A total of 4700 fast food items were sold during the month.
a.) How many were fsh?
b.) How many were french fries?
Your Answer:
For this problem:
a.
... [Show More] For Fish there were: 4700 ( .28)=1316, hovewer for Franch Fries they were: 4700 (.40)=1880
a.) Fish : 4700(.28) = 1316
b.) French Fries: 4700(.40) = 1880
Queston 2
10 / 10 pts
You may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
Consider the following data:
19 46 74 40 44 65 33 76 50 58 31 37 70 41 61 51 56 73 48 55
Find the 40th percentle of this data.
Your Answer:
For this problem:n= 20, since there are 20 numbers.
The numbers in order are: 19 31 33 37 40 41 44 46 48 50 51 55 56 58 61 65 70 74 76
Since we are looking for 40th percentle I must calculate:
Finally, 40 th percentle is the 8th abservaton, where the 8th abservaton is number 46.
There are a total of twenty numbers, so n= 20. In order to fnd the percentles, we must put the
numbers in ascending order:
19 31 33 37 40 41 44 46 48 50 51 55 56 58 61 65 70 73 74 76
For the 40th percentle:
Therefore, the 40th percentle index for this data set is the 8th observaton. In the list above, the 8th
observaton is 46.
Queston 3
10 / 10 pts
You may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
In a tri-state conference, 40% atendees are from California, 10% from Oregon, and 50% from
Washington. As it turns out 8 % of the atendees from California, 11% of the atendees from Oregon, and
13% of the atendees from Washington came to the conference by train. If an atendee is selected at
random and found to have arrived by train, what is the probability that the person is from California?
Your Answer:
For this problem:
P(TrainICA)=.08
P(TrainIOR)=.11
P(Train)IWA)=.13
Wheras: P(CA)=.40; P(OR)=.10 and P(WA)=.50. The queston is What is the probability that the person is
from CA( California)?P(Train│C)=.08.. P(Train│O)=.11..
P(Train│W)=.13..
P(C)=.40,P(O)=.10,P(W)=.50.
We want to fnd P(C│Train), so use:
Queston 4
5 / 10 pts
You may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
Standard Normal Table (Links to an external site.)
Find each of the following probabilites:
a. Find P(Z ≤ .17).
b. Find P(Z ≥ -.34) .
c. Find P(-1.14 ≤ Z ≤ 0.55).
Your Answer:
For this problem:
While using a normal table for all of the problem mentned:
a. The answer is: 0.5675
b. The answer is:
c. The answer is:a.
P(Z ≤ .17)=.56749.
b.
P(Z ≥ -.34)=1- .36693= .63307.
c.
P(-1.14 ≤ Z ≤ 0.55)= .70884- .12714=.5817 .
As noted in earlier exams, you should use the values as they are given in the charts and tables provided.
Queston 5
10 / 10 pts
You may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
Standard Normal Table (Links to an external site.)
Suppose that you are atemptng to estmate the annual income of 1200 families. In order to use the
infnite standard deviaton formula, what sample size, n, should you use?
Your Answer:
For this problem:
I must have :
Therefore the smaple size, less/eaqual than 60
In order to use infnite standard deviaton formula, we should have:n≤0.05(1200)
n≤60
So, the sample size should be less than 60.
Queston 6
10 / 10 pts
You may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
Standard Normal Table (Links to an external site.)
T Table (Links to an external site.)
A shipment of 350 new blood pressure monitors have arrived. Tests are done on 60 of the new monitors
and it is found that 8 of the 60 give incorrect blood pressure readings. Find the 80% confdence interval
for the proporton of all the monitors that give incorrect readings.
Answer the following questons:
1. Multple choice: Which equaton would you use to solve this problem?
A.
B.
C.
D.E.
2. List the values you would insert into that equaton.
3. State the fnal answer to the problem
Your Answer:
1. The answer is: E
2. I must calculate samples that are defectve:
, meaning P=.1333, while estmatn: p=.1333; n=60 (monitors tested);z=1.28, based on
80% confdence and N=350
3. As asked to state the fnal asnwer to the problem is: While calculatng the problem using formula (E)
the answer is:
.1333 .0512, therefore the defectve proporton is between (.0821 and .1845)
We have a fnite populaton, so we will use Case 2:
E.
The proporton of the sample that are defectve is 8/60 = .1333 so we set P=.1333. As we mentoned
previously, we estmate p by P. So, p=.1333. A total of 60 monitors were tested, so n=60. Based on a
confdence limit of 80 %, we fnd in table 6.1 that z=1.28. The total number of monitors is 350, so set
N=350. Now, we can substtute all of these values into our equaton:
.1333± .0512
So the proporton of the total that are defectve is between .0821 and .1845.
Queston 7
10 / 10 ptsYou may fnd the following fles helpful throughout the exam:
Statstcs_Equaton_Sheet (Links to an external site.)
Standard Normal Table (Links to an external site.)
It is recommended that pregnant women over eighteen years old get 85 milligrams of vitamin C each
day. The standard deviaton of the populaton is estmated to be 12 milligrams per day. A doctor is
concerned that her pregnant patents are not getng enough vitamin C. So, she collects data on 45 of her
patents and fnds that the mean vitamin intake of these 45 patents is 81 milligrams per day. Based on a
level of signifcance of α = .02, test the hypothesis.
Your Answer:
For this problem:
As realized this is lef tailed test, meaning P(Z [Show Less]