Download Sampling Distribution Practice Problems Solutions

Sampling Distribution Practice Problems Solutions
1. The company JCrew advertises that 95% of its online orders ship within two working
days. You select a random sample of 200 of the 10,000 orders received over the past month to
audit. The audit reveals that 180 of these orders shipped on time. This is a categorical problem
(sample proportion).
a. What is the sample proportion of orders shipped on time?
� = 180/200 = 0.9
𝒑𝒑
b. Does the sample data satisfy conditions necessary for the sample proportion to follow an
approximately normal distribution?
np >= 5, 10, or 15 and n(1-p) >= 5, 10, or 15
� = 0.9, p=0.95 (“real” number that ship on time according to JCrew)
n=200, 𝒑𝒑
(200)(0.95) = 190
(200)(1-0.95) = 10
Yes, conditions are satisfied
c. What is the mean and standard error (SE) of the sample distribution assuming normal?
mean = p = 0.95
SE = square root of [p(1-p)/n] = square root of [(0.95)(1-0.95)/200] = 0.015
d. If JCrew really ships 95% of its orders on time, what is probability that the proportion in a
random sample of 200 orders is as small or smaller as the proportion in the audit?
� <= 0.9) when n=200. P(Z <=(𝒑𝒑
� - p)/SE) = P(Z <=(0.9 - 0.95/0.015) = P(Z <= -3.33). Closest to
P(𝒑𝒑
this on from the Standard Normal Table is -3.09, P = 0.001. Thus, P(Z <= -3.33) is less than
0.001.
e. If we treated the problem as a binomial, how would the problem be set up? That is, what
would we want to find the probability of?
Trials would be 200, successes would be 180. P(X <= 180).
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2. A Gallup poll found that for those Americans who have lost weight, 31% believed the most
effective strategy involved exercise. What is the probability that from random sample of 300
Americans the sample proportion falls between 29% and 33%? Categorical (sample proportion).
n = 300, p = 0.31
np = (300)(0.31) = 90
n(1-p) = (300)(1-0.31) = 210
Both >= 5, 10 or 15
� <= 0.33) = P(𝒑𝒑
� <= 0.33) – P(𝒑𝒑
� <= 0.29)
P(0.29 <= 𝒑𝒑
SE = square root of [(0.31)(0.69)/300] = 0.0267
P(Z<= (0.33 – 0.31)/0.0267) – P(Z <= (0.29 – 0.31)/0.0267)
P(Z<= 0.75) – P(Z <= - 0.75)
From the Standard Normal Table, we find these two probabilities to be:
0.7734 – 0.2266 = 0.5468
3. A bottling company uses a machine to fill the bottles with olive oil. The bottles are designed
to contain 475 milliliters (ml). In fact, the contents vary according to a normal distribution with
a mean of 473 ml and standard deviation of 3 ml. Quantitative (sample mean).
a. What is the distribution, mean, and standard error of the sample mean of six randomly
selected bottles?
Distribution is normal, sample mean should be normal with a mean = μ = 473, SE =
sigma/square root of n = 3/√6 = 1.22.
b. What is the probability that the mean of six bottles is less than 470 ml?
� < 470) = P(Z < (𝒙𝒙
� – μ)/SE) = P(Z < (470-473)/1.22) = P(Z < - 0.246). From the Standard
P(𝒙𝒙
Normal Table we find P(Z < - 0.246) = 0.0069.
c. What is probability that the mean of six bottles is more than 475 ml?
� > 475) = P(Z > (𝒙𝒙
� – μ)/SE) = P(Z > (475-473)/1.22) = P(Z > 1.64 ) = 1 – P(Z < 1.64). From the
P(𝒙𝒙
Standard Normal Table for P(Z < 1.64) 0.9495 resulting in final probability of 1 – 0.9495 =
0.0505
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4. Penn State Fleet which operates and manages car rentals for Penn State employees found
that the tire lifetime for their vehicles has a mean of 50,000 miles and standard deviation of
3500 miles. Quantitative (sample mean).
a. What would be the distribution, mean and standard error mean lifetime of a random sample
of 50 vehicles?
Mean = μ = 50,000, sigma = 3500.
We can’t assume this is normal so have to go for the second assumption. Our sample will
have to be larger than 30. Our sample is 50, so this works.
n = 50 which is > 30, so we can assume sample mean to be approximately normal with mean
= μ = 50,000 and SE = sigma/sr of n = 3500/√50 = 495.
b. What is the probability that the sample mean lifetime for these 50 vehicles exceeds 52,000?
� > 52,000) = P(Z > (𝒙𝒙
� – μ)/SE) = P(Z > (52,000 - 50,000)/495) = P(Z > 4.04). From the
P(𝒙𝒙
Standard Normal Table we it only goes to 3.09, cumulative probability of 0.999. P(Z > 4.04) = 1
- 0.999 = 0.001. P(Z > 4.04) is less than 0.001.
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