Probability Distributions - Solutions
1 In each part, indicate, (1) whether the variable is discrete or continuous AND (2) whether it is
binomial or not AND (3) if it is binomial, give values for n and p.
a. Number of times a “head” is flipped in 10 flips of a coin
1. Discrete
2. Binomial
3. n = 10, p = 0.5
b. Time to complete a 30 question multiple choice test
1. Continuous 2. Not Binomial since continuous
c. Number of correct answers on a 30 question multiple choice test for somebody who randomly
guesses at every question. There are four answer choices for each question,
1. Discrete
2. Binomial – counting correct answers 3. n = 30, p = 0.25
d. A woman buys a lottery ticket every week for which the probability of winning anything at all
is 1/10. She continues to buy them until she has won 3 times. X = the number of tickets she buys.
1. Discrete
2. Not binomial since the number of trials is not fixed
2 In Stat 200 last year, students were asked to rate how much they liked various kinds of music on
a scale of 1 (don’t like at all) to 6 (like very much). