 1 25 30 4 10.As a member of a marketing team, you have been tasked with determining the number of DVDs that people have rented over the past six months. Their responses are summarized in the relative frequency table below. What is the cumulative relative frequency of customers who rent 27 or fewer DVDs? ________________________________________ Provide your answer below: .85 ________________________________________ Number of DVDs Rented Relative Frequency Cumulative Frequency 10-15 0.30 0.30 16-21 0.35 0.65 22-27 0.20 28-33 0.15 1.00 11. Given the relative frequency table below, which of the following is the corresponding cumulative relative frequency table? "Value " " Frequency " "4 " " 0.28 " "5 " " 0.24 " "6 " " 0.04 " "7 " " 0.2 " "8 " " 0.24 " ________________________________________ Select the correct answer below: ________________________________________ Value Frequency 4 0.28 5 0.6 6 0.68 7 0.84 8 1 Value Frequency 4 0.28 5 0.6 6 0.64 7 0.84 8 1 Value Frequency 4 0.28 5 0.52 6 0.56 7 0.76 8 1 Value Frequency 4 0.28 5 0.56 6 0.6 7 0.8 8 1 12. Several executives were asked how many suits they own. The results are tabulated in the following frequency table. Which histogram accurately summarizes the data? "Value " " Frequency " "8 " " 6 " "9 " " 5 " "10 " " 3 " "11 " " 5 " "12 " " 3 " "13 " " 2 " ________________________________________ Select the correct answer below: ________________________________________ 13. Describe the shape of the given histogram. ________________________________________ Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed Bimodal 14. Several people were asked to report the number of hours of sleep they average per night. The results are shown in the histogram below. How many of those people average between 4.5 and 6.5 hours of sleep per night? Provide your answer below: 11 ________________________________________ $$ 15. Describe the shape of the given histogram. ________________________________________ Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed bimodal 16. The histogram below represents the prices of digital SLR camera models at a store. Describe the shape of the distribution. Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed Bimodal 17. Given the following histogram for a set of data, how many values in the data set are between 5.5 and 8.5? ________________________________________ Provide your answer below: 17 18. Describe the shape of the given histogram. ________________________________________ Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed Bimodal 19. A professor gave students a test, and the distribution of the scores of the students is shown in the histogram below. What shape does the distribution have? ________________________________________ Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed Bimodal 20. The author of a book wants to know what price his book is being sold for. He gets the price from all the bookstores in a city and creates a histogram of the results. What is the shape of the distribution? ________________________________________ Select the correct answer below: ________________________________________ uniform unimodal and symmetric unimodal and left-skewed unimodal and right-skewed bimodal 21. Given the following histogram for a set of data, how many values in the data set are between 7.5 and 9.5? ________________________________________ Provide your answer below: 4 22. Gail is a car salesperson, who keeps track of her sales over time. The line graph below shows the data for the number of cars she sells per week. At what week were her sales 8? Do not include the unit in your answer. ________________________________________ Provide your answer below: 5 23. Porter is keeping track of the total number of books he has read over time. The line graph below shows the data. How many books did Porter read from month 2 to 5? Do not include the unit in your answer. ________________________________________ Provide your answer below: 11-4 = 7 24. The bar graph below shows the number of men and women in different classes. How many total students are in the computer science class? Do not include the units in your answer. ________________________________________ Provide your answer below: 29 25. The bar graph below shows the number of men and women in different classes. A side-by-side bar graph has a horizontal axis labeled Classes with groups Chemistry and Law and a vertical axis labeled Students from 0 to 14 in increments of 2. There are two vertical bars over each horizontal axis label, with the bar on the left representing men and the bar on the right representing women. The heights of the bars are as follows, with the horizontal axis label listed first and the bar heights listed second from left to right: Chemistry, 14 and 11; and Law, 12 and 13. How many total students are in the Chemistry class? Do not include the units in your answer. ________________________________________ Provide your answer below: 25 26. An accounting manager is conducting research on how many times each accountant in the office checks their work. The following table shows the number of times each accountant checks their work. Check Work Frequency None 6 Once 3 Twice 2 Three times 0 Four times 0 27. The accounting manager encourages accountants to check their work at least two times to ensure there are no calculation errors. According to the data above, should the accounting manager be concerned about errors made in the work conducted at the office? ________________________________________ Select the correct answer below: ________________________________________ No, the accounting manager should not be concerned because everyone in the office is checking their work at least two times. No, the accounting manager should not be concerned because everyone in the office is checking their work at least three times. Yes, the accounting manager should be concerned because no one in the office is checking their work. Yes, the accounting manager should be concerned because a majority of employees are not checking their work. 28. Jackie invited her friends over for a movie night. She asked each of her friends coming over about their favorite movie snack. The following table shows the favorite movie snacks of her friends. Movie Snack Frequency Popcorn 14 Nachos 3 Candy 8 Pizza 6 Chips 3 Ice Cream 2 29. If Jackie can only get two movie snacks for a movie night with her friends, which two movie snacks should she purchase? ________________________________________ Select the correct answer below: ________________________________________ According to the data, Jackie should purchase popcorn and candy. According to the data, Jackie should purchase nachos and candy. According to the data, Jackie should purchase pizza and chips. According to the data, Jackie should purchase ice cream and chips. 30. Marc is keeping track of the total number of movies he has watched over time. The line graph below shows the data where the number of movies corresponds to the number of movies that had been watched at the beginning of the week shown on the horizontal axis. How many movies did Marc watch between the beginning of week 1 and the beginning of week 5? Do not include the unit in your answer. ________________________________________ Provide your answer below: 7 31. Josslyn is a car salesperson who keeps track of her sales over time. The line graph below shows how many cars she sells per week. What was the change in cars sold from week 2 to 6? Do not include the unit in your answer. ________________________________________ Provide your answer below: -8 (negative 8) 32. A set of data is summarized by the stem and leaf plot below. Stem1234Leaf1356671123348900002367788811122334567777 ________________________________________ Provide your answer below: ________________________________________ There are 8 values in the data set which are greater than or equal to 20 and less than or equal to 29. There are 6 values in the data set which are greater than or equal to 10 and less than or equal to 19. There are 14 values in the data set which are greater than or equal to 40 and less than or equal to 49. 33. A set of data is summarized by the stem and leaf plot below. Stem1234Leaf000011223445899990123667788890000001345567999923335567777889 Which of the following statements are true? Select all correct answers. ________________________________________ Select all that apply: ________________________________________ • The value 18 appears 0 times in the data set. • ________________________________________ • The value 39 appears 4 times in the data set. • ________________________________________ • The value 21 appears 2 times in the data set. • ________________________________________ • The value 37 appears 0 times in the data set. • ________________________________________ • The value 42 appears 1 time in the data set. • ________________________________________ • The value 22 appears 1 time in the data set. 34. A set of data is summarized by the stem and leaf plot below. Stem12Leaf11223344667789126899 ________________________________________ Provide your answer below: ________________________________________ There are 6 values in the data set which are greater than or equal to 20 and less than or equal to 29. There are 14 values in the data set which are greater than or equal to 10 and less than or equal to 19. 35. [Show Less]
- Exam
- $10.45
- 0
- 917
MATH 225N Week 5 Assignment: Evaluating Probability Using the Normal Distribution.
MATH 225N Week 5 Assignment: Evaluating Probability Using the Normal Distribution. 1. Ms. Wilson's math test scores are normally distributed with a me... [Show More] an score of 73 (μ) and a standard deviation of 5 (σ). Using the Empirical Rule, about 99.7% of the scores lie between which two values? The Empirical Rule says that 99.7% of the data lies within three standard deviations of the mean. The standard deviation is 5. So, the data that lie within three standard deviations of 73 (between −3σ and 3σ) will be the data that lie in the range that is (5)(3)=15 units less than the mean (73) and more than the mean (73). So, the values 73−15=58 and 73+15=88 are within three standard deviations of the mean. About 99.7% of the x-values lie between 58 and 88. 2. Mrs. Miller's geometry test scores are normally distributed with a mean score of 70 (μ) and a standard deviation of 3 (σ). Using the Empirical Rule, about 95% of the scores lie between which two values? The Empirical Rule says that 95% of the data lies within two standard deviations of the mean. The standard deviation is 3. So, the data that lie within two standard deviations of 70 (between −2σ and 2σ) will be the data that lie in the range that is (3)(2)=6 units less than the mean (70) and more than the mean (70). So, the values 70−6=64 and 70+6=76 are within two standard deviations of the mean. About 95% of the x-values lie between 64 and 76. 3. Ms. Wilson's statistics test scores are normally distributed with a mean score of 72 (μ) and a standard deviation of 3 (σ). Using the Empirical Rule, about 68% of the scores lie between which two values? 69 and 75. 4. Mr. Karly's math test scores are normally distributed with a mean score of 87 (μ) and a standard deviation of 4 (σ). Using the Empirical Rule, about 99.7% of the data values lie between which two values? 75−99 5. In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches. Suppose X, height in inches of adult women, follows a normal distribution. Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches? 16% 6. The number of pages per book on a bookshelf is normally distributed with mean 189 pages and standard deviation 20 pages. What is the probability that a randomly selected book has greater than 229 pages? Use the empirical rule 2.5% 7. After collecting the data, Peter finds that the standardized test scores of the students in a school are normally distributed with mean 85 points and standard deviation 3 points. Use the Empirical Rule to find the probability that a randomly selected student's score is greater than 76 points. 99.85%. [Show Less]
- Exam
- $5.45
- 0
- 728
MATH 225N Week 5 Assignment: Central Limit Theorem for Means
MATH 225N Week 5 Assignment: Central Limit Theorem for Means. 1. Question A family of statisticians is trying to decide if they can afford for their chil... [Show More] d to play youth baseball. The cost of joining a team is normally distributed with a mean of $750 and a standard deviation of $185 . If a sample of 40 teams is selected at random from the population, select the expected mean and standard deviation of the sampling distribution below. Correct answer: σx¯=$29.25 μx¯=$750 The standard deviation of the sampling distribution σx¯=σn−−√=$18540−−√=$29.25 When the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$750 . Question A cupcake baker is planning a supplies order and needs to know how much flour he needs. He knows that his recipes use an average of 100 grams of flour, normally distributed, with a population standard deviation of 15 grams. If he is consulting a sample size of 30 recipes, select the mean and standard deviation of the sampling distribution to help him order his supplies from the options below. σx¯=2.74 grams μx¯=100 grams The standard deviation of the sampling distribution is σx¯=σn−−√=1530−−√=2.74 grams Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=100 grams. Question A head librarian for a large city is looking at the overdue fees per user system-wide to determine if the library should extend its lending period. The average library user has $19.67 in fees, with a standard deviation of $7.02 . The data is normally distributed and a sample of 72 library users is selected at random from the population. Select the expected mean and standard deviation of the sampling distribution from the options below. Correct answer: σx¯=$0.83 μx¯=$19.67 The standard deviation of the sampling distribution is σx¯=σn−−√=$7.0272−−√=$0.83 When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$19.67 . 2. Question A well known social media company is looking to expand their online presence by creating another platform. They know that they current average 2,500,000 users each day, with a standard deviation of 625,000 users. If they randomly sample 50 days to analyze the use of their existing technology, identify each of the following, rounding to the nearest whole number if necessary: We are given population mean μ=2,500,000 and population standard deviation σ=625,000 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=50 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=2,500,000 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=625,00050−−√≈88,388 3. Question A bank is reviewing its risk management policies with regards to mortgages. To minimize the risk of lending, the bank wants to compare the typical mortgage owed by their clients against other homebuyers. The average mortgage owed by Americans is $306,500 , with a standard deviation of $24,500 . Suppose a random sample of 150 Americans is selected. Identify each of the following, rounding your answers to the nearest cent when appropriate: • 1306500$306500$306500 • 224500$24500$24500 • 3150$150$150 • 4306500$306500$306500 • $2000.42 We are given population mean μ=$306,500 and population standard deviation σ=$24,500 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=150 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=$306,500 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=$24,500150−−−√=$2,000.42 4. Question The average time it takes a certain brand of ibuprofen to start working is 25 minutes, with a standard deviation of 13 minutes, distributed normally. A pharmacist randomly samples 20 pills from this brand, because she is researching different brands in order to find the quickest acting ibuprofen to recommend to her customers. Identify the following to help her make her recommendations, rounding to the nearest hundredth if necessary: We are given that the population mean is μ=25 minutes and that the population standard deviation σ=13 minutes, distributed normally. We want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=20 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=25 minutes To find the standard deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=1320−−√=2.91 minutes 5. Question Major league baseball recruiters are analyzing college players as potential draft choices. In a survey of college baseball players, the recruiters found that they hit an average of 13 home runs per season, with a standard deviation of 5 . Suppose a random sample of 45 baseball players is selected. Identify each of the following and remember to round to the nearest whole number: We are given population mean μ=13 and population standard deviation σ=5 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=45 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=13 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=545−−√≈1 6. Question The average credit card debt owed by Americans is $6375 , with a standard deviation of $1200 . Suppose a random sample of 36 Americans is selected. Identify each of the following: We are given population mean μ=6375 and population standard deviation σ=1200 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=36 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=6375 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=120036−−√=200 7. Question The heights of all basketball players are normally distributed with a mean of 72 inches and a population standard deviation of 1.5 inches. If a sample of 15 players are selected at random from the population, select the expected mean of the sampling distribution and the standard deviation of the sampling distribution below. Correct answer: σx¯=0.387 inches μx¯=72 inches The standard deviation of the sampling distribution σx¯=σn√=1.51√5=0.387 inches. Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=72 inches 8. Question The owners of a baseball team are building a new baseball field for their team and must determine the number of seats to include. The average game is attended by 6,500 fans, with a standard deviation of 450 people. Suppose a random sample of 10 games is selected to help the owners decide the number of seats to include. Identify each of the following and be sure to round to the nearest whole number: We are given population mean μ=6,500 and population standard deviation σ=450 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=10 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=6,500 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=45010−−√≈142 9. Question The Washington Wheat Farmers Club is studying the impact of rising grain prices on their members' planting habits. The club members produce an average of 150 million bushels of wheat per year, with a standard deviation of 18 million bushels. The club takes a random sample of 35 years to create a statistical study. Identify each of the following, rounding to the nearest hundredth when necessary: We are given population mean μ=150 and population standard deviation σ=18 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=36 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=150 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=18/35−−√≈3.04 10. Question Given the plot of normal distributions A and B below, which of the following statements is true? Select all correct answers. A curve labeled A rises to a maximum near the left of the horizontal axis and the falls. Another curve labeled B rises to a maximum to the right of and below curve A and falls. Correct answer: B has the larger mean. B has the larger standard deviation. Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. Because B is farther to the right than A, the mean of B is greater than the mean of A. Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. Because B is shorter and more spread out than A, we find that B has the larger standard deviation. 11. Question The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the largest standard deviation. A figure consists of three curves along a horizontal axis, labeled Upper A, Upper B and Upper C. Curve Upper A is farthest to the right, curve Upper B is tall and skinny, and curve Upper C is farthest to the left. Correct answer: C Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. The distribution that is the most spread out is C, so that has the largest standard deviation. 12. Question Given the plot of normal distributions A and B below, which of the following statements is true? Select all correct answers. A figure consists of two curves labeled Upper A and Upper B. The curve Upper A is tall and evenly spread out from the center and the curve Upper is B is shorter and more spread out than A. Correct answer: The means of A and B are equal. B has the larger standard deviation. Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. Because A and B are centered at the same point, their means are equal. Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. Because B is shorter and more spread out than A, we find that B has the larger standard deviation. 13. Which of the following lists of data has the smallest standard deviation? ________________________________________ ________________________________________ 12, 12, 8, 12, 11, 12, 12, 9, 11, 12 14. Which of the following lists of data has the smallest standard deviation? 17 , 19 , 17 , 18 , 17 , 16 , 16 , 16 , 17 , 20 15. Question Given the plot of normal distributions A and B below, which of the following statements is true? Select all correct answers. A figure consists of two curves labeled Upper A and Upper B. Curve Upper A is shorter and more spread out than curve Upper B, and the curve Upper B is taller and farther to the right than curve Upper A. Correct answer: B has the larger mean. A has the larger standard deviation. Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. Because B is farther to the right than A, the mean of B is greater than the mean of A. Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. Because A is shorter and more spread out than B, we find that A has the larger standard deviation. 16. Question The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest standard deviation. A figure consists of three curves along a horizontal axis, labeled Upper A, Upper B and Upper C. Curve Upper A is evenly spread out, curve Upper B is tall and the least spread out, and curve Upper C is short and more evenly spread out from the center. B Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. The distribution that is the tallest and least spread out is B, so that has the smallest standard deviation. 17. Question The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest standard deviation. A figure consists of three curves along a horizontal axis, labeled Upper A, Upper B and Upper C. Curve Upper A is farthest to the left from the center, curve Upper B is evenly spread out to the right from the center, and curve Upper C is tall and the least spread out. C Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. The distribution that is the tallest and least spread out is C, so that has the smallest standard deviation. 18. Question The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest mean. A curve labeled B rises to a maximum and then falls. A curve labeled A rises to a maximum below and to the right of A and then falls. A curve labeled C rises to a maximum to the right of and below the maximum of A. B Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. The distribution that is farthest to the left is B, so that has the smallest mean. 19. Question The graph below shows the graphs of several normal distributions, labeled A, B, and C, on the same axis. Determine which normal distribution has the smallest mean. A figure consists of three curves along a horizontal axis, labeled Upper A, Upper B and Upper C. Curve Upper A is farthest to the left, curve Upper B is farthest to the right, and curve Upper C is tall and skinny. ________________________________________ A 20. A head football coach is concerned about the weight gain of some of his players. He finds that the weight of all football players is normally distributed with a mean of 250 pounds and a population standard deviation of 54 pounds. If the coach selects a random sample of 10 players from the population, identify the expected mean and the standard deviation of the sampling distribution below. σx¯=17.08 pounds μx¯=250 pounds 21. Question A businesswoman wants to open a coffee stand across the street from a competing coffee company. She notices that the competing company has an average of 170 customers each day, with a standard deviation of 45 customers. Suppose she takes a random sample of 31 days. Identify the following to help her decide whether to open her coffee stand, rounding to the nearest whole number when necessary: We are given population mean μ=170 and population standard deviation σ=45 , and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=31 . By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=170 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=45/31−−√≈8 22. [Show Less]
- Exam
- $10.45
- 0
- 194
MATH 225N Week 5 Assignment: Applications of the Normal Distribution
MATH 225N Week 5 Assignment: Applications of the Normal Distribution Question Sugar canes have lengths, X , that are normally distributed with mean 365... [Show More] .45 centimeters and standard deviation 4.9 centimeters. What is the probability of the length of a randomly selected cane being between 360 and 370 centimeters? • Round your answer to four decimal places. ________________________________________ The mean is μ=365.45 , and the standard deviation is σ=4.9 . As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 370 for X , 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.8234 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 360 for X , 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.1330 . Now subtract, 0.8234−0.1330=0.6904 . Thus, the probability of the length of a randomly selected cane being between 360 and 370 centimeters is 0.6904 . Question The number of miles a motorcycle, X, will travel on one gallon of gasoline is modeled by a normal distribution with mean 44 and standard deviation 5. If Mike starts a journey with one gallon of gasoline in the motorcycle, find the probability that, without refueling, he can travel more than 50 miles. • Round your answer to four decimal places. The mean is μ=44, and the standard deviation is σ=5. The probability that Mike can travel, without refueling, more than 50 miles is shown below. A normal curve is over a horizontal axis and is divided into 3 regions. Vertical line segments extend from the horizontal axis to the curve at the mean, 44 and at 50. The right region is shaded. First find the probability to the left of 50 and subtract from 1. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 50 for X, 44 for Mean, 5 for Standard_dev, and TRUE for Cumulative. This probability, rounded to four decimal places, is 0.8849. Now subtract, 1−0.8849=0.1151. Thus, the desired probability is P(X>50)=0.1151. Question A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long. • Round your answer to four decimal places. The mean is μ=14, and the standard deviation is σ=9–√=3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 12 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative. The probability, rounded to four decimal places, is P(X<12)≈0.2525. The desired probability is P(X≥12), so subtract from 1 to get P(X≥12)=1−0.2525=0.7475. 1. An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100 and a standard deviation of 15, what is the minimum IQ required for admission into the organization? Use Excel, and round your answer to the nearest integer: 126 2. The top 5% of applicants on a test will receive a scholarship. If the test scores are normally distributed with a mean of 600 and a standard distribution of 85, how low can an applicant score to still qualify for a scholarship? Use Excel, and round your answer to the nearest integer. 740 -Here, the mean, μ, is 600 and the standard deviation, σ, is 85. Let x be the score on the test. As the top 5% of the applicants will receive a scholarship, the area to the right of x is 5%=0.05. So the area to the left of x is 1−0.05=0.95. Use Excel to find x. -Open Excel. Click on an empty cell. Type =NORM.INV(0.95,600,85) and press ENTER. -The answer rounded to the nearest integer, is x≈740. Thus, an applicant can score a 740 and still be in the top 5% of applicants on a test in order to receive a scholarship. 3. The weights of oranges are normally distributed with a mean of 12.4 pounds and a standard deviation of 3 pounds. Find the minimum value that would be included in the top 5% of orange weights. Use Excel, and round your answer to one decimal place. 17.3 -Here, the mean, μ, is 12.4 and the standard deviation, σ, is 3. Let x be the minimum value that would be included in the top 5% of orange weights. The area to the right of x is 5%=0.05. So, the area to the left of x is 1−0.05=0.95. Use Excel to find x. -1. Open Excel. Click on an empty cell. Type =NORM.INV(0.95,12.4,3) and press ENTER. -The answer, rounded to one decimal place, is x≈17.3. Thus, the minimum value that would be included in the top 5% of orange weights is 17.3 pounds 4. Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean of μ=81 points and a standard deviation of σ=4 points. The middle 50% of the exam scores are between what two values? Use Excel, and round your answers to the nearest integer. 78, 84 The probability to the left of x1 is 0.25. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.25,81,4) and press ENTER. Rounding to the nearest integer, x1≈78. The probability to the left of x2 is 0.25+0.50=0.75. