Sampling Distributions – Solutions
1 Suppose the amount students at PSU spent on textbooks this semester is a normal random
variable with mean μ= $360 and standard deviation σ = $90.
a. Use the empirical rule for bell-shaped data to determine intervals that will contain about 68%,
95% and 99.7% of amounts spent on textbooks by PSU students. Show work for each
interval.
68% will be in interval 360± 90, or $270 to $450
95% will be in interval 360± (2×90), or $180 to $540
99.7% will be in interval 360± (3×90), or $90 to $630
b. With Software, find the cumulative probability for $300. [Note: The notation for the
probability you’re finding is P(X ≤ $300).]
x
300
P( X <= x )
0.252493
c. Write a sentence interpreting the value found in the previous part that would be understood
by somebody with no training in statistics.
About 25% of PSU students spent less than $300 on textbooks this
semester.
d. What proportion of students spent more than $535? (Start like you did for part c, but then
you’ll have to do a “by hand” calculation to determine the final answer. [Note: The notation
for the probability you’re finding is P(X > $535).]
Cumulative for $535 = 0.974079, so answer = 1 - .974049 = .025921
e. What is the probability that a randomly selected student will have spent between $300 and
$535? Show work. [Note: You’ll be able to utilize information from parts c and d.]
Find as difference in cumulative probabilities from parts c and d: = 0.974079 – 0.252493 =
.721586
f.
Calculate a z-score for $400.
z = (400 – 360)/90 = 40/90 = 0.44
g. Use standard normal table of the text to determine the cumulative probability for the z-score
you calculated in the previous part. If you don’t have the book with you, use the link to the
table within the Week 6 folder.
Answer = 0.6700
h. Write a sentence interpreting the value found in the previous part that would be understood
by somebody with no training in statistics.
About 67% of Penn State students spent less than $400 on textbooks.
i.
Use Software to find the probability a randomly selected student spent less than $400. If you
didn’t get the same answer as you did in part h, figure out why. The answers should be the
same.
0.6716
j.
What proportion of students spent more than $400 on textbooks? Show any work.
1−.6716 = .3284
k. Find the probability an amount spent is less than or equal to $240. Solve the problem using
standard normal table. Show calculation of the z-score as part of your solution. (You can
verify your answer using Software if you wish.)
z = (240 – 360)/90 = –120/90 = –1.33, answer = 0.0918
l.
With Software, find the 75th percentile of amounts spent on textbooks. This will involve
finding the Inverse Cumulative Probability!
P( X <= x )
0.75
x
420.704
About $420.70
m. Write a sentence interpreting the value found in the previous part that would be understood
by somebody with no training in statistics.
About 75% of PSU students spent less than $420.70 (and about 25% spent more).
n. In standard normal table search for the cumulative probability most near 0.75. You’re looking
“inside” the table where the probabilities are. What is the z-score with this cumulative
probability?
z = 0.67 Look inside table where the probabilities are; locate 0.7500 (or the closest
value to 0.7500), then identify z.
o. Refer to the previous part. Determine the amount spent on textbooks having the z-score found
in the previous part. Show work. [Note: Compare your answer to the answer found in part m.
If the answers differ by a lot, something’s not right.]
Using z = (observed – mean), we want to solve for Observed.
Std. Dev
(0.67)(90) + 360 = $420.30 , which comes from solving 0.67 
observed  360
90
2 The term sampling frame refers to the group that actually had a chance to get into the sample.
Ideally, this is the same as the population of interest, but sometimes it isn’t. In the following
situation, describe the population, the sampling frame, the sample, the parameter of interest, and
the statistic.
A Gallup Poll is done using random digit dialing to reach individuals in households with
land-line telephones. The purpose is to estimate the proportion of U.S. adults who favor stronger
gun control laws. One-thousand persons are sampled, and 63% favor stronger gun control.
a. Population = all U.S. adults
b. Sampling frame = adults in the households with land-line telephones
c. Parameter = proportion of U.S. adults favoring stronger gun control
d. Sample = the 1,000 surveyed
e. Statistic =63%, the sample percent