MATH 225N Week 5 Assignment: Understanding Normal Distribution
1. Lexie averages 149 points per bowling game with a standard deviation of 14 points.
... [Show More] Suppose Lexie's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(149,14).
Suppose Lexie scores 186 points in the game on Tuesday. The z-score when x = 186 is 2.643 - no response given. The mean is 149 - no response given.
This z-score tells you that x = 186 is 2.643- no response given standard deviations to the right of the mean.
The z-score can be found using this formula:
z=x−μσ=186−149/14=3714≈2.643
The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, scoring 186 points is 2.643 standard deviations away from the mean. A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem: μ=149 points in the game.
2. Suppose X∼N(18,2), and x=22. Find and interpret the z-score of the standardized normal random variable.
The z-score when x=22 is . The mean is .
This z-score tells you that x=22 is standard deviations to the right of the mean.
3. Suppose X∼N(12.5,1.5), and x=11. Find and interpret the z-score of the standardized normal random variable.
X is a normally distributed random variable with μ=12.5 (mean) and σ=1.5 (standard deviation). To calculate the z-score,
z=x−μσ=11−12.51.5=−1.51.5=−1
This means that x=11 is one standard deviation (1σ) below or to the left of the mean. This makes sense because the standard deviation is 1.5. So, one standard deviation would be (1)(1.5)=1.5, which is the distance between the mean (μ=12.5) and the value of x (11).
5. 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
6. Suppose X∼N(13.5,1.5), and x=9. Find and interpret the z-score of the standardized normal random variable.-3,13.5,3
7.
Suppose X∼N(10,0.5), and x=11.5. Find and interpret the z-score of the standardized normal random variable.
3,10,3
This means that x=11.5 is three standard deviations (3σ) above or to the right of the mean. This makes sense because the standard deviation is 0.5. So, three standard deviations would be (3)(0.5)=1.5, which is the distance between the mean (μ=10) and the value of x (11.5).
8. Annie averages 23 points per basketball game with a standard deviation of 4 points. Suppose Annie's points per basketball game are normally distributed. Let X= the number of points per basketball game. Then X∼N(23,4).
2.75,23,2.75
9. Suppose X∼N(9,1.5), and x=13.5. Find and interpret the z-score of the standardized normal random variable.
3,9,3
10. Rosetta averages 148 points per bowling game with a standard deviation of 14 points. Suppose Rosetta's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(148,14).
2.714,148,2.714
11. Suppose X∼N(5.5,2), and x=7.5.
This means that x=7.5 is one standard deviation (1σ) above or to the right of the mean, μ=5.5.
12. 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 _______. This z-score tells you that x=10 is _______ standard deviations to the _______(right /left) of the mean, _______.
"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."
13. John averages 58 words per minute on a typing test with a standard deviation of 11 words per minute. Suppose John's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then X∼N(58,11).
1.272,58,1.272
14. Suppose X∼N(16.5,0.5), and x=16.
-1,16.5,1
15. Gail averages 64 words per minute on a typing test with a standard deviation of 9.5 words per minute. Suppose Gail's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then X∼N(64,9.5).
The z-score can be found using this formula:
z=x−μσ=89−64/9.5=259.5≈2.632
The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, typing 89 points is 2.632 standard deviations away from the mean. A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem: μ=64 words per minute in a typing test. 2.632,64, 2.632
16. William averages 58 words per minute on a typing test with a standard deviation of 10.5 words per minute. Suppose William's words per minute on a typing test are normally distributed. Let X= the number of words per minute on a typing test. Then X∼N(58,10.5).
2.571,58,2.571
17. Hugo averages 22 points per basketball game with a standard deviation of 4 points. Suppose Hugo's points per basketball game are normally distributed. Let X= the number of points per basketball game. Then X∼N(22,4).
-3.75,22,3.75
18. 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. Let x=68, the height of a woman who is 5' 8" tall. Find and interpret the z-score of the standardized normal random variable.
This means that x=68 is one standard deviation (1σ) above or to the right of the mean, μ=64.
19. Suppose X∼N(16.5,2), and x=18.5.
1,16.5,1
20.
Suppose X∼N(6.5,1.5), and x=3.5.
-2,6.5,2
21. Suppose X∼N(20,2), and x=26.
3,20,3
22. Isabella averages 152 points per bowling game with a standard deviation of 14.5 points. Suppose Isabella's points per bowling game are normally distributed. Let X= the number of points per bowling game. Then X∼N(152,14.5).
2.414,152,2.414 [Show Less]