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.75,81,4) and press ENTER. Rounding to the nearest integer, x2≈84. Thus, the middle 50% of the exam scores are between 78 and 84. 5. The number of walnuts in a mass-produced bag is modeled by a normal distribution with a mean of 44 and a standard deviation of 5. Find the number of walnuts in a bag that has more walnuts than 80% of the other bags. Use Excel, and round your answer to the nearest integer. 48 Here, the mean, μ, is 44 and the standard deviation, σ, is 5. Let x be the number of walnuts in the bag. The area to the left of x is 80%=0.80. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.80,44,5) and press ENTER. The answer, rounded to the nearest integer, is x≈48. Thus, there are 48 walnuts in a bag that has more walnuts than 80% of the other bags. The weight of bags of green landscaping gravel, X, is modeled by normal distribution with a mean 26.7 kilograms and standard deviation 0.3 kilogram. Determine the probability that a randomly selected bag of green gravel will weigh less than 26 kilograms. • Round your answer to four decimal places. Answer Explanation Correct answers: • 0 point 0 0 9 8 $0.0098$0.0098 The mean is μ=26.7, and the standard deviation is σ=0.3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 26 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<26)≈0.0098. Question The average speed of a car on the highway is 85 kmph with a standard deviation of 5 kmph. Assume the speed of the car, X, is normally distributed. Find the probability that the speed is less than 80 kmph. • Round your answer to four decimal places. • The mean is μ=85, and the standard deviation is σ=5. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 80 for X, 85 for Mean, 5 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<80)≈0.1587. Question The time, X minutes, taken by Tim to install a satellite dish is assumed to be a normal random variable with mean 127 and standard deviation 20. Determine the probability that Tim will takes less than 150 minutes to install a satellite dish. • Round your answer to four decimal places. ________________________________________ The mean is μ=127, and the standard deviation is σ=20. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 150 for X, 127 for Mean, 20 for standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<150)≈0.8749. Question The average number of acres burned by forest and range fires in a county is 4,500 acres per year, with a standard deviation of 780 acres. The distribution of the number of acres burned is normal. What is the probability that between 3,000 and 4,800 acres will be burned in any given year? Round your answer to four decimal places. ---The mean is μ=4500 and the standard deviation is σ=780 . Because the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of the NORMDIST function, enter 4800 for X , 4500 for mean, 780 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.6497 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of the NORMDIST function, enter 3000 for X , 4500 for mean, 780 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.0272 . Now subtract: 0.6497−0.0272=0.6225 . Thus, the probability that between 3,000 and 4,800 acres will be burned in any given year is 0.6225 . The weight of bags of green landscaping gravel, X , is modeled by normal distribution with a mean 26.7 kilograms and standard deviation 0.3 kilogram. Determine the probability that a randomly selected bag of green gravel will weigh between 26.5 and 27.5 kilograms. • Round your answer to four decimal places. The mean is μ=26.7 , and the standard deviation is σ=0.3 . As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 27.5 for X , 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.9962 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 26.5 for X , 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.2525 . Now subtract, 0.9962−0.2525=0.7437 . Thus, the probability that a randomly selected bag of green gravel will weigh between 26.5 and 27.5 kilograms is 0.7437 . Question Suppose that the weight, X, in pounds, of a 40-year-old man is a normal random variable with mean 147 and standard deviation 16. Calculate P(X<185). • Round your answer to four decimal places. Correct answers: • P(X<185)= 0 point 9 9 1 2 $\text{P(X<185)=}0.9912$P(X<185)=0.9912 The mean is μ=147, and the standard deviation is σ=16. 1. Open Excel, click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 185 for X, 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<185)≈0.9912. A certain type of mango that weighs over 450 grams is considered to be large in size. The weights for this certain type of mango, X grams, are normally distributed with a mean of 400 and a standard deviation of 20. Find the probability that a mango of this type, selected at random, will not be considered as large. • Round your answer to four decimal places. The mean is μ=400, and the standard deviation is σ=20. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 450 for X, 400 for Mean, 20 for standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<450)≈0.9938. Question Suppose that the weight, X , in pounds, of a 40 -year-old man is a normal random variable with mean 147 and standard deviation 16 . Calculate P(120≤X≤153) . • Round your answer to four decimal places. Correct answers: The mean is μ=147 , and the standard deviation is σ=16 . As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 153 for X , 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.6462 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 120 for X , 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.0458 . Now subtract, 0.6462−0.0458=0.6004 . Thus, P(120≤X≤153)=0.6004 . Question A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and standard deviation 3 centimeters. Calculate the probability that the length of a component lies between 19 and 21 centimeters. • Round your answer to four decimal places. Answer Explanation Correct answers: • 00 point 0 3 8 0$0.0380$0.0380 The mean is μ=14 , and the standard deviation is σ=3 . As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx . 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 21 for X , 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X . Thus, the answer, rounded to four decimal places, is 0.9902 . 5. Click on any other empty cell. Click Insert function, fx . 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 19 for X , 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X . Thus, the answer, rounded to four decimal places, is 0.9522 . Now subtract, 0.9902−0.9522=0.0380 . Thus, the probability that the length of a component lies between 19 and 21 centimeters is 0.0380 . 6. A firm’s marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 million and a standard deviation of $250,000. Determine the sales level that has only a 3% chance of being exceeded next year. Use Excel, and round your answer to the nearest dollar.$3,670,198 Here, the mean, μ, is 3.2 million =3,200,000 and the standard deviation, σ, is 250,000. Let x be sales for next year. To determine the sales level that has only a 3% chance of being exceeded next year, the area to the right of x is 0.03. So the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,3200000,250000) and press ENTER. The answer, rounded to the nearest dollar, is x≈3,670,198. Thus, the sales level that has only a 3% chance of being exceeded next year is $3,670,198. 7. Suppose that the weight of navel oranges is normally distributed with a mean of μ=6 ounces and a standard deviation of σ=0.8 ounces. Find the weight below that one can find the lightest 90% of all navel oranges. Use Excel, and round your answer to two decimal places. 7.03 Here, the mean, μ, is 6 and the standard deviation, σ, is 0.8. The area to the left of x is 90%=0.90. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.90,6,0.8) and press ENTER. The answer, rounded to two decimal places, is x≈7.03. Thus, navel oranges that weigh less than 7.03 ounces compose the lightest 90% of all navel oranges. 8. A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 47,500 miles and a standard deviation of 3,000 miles. What mileage would correspond to the the highest 3% of the tires? Use Excel, and round your answer to the nearest integer. 53,142 Here, the mean, μ, is 47,500 and the standard deviation, σ, is 3,000. Let x be the minimum number of miles for a tire to be in the top 3%. The area to the right of x is 3%=0.03. So, the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,47500,3000) and press ENTER. The answer, rounded to the nearest integer, is x≈53,142. Thus, the approximate number of miles for the highest 3% of the tires is 53,142 miles. 1. The average credit card debt owed by Americans is $6375, with a standard deviation of $1200. Suppose a random sample of 36 Americans is selected. Identify each of the following: 1. 6375 2. 1200 3. n=36 4. 6375 5. 200 2. The heights of all basketball players are normally distributed with a mean of 72 inches and a population standard deviation of 1.5 inches. If a sample of 15 players are selected at random from the population, select the expected mean of the sampling distribution and the standard deviation of the sampling distribution below. σx¯=0.387 inches μx¯=72 inches The standard deviation of the sampling distribution σx¯=σn√=1.51√5=0.387inches. Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=72 inches. 1. After collecting the data, Peter finds that the standardized test scores of the students in a school are normally distributed with mean 85 points and standard deviation 3 points. Use the Empirical Rule to find the probability that a randomly selected student's score is greater than 76 points. Provide the final answer as a percent rounded to two decimal places. 99.85% Notice that 76 points is 3 standard deviations less than the mean. Based on the Empirical Rule, approximately 99.7% of the scores are within 3 standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the scores are less than the score that is 3 standard deviations below the mean. Alternatively, 99.85% of the scores are greater than the score that is 3 standard deviations below the mean. Therefore, the probability that a randomly selected student's score is greater than 76 points is approximately 99.85%. 2. After collecting the data, Christopher finds that the total snowfall per year in Reamstown is normally distributed with mean 94 inches and standard deviation 14 inches. Which of the following gives the probability that in a randomly selected year, the snowfall was greater than 52 inches? Use the empirical rule Provide the final answer as a percent rounded to two decimal places. 99.85% 3. The College Board conducted research studies to estimate the mean SAT score in 2016 and its standard deviation. The estimated mean was 1020 points out of 1600 possible points, and the estimated standard deviation was 192 points. Assume SAT scores follow a normal distribution. Using the Empirical Rule, about 95% of the scores lie between which two values? 636 to 1404 4. After collecting the data, Kenneth finds that the body weights of the forty students in a class are normally distributed with mean 140 pounds and standard deviation 9 pounds. Use the Empirical Rule to find the probability that a randomly selected student has a body weight of greater than 113 pounds. Provide the final answer as a percent rounded to two decimal places. 99.85% 5. Mrs. Miller's science test scores are normally distributed with a mean score of 77 (μ) and a standard deviation of 3 (σ). Using the Empirical Rule, about 68% of the scores lie between which two values? 74-80 6. Brenda has collected data to find that the finishing times for cyclists in a race has a normal distribution. What is the probability that a randomly selected race participant had a finishing time of greater than 154 minutes if the mean is 143 minutes and the standard deviation is 11 minutes? Use the empirical rule. 16% Notice that 154 minutes is one standard deviation greater than the mean. Based on the Empirical Rule, 68% of the finishing times are within one standard deviation of the mean, so 100%−68%=32% of the finishing times are outside of one standard deviation from the mean in either direction. Since the normal distribution is symmetric, half of the 32% will be less than the mean and half will be greater, so 16% of the finishing times are greater than one standard deviation more than the mean. 7. Suppose X∼N(20,2), and x=26. Find and interpret the z-score of the standardized normal random variable. 3, 20, 30 8. Isabella averages 17 points per basketball game with a standard deviation of 4 points. Suppose Isabella's points per basketball game are normally distributed. Let X= the number of points per basketball game. Then X∼N(17,4). 3, 17, 3 9. Suppose X∼N(6.5,1.5), and x=3.5. Find and interpret the z-score of the standardized normal random variable. -2, 6.5, 2 For example, let's say the mean of a normal distribution is 5 and the standard deviation is 3. Then the value 8 is one standard deviation above the mean, because 5+1⋅3=8. The value 11 is two standard deviations above the mean, because 5+2⋅3=11. The value 14 is three standard deviations above the mean, because 5+3⋅3=14. (If the value is below the mean, then we would subtract the value of standard deviation from the mean.) 10. Suppose X∼N(5.5,2), and x=7.5. Find and interpret the z-score of the standardized normal random variable. This means that x=7.5 is one standard deviation (1σ) above or to the right of the mean, μ=5.5. 11. Jerome averages 16 points a game with a standard deviation of 4 points. Suppose Jerome's points per game are normally distributed. Let X = the number of points per game. Then X∼N(16,4). "Suppose Jerome scores 10 points in the game on Monday. The z-score when x=10 is z=−1.5. This z-score tells you that x=10 is 1.5 standard deviations to the left of the mean, 16." 12. Josslyn was told that her score on an aptitude test was 3 standard deviations above the mean. If test scores were approximately normal with μ=79 and σ=9, what was Josslyn's score? Do not include units in your answer. For example, if you found that the score was 79 points, you would enter 79. 106 13. Marc's points per game of bowling are normally distributed with a standard deviation of 13 points. If Marc scores 231 points, and the z-score of this value is 4, then what is his mean points in a game? Do not include the units in your answer. For example, if you found that the mean is 150 points, you would enter 150. 179 We can think of this conceptually as well. We know that the z-score is 4, which tells us that x=231 is four standard deviations to the right of the mean, and each standard deviation is 13. So four standard deviations is (4)(13)=52 points. So, now we know that 231 is 52 units to the right of the mean. (In other words, the mean is 52 units to the left of x=231.) So the mean is 231−52=179. 14. Floretta's points per basketball game are normally distributed with a standard deviation of 4 points. If Floretta scores 10 points, and the z-score of this value is −4, then what is her mean points in a game? Do not include the units in your answer. For example, if you found that the mean is 33 points, you would enter 33. 26 We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=10 is four standard deviations to the left of the mean, and each standard deviation is 4. So four standard deviations is (−4)(4)=−16 points. So, now we know that 10 is 16 units to the left of the mean. (In other words, the mean is 16 units to the right of x=10.) So the mean is 10+16=26. 15. Jamie was told that her score on an aptitude test was 3 standard deviations below the mean. If test scores were approximately normal with μ=94 and σ=6, what was Jamie's score? Do not include units in your answer. For example, if you found that the score was 94 points, you would enter 94. 76 We can think of this conceptually as well. We know that the z-score is −3, which tells us that x is three standard deviations to the left of the mean, 94. So we can think of the distance between 94 and the x-value as (3)(6)=18. So Jamie's score is 94−18=76. 16. A normal distribution is observed from the number of points per game for a certain basketball player. If the mean is 16 points and the standard deviation is 2 points, what is the probability that in a randomly selected game, the player scored between 12 and 20 points? Use the empirical rule Provide the final answer as a percent. 95% 17. A random sample of vehicle mileage expectancies has a sample mean of x¯=169,200 miles and sample standard deviation of s=19,400 miles. Use the Empirical Rule to estimate the percentage of vehicle mileage expectancies that are more than 188,600 miles. 16% Since the mean is 169,200, a mileage of 188,600 is 188,600−169,200=19,400 miles more than the mean, which is one standard deviation. By the Empirical Rule, we know that about 68% of the data lies within one standard deviation of the mean. Therefore, 100%−68%=32% of the data lie more than one standard deviation away from the mean. Since the distribution of vehicle expectancies is symmetric, about half of the remaining 32% will lie to either extreme. So about 16% of vehicle expectancies are more than 188,600 miles. 18. A random sample of lobster tail lengths has a sample mean of x¯=4.7 inches and sample standard deviation of s=0.4 inches. Use the Empirical Rule to determine the approximate percentage of lobster tail lengths that lie between 4.3 and 5.1 inches. Round your answer to the nearest whole number (percent). 68% 19. A random sample of SAT scores has a sample mean of x¯=1060 and sample standard deviation of s=195. Use the Empirical Rule to estimate the approximate percentage of SAT scores that are less than 865. Round your answer to the nearest whole number (percent). 16% 20. The number of pages per book on a bookshelf is normally distributed with mean 248 pages and standard deviation 21 pages. Using the empirical rule, what is the probability that a randomly selected book has less than 206 pages? 2.5% The Empirical Rule states: For a normal distribution, nearly all of the data will fall within 3 standard deviations of the mean. The empirical rule can be broken down into three parts: 68% of data falls within one standard deviation from the mean (between −1σ and 1σ). 95% fall within two standard deviations from the mean (between −2σ and 2σ). 99.7% fall within three standard deviations from the mean (between −3σ and 3σ). The empirical rule is also known as the 68−95−99.7 Rule. 21. Mr. Karly's math test scores are normally distributed with a mean score of 87 (μ) and a standard deviation of 4 (σ). Using the Empirical Rule, about 99.7% of the data values lie between which two values? 75-99% 22. In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches. Suppose X, height in inches of adult women, follows a normal distribution. Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches? 16% 23. A normal distribution is observed from the number of points per game for a certain basketball player. The mean for this distribution is 20 points and the standard deviation is 3 points. Use the empirical rule for normal distributions to estimate the probability that in a randomly selected game the player scored less than 26 points. Provide the final answer as a percent rounded to one decimal place. 97.5% 24. A normal distribution is observed from the number of points per game for a certain basketball player. If the mean is 15 points and the standard deviation is 3 points, what is the probability that in a randomly selected game, the player scored greater than 24 points? Use the empirical rule 15% 25. The College Board conducted research studies to estimate the mean SAT score in 2016 and its standard deviation. The estimated mean was 1020 points out of 1600 possible points, and the estimated standard deviation was 192 points. Assume SAT scores follow a normal distribution. Using the Empirical Rule, about 95% of the scores lie between which two values? 636-1404 26. The typing speeds for the students in a typing class is normally distributed with mean 44 words per minute and standard deviation 6 words per minute. What is the probability that a randomly selected student has a typing speed of less than 38 words per minute? Use the empirical rule Provide the final answer as a percent. If necessary round the percent to the nearest whole number. 16% Notice that 38 words per minute is one standard deviation less than the mean. Based on the Empirical Rule, 68% of the typing speeds are within one standard deviation of the mean. Since the normal distribution is symmetric, this implies that 16% of the typing speeds are less than one standard deviation less than the mean. 27. Nick has collected data to find that the body weights of the forty students in a class has a normal distribution. What is the probability that a randomly selected student has a body weight of greater than 169 pounds if the mean is 142 pounds and the standard deviation is 9 pounds? Use the empirical rule. .15% Notice that 169 pounds is three standard deviations greater than the mean. Based on the Empirical Rule, 99.7% of the body weights are within three standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the body weights are greater than three standard deviation more than the mean. The pink section represents the 68% of the data that lies within one standard deviation of the mean. This section is split in half, with 34% to the left of the mean and 34% to the right of the mean. The smaller red sections (marked with 13.5%), along with the inner pink sections, represent the 95% of the data that lies within two standard deviations of the mean. (If we add the four sections together, we will get 95%: 13.5%+34%+34%+13.5%=95%.) The small blue sections (marked with 2.35%), along with the four inner sections, represent the 99.7% of the data that lies within three standard deviations of the mean. (If we add the sections together, we will get 99.7%: 2.35%+13.5%+34%+34%+13.5%+2.35%=99.7%.) *Remember: 50% of the data lie above the mean, and 50% of the data lie below the mean in normal distributions. So, if we have a normally distributed data set, with a mean of 10.5, then we can say that 50% of the data is less than 10.5 and 50% of the data is greater than 10.5. 28. The times to complete an obstacle course is normally distributed with mean 73 seconds and standard deviation 9 seconds. What is the probability using the Empirical Rule that a randomly selected finishing time is less than 100 seconds? 99.85% 29. After collecting the data, Douglas finds that the finishing times for cyclists in a race is normally distributed with mean 149 minutes and standard deviation 16 minutes. What is the probability that a randomly selected race participant had a finishing time of less than 165 minutes? Use the empirical rule 84% 30. Charles has collected data to find that the total snowfall per year in Reamstown has a normal distribution. Using the Empirical Rule, what is the probability that in a randomly selected year, the snowfall was less than 87 inches if the mean is 72 inches and the standard deviation is 15 inches? 84% 31. Christopher has collected data to find that the total snowfall per year in Laytonville has a normal distribution. What is the probability that in a randomly selected year, the snowfall was greater than 53 inches if the mean is 92 inches and the standard deviation is 13 inches? Use the empirical rule Notice that 53 inches is three standard deviations less than the mean. Based on the Empirical Rule, 99.7% of the yearly snowfalls are within three standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the yearly snowfalls are less than three standard deviations below the mean. Alternatively, 99.85% of the yearly snowfalls are greater than three standard deviations below the mean. 32. The times to complete an obstacle course is normally distributed with mean 87 seconds and standard deviation 7 seconds. What is the probability that a randomly selected finishing time is greater than 80 seconds? Use the empirical rule Alternatively, 84% of the finishing times are greater than one standard deviation below the mean. 33. 1. Sugar canes have lengths, X, that are normally distributed with mean 365.45 centimeters and standard deviation 4.9centimeters. What is the probability of the length of a randomly selected cane being between 360 and 370 centimeters? • Round your answer to four decimal places. .6904 2. On average, 28 percent of 18 to 34 year olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a random variable X, which has a standard deviation of five percent. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32. • Round your answer to four decimal places. .7881 • The mean is μ=28, and the standard deviation is σ=5. • 1. Open Excel and click on any empty cell. Click Insert function, fx. • 2. Search for NORMDIST in the search for a function dialog box and click GO. • 3. Make sure NORMDIST is on top in select a function. Then click OK. • 4. In the function arguments of the NORMDIST function, enter 32 for X, 28 for Mean, 5 for Standard_dev, and TRUE for Cumulative. • Thus, the answer, rounded to four decimal places, is P(X<32)≈0.7881. 3. In a survey of men aged 20-29 in a country, the mean height was 73.4 inches with a standard deviation of 2.7 inches. Find the minimum height in the top 10% of heights. • Use Excel, and round your answer to one decimal place. 76.9 4. Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean of μ=81 points and a standard deviation of σ=4 points. The middle 50% of the exam scores are between what two values? • Use Excel, and round your answers to the nearest integer. 78-84 [Show Less]
- Exam
- $10.45
- 0
- 321
MATH 225N Week 7 Discussion: Hypothesis Testing
MATH 225N Week 7 Discussion: Hypothesis Testing Initial Post Instructions Describe a hypothesis test study that would help your work or conclusions in so... [Show More] me way. Describe what variable would be tested and what would be your guess of the value of that variable. Then include how the result, if the null were rejected or not, might change your conclusions or actions in some way. NB: 2 Answers Displayed Answer 1 We interpret data by assuming a specific structure or outcome and use statistical methods to confirm or reject the assumption. According to Opens tax, the process begins with a model, is build. This theory, if it has validity, will lead to predictions; what we call hypothesis. (Openstax, pg. 381-382) There are two types of statistical hypotheses. Null hypothesis, denoted by H0, is usually the hypothesis that sample observations result purely from chance. Alternative Hypotheses, is denoted by H1 or Ha, is the hypothesis that sample observations and are influenced by some non-random cause. (Star Trek)A hypothesis test can result in one of two way, either rejecting the null hypothesis or claim or by not rejecting the null hypothesis. The variable that I feel would help my work in some way would be intramuscular antibiotics shorten length of illness to 3 days. For my chosen variable, the null hypothesis is represented as H0 = 3 and the alternative hypothesis, intravenous antibiotics do not shorten illness length to 3 days, as Ha ≠ 3. I chose this variable and value based on my husband’s recent experience with an intramuscular antibiotic used to treat his strep throat. He received an injection on 2-10-2019 and today, 2-12-2019, he states feeling back to normal without the presence of a fever, sore throat, or fatigue. Considering all the various types of infectious diseases processes there are, my hypothesis would likely result in a rejection of the null hypothesis. Some infections are quite serious and remain present in a person’s body for longer periods of time. There are also some individuals that avoid seeking medical attention when a change in health status appears, resulting in the ability of an infection to colonize for a longer period. If I were to conduct this hypothesis test, I was specifically focus on cases of strep throat and I would also limit my study population to those individuals that sought medical attention when symptoms first presented. By controlling these two areas, there is a higher chance of obtaining accurate results and determining the validity of my claim. References What Is Hypothesis Testing? (2019, December 5). Retrieved from http://stattrek.com/hypothesis-test/hypothesis-testing.aspx (Links to an external site.) Holmes, A., Illowsky, B., & Dean, S. (2017). Introductory Business Statistics. Houston, TX: OpenStax CNX. Retrieved from https://openstax.org/details/books/introductory-business-statistics Answer 2 I could use what we learned this week to test a hypothesis regarding whether a medication helps to reduce length of hospital stay for patients after cardiac surgery. I reviewed a study from 2006 that researched length of stay after cardiac surgery. In a sample of 1620 patients, the mean LOS for female patients with an age >65 was 12 days. The average length of stay can be drastically altered by many things, most commonly postoperative atrial fibrillation and pleural effusions (Rosborough, 2006). I will use the following data to determine if adding a new medication reduces length of hospital stay. The fictional new medication will be called (Med-X). For statistical example I will use the following fictional values: Established length of stay: 5.5 Days SD: 1.5 Days Established length of stay after Med-X added: 4.7 Days: Sample Size – 25 Patients Significance level: 95% With this data I would write a hypothesis as such: Null Hypothesis: Mean length of stay = 5.5 Days Alternative Hypothesis: Mean length of stay < 5.5 Days The critical value for this hypothesis would be -1.64. I would calculate a test statistic with the following equation: Zc = (4.7 – 5.5) / (1.5 / Square root of 25). This calculates to -2.67. Because this value is less than the critical value (-1.64), and it is a left-tailed test, I could say with 95% confidence that Med-X does reduce average length of stay after open heart surgery. I could then implement Med-X postoperatively for all cardiac surgery patients. This is not to say with absolute certainty that this is correct; a type one error could be made. A type one error would result in rejecting the null hypothesis when in fact it is true. If this were the case I could incorrectly assume average LOS = 4.7 when in fact the null hypothesis is true and average LOS = 5.5. This may result in planning for shorter stays and being inadequately supplied and staffed for the patient’s actual length of stay. [Show Less]
- Exam
- $6.45
- 1
- 774
MATH 225N Week 7 Assignment Conduct a Hypothesis Test for Proportion – P-Value Approach
MATH 225N Week 7 Assignment Conduct a Hypothesis Test for Proportion – P-Value Approach Question A college administrator claims that the proportion o... [Show More] f students that are nursing majors is greater than 40%. To test this claim, a group of 400 students are randomly selected and its determined that 190 are nursing majors. The following is the setup for this hypothesis test: H0:p=0.40 Ha:p>0.40 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve: Correct answers: • P-value=0.001 Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: 1. Determine if the hypothesis test is left tailed, right tailed, or two tailed. 2. Compute the value of the test statistic. 3. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0 If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0 If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a right tailed test and the test statistic, rounding to two decimal places, is z=0.475−0.400.40(1−0.40)400‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈3.06. Thus the p-value is the area under the Standard Normal curve to the right of a z-score of 3.06. From a lookup table of the area under the Standard Normal curve, the corresponding area is then 1 - 0.999 = 0.001. z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 3.0 0.999 0.999 0.999 0.999 0.999 0.999 0.999 0.999 0.999 0.999 Required p-value = 0.001 Explanation: Formula to calculate the test statistic z is z=((1−p)∗p)/np^−p where p^=x/n=190/400=0.475,p=0.40,n=400 →((1−0.4)∗0.4)/4000.475−0.4 →0.0244950.075 ⇒3.06 P(z>3.06) = 1-P(z<3.06) ⇒ 1 - 0.999 [Find 3.0 in row and 0.06 in column in above table] ⇒ 0.001 Hence, p-value is 0.001 Determine the p-value for a hypothesis test for proportion Question A police officer claims that the proportion of accidents that occur in the daytime (versus nighttime) at a certain intersection is 35%. To test this claim, a random sample of 500 accidents at this intersection was examined from police records it is determined that 156 accidents occurred in the daytime. The following is the setup for this hypothesis test: H0:p = 0.35 Ha:p ≠ 0.35 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve: Perfect. Your hard work is paying off 😀 Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: 1. Determine if the hypothesis test is left tailed, right tailed, or two tailed. 2. Compute the value of the test statistic. 3. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0 If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0 If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a two tailed test and the test statistic, rounding to two decimal places, is z=0.312−0.350.35(1−0.35)500‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈1.78. Thus the p-value is the area under the Standard Normal curve to the left of a z-score of -1.78, plus the area under the Standard Normal curve to the right of a z-score of 1.78 z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 1.7 0.045 0.044 0.043 0.042 0.041 0.040 0.039 0.038 0.038 0.037 From a lookup table of the area under the Standard Normal curve, the corresponding area is then 2(0.038) = 0.076. Make a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach Question A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher administers the test to 200 random students and determines that 151 students pass the exam. The following is the setup for this hypothesis test: H0:p=0.80 Ha:p>0.80 In this example, the p-value was determined to be 0.944. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) Correct! You nailed it. To come to a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach, the first step is to compare the p-value from the sample data with the level of significance. The decision criteria is then as follows: If the p-value is less than or equal to the given significance level, then the null hypothesis should be rejected. So, if p≤α, reject H0; otherwise fail to reject H0. When we have made a decision about the null hypothesis, it is important to write a thoughtful conclusion about the hypotheses in terms of the given problem's scenario. Assuming the claim is the null hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to reject the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to reject the claim. Assuming the claim is the alternative hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to support the claim. if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to support the claim In this example the p-value = 0.944. We then compare the p-value to the level of significance to come to a conclusion for the hypothesis test. In this example, the p-value is greater than the level of significance which is 0.05. Since the p-value is greater than the level of significance, the conclusion is to fail to reject the null hypothesis. Make a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach Question A police office claims that the proportion of people wearing seat belts is less than 65%. To test this claim, a random sample of 200 drivers is taken and its determined that 126 people are wearing seat belts. The following is the setup for this hypothesis test: H0:p=0.65 Ha:p<0.65 In this example, the p-value was determined to be 0.277. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) Perfect. Your hard work is paying off 😀 Correct answer: The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to support the claim. To come to a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach, the first step is to compare the p-value from the sample data with the level of significance. The decision criteria is then as follows: If the p-value is less than or equal to the given significance level, then the null hypothesis should be rejected. So, if p≤α, reject H0; otherwise fail to reject H0. When we have made a decision about the null hypothesis, it is important to write a thoughtful conclusion about the hypotheses in terms of the given problem's scenario. Assuming the claim is the null hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to reject the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to reject the claim. Assuming the claim is the alternative hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to support the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to support the claim. In this example the p-value = 0.277. We then compare the p-value to the level of significance to come to a conclusion for the hypothesis test. In this example, the p-value is greater than the level of significance which is 0.05. Since the p-value is greater than the level of significance, the conclusion is to fail to reject the null hypothesis. Determine the p-value for a hypothesis test for proportion Question A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher administers the test to 200 random students and determines that 151 students pass the exam. The following is the setup for this hypothesis test: H0:p=0.80 Ha:p>0.80 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve: Great work! That's correct. Correct answers: • P-value=0.944 Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: 1. Determine if the hypothesis test is left tailed, right tailed, or two tailed. 2. Compute the value of the test statistic. 3. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0 If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0 If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a right tailed test and the test statistic, rounding to two decimal places, is z=0.755−0.800.80(1−0.80)200‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈−1.59. Thus the p-value is the area under the Standard Normal curve to the right of a z-score of -1.59. From a lookup table of the area under the Standard Normal curve, the corresponding area is then 1 - 0.056 = 0.944. Determine the p-value for a hypothesis test for proportion Question A business owner claims that the proportion of take out orders is greater than 25%. To test this claim, the owner checks the next 250 orders and determines that 60 orders are take out orders. The following is the setup for this hypothesis test: H0:p=0.25 Ha:p>0.25 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve: Perfect. Your hard work is paying off 😀 Correct answers: • P-value=0.643 Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: 1. Determine if the hypothesis test is left tailed, right tailed, or two tailed. 2. Compute the value of the test statistic. 3. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0 If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0 If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a right tailed test and the test statistic, rounding to two decimal places, is z=0.24−0.250.25(1−0.25)250‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√≈−0.37. Thus the p-value is the area under the Standard Normal curve to the right of a z-score of -0.37. From a lookup table of the area under the Standard Normal curve, the corresponding area is then 1 - 0.357 = 0.643 Make a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach Question A police officer claims that the proportion of accidents that occur in the daytime (versus nighttime) at a certain intersection is 35%. To test this claim, a random sample of 500 accidents at this intersection was examined from police records it is determined that 156 accidents occurred in the daytime. The following is the setup for this hypothesis test: H0:p = 0.35 Ha:p ≠ 0.35 In this example, the p-value was determined to be 0.075. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) Correct answer: The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to reject the claim. To come to a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach, the first step is to compare the p-value from the sample data with the level of significance. The decision criteria is then as follows: If the p-value is less than or equal to the given significance level, then the null hypothesis should be rejected. So, if p≤α, reject H0; otherwise fail to reject H0. When we have made a decision about the null hypothesis, it is important to write a thoughtful conclusion about the hypotheses in terms of the given problem's scenario. Assuming the claim is the null hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to reject the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to reject the claim. Assuming the claim is the alternative hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to support the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to support the claim. In this example the p-value = 0.075. We then compare the p-value to the level of significance to come to a conclusion for the hypothesis test. In this example, the p-value is greater than the level of significance which is 0.05. Since the p-value is greater than the level of significance, the conclusion is to fail to reject the null hypothesis.Yes that's right. Keep it up! Make a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach Question A medical researcher claims that the proportion of people taking a certain medication that develop serious side effects is 12%. To test this claim, a random sample of 900 people taking the medication is taken and it is determined that 93 people have experienced serious side effects. . The following is the setup for this hypothesis test: H0:p=0.12 Ha:p≠0.12 In this example, the p-value was determined to be 0.124. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%) Perfect. Your hard work is paying off 😀 Correct answer: The decision is to fail to reject the Null Hypothesis. The conclusion is that there is not enough evidence to reject the claim. To come to a conclusion and interpret the results for a hypothesis test for proportion using the P-Value Approach, the first step is to compare the p-value from the sample data with the level of significance. The decision criteria is then as follows: If the p-value is less than or equal to the given significance level, then the null hypothesis should be rejected. So, if p≤α, reject H0; otherwise fail to reject H0. When we have made a decision about the null hypothesis, it is important to write a thoughtful conclusion about the hypotheses in terms of the given problem's scenario. Assuming the claim is the null hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to reject the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to reject the claim. Assuming the claim is the alternative hypothesis, the conclusion is then one of the following: • if the decision is to reject the null hypothesis, then the conclusion is that there is enough evidence to support the claim. • if the decision is to fail to reject the null hypothesis, then the conclusion is that there is not enough evidence to support the claim. In this example the p-value = 0.124. We then compare the p-value to the level of significance to come to a conclusion for the hypothesis test. In this example, the p-value is greater than the level of significance which is 0.05. Since the p-value is greater than the level of significance, the conclusion is to fail to reject the null hypothesis. [Show Less]
- Exam
- $10.45
- 0
- 579
MATH 225N Week 8 Assignment: Correlation and Causation
MATH 225N Week 8 Assignment: Correlation and Causation Understand the difference between correlation and causation Question True or False: The more sa... [Show More] mples taken in a scientific study, the longer the amount of time it will take to complete the research on the samples. Although there are other factors that affect study time, such as experience and equipment, increasing the number of samples will cause an increase in research time. Correct! You nailed it. Understand the difference between correlation and causation Question Suppose that a large controlled experiment tests whether caffeine improves reaction times. A very large number of randomly selected participants are randomly given identical-seeming pills with varying doses of caffeine (including none) and then given tests of reaction times under the same conditions. The experiment finds a strong negative correlation between caffeine dose and reaction time. (Note that lower reaction times are better.) Correct answer: There is evidence that caffeine causes lower (better) reaction times. Correlation alone does not prove causation, but this scenario provides more evidence than just correlation between two variables. Since the data were obtained from an appropriately randomized controlled experiment, a correlation can be used as evidence of a causal relationship. Since all other variables were controlled, there is no third variable that could be associated with caffeine that actually causes differences in reaction times. Yes that's right. Keep it up! Interpret the slope and y-intercept of the least squares regression line Question Which of the following situations could have a regression line with a negative y-intercept? ________________________________________ Select all that apply: Great work! That's correct. ________________________________________ • The present value of an insurance policy as a function of the age of the ensured person if the person purchases the policy when he or she is 50 years old. • ________________________________________ • The number of millions of barrels of oil in an oil field as a function of the number of years the oil field has been in production. • ________________________________________ • The value of a Fortune-500 company, founded in the year 1990, as a function of the number of years after the year 2010. • ________________________________________ • The value of a rail locomotive as a function of the number of years since it was put into service. Interpret the slope and y-intercept of the least squares regression line Question Which of the following data sets or plots could have a regression line with a negative slope? Select all that apply. Perfect. Your hard work is paying off 😀 ________________________________________ • The number of tons of trash in a landfill as a function of the number of years since the landfill was built. • ________________________________________ • The number of loads of trash a dump truck hauls per month as a function of the number of years since the dump truck was built. • ________________________________________ • The number of tons of trash produced in a city as a function of the population. • ________________________________________ • The number of minutes you will wait to drop off your trash at a landfill as a function of the number of dump trucks at the landfill Correct answer: The slope is related to the increase or decrease of the dependent variable as a function of the independent variable. If the dependent variable can decrease, then the slope can be negative. Understand the difference between correlation and causation Question Suppose that data collected from police reports of motor vehicle crashes show a moderate positive correlation between the speed of the motor vehicle at the time of the crash and the severity of injuries to the driver. Answer the following question based only on this information. True or false: It can be concluded that the faster a motor vehicle is traveling at the time of a crash, the more severe the injuries to the driver are. Well done! You got it right.Correlation does not prove causation. The provided information shows that there is a positive association between speed and severity of injuries, but that information alone is not sufficient to conclude that greater speed causes more severe injuries. Based only on this information, there could be a third variable associated with speed that causes more severe injuries in crashes. That said, the statement "the faster a motor vehicle is traveling at the time of a crash, the more severe the injuries to the driver are" does not imply a causal relationship between speed and severity of injuries. Understand the difference between correlation and causation Question An experiment finds that runners with the longest legs complete a mile run in less time. Identify the relation between leg length and time of a mile run. Leg length and time of a mile run are negatively correlated. A longer leg length is associated with a lower time of a mile run, which implies a negative relationship. There would need to be more evidence to prove causation.Yes that's right. Keep it up! Interpret the slope and y-intercept of the least squares regression line Question Which of the following situations would have data sets or plots that could have a regression line with a positive slope? Select all that apply. Great work! That's correct. Correct answer: The cumulative number of ships launched by ship builders as a function of the number of months since the start of the year The number of cars produced as a function of the number of days since the end of the last strike. The total shipping orders that are backlogged as a function of the number of ships in drydock. The slope is related to the increase or decrease of the dependent variable as a function of the independent variable. If the dependent variable can increase, then the slope can be positive, such as with the cumulative number of ships launched. Interpret the slope and y-intercept of the least squares regression line Question Which of the following situations could produce data sets or plots would have a regression line with a negative slope? Select all that apply. Well done! You got it right. Correct answer: The number of miles a ship travels each year as a function of the number of years since it was launched. the number of voyages a ship can make per year as a function of the length (in days) of each voyage. The slope is related to the increase or decrease of the dependent variable as a function of the independent variable. If the dependent variable can decrease, then the slope can be negative, such as with the number of cats born each year. . \ [Show Less]
- Exam
- $5.45
- 0
- 380
Purchase the bundle to get the full access instantly
- 3 Million+ Study Documents
- 100% Money Back Guarantee
- Immediate Download Access
Add To Cart
Reviews 0
Bundle Details
$16.45
- Trusted by 800,000+ Students
- 24/7 Money Back Guarantee
- Download is directly available
309
0
Seller Details
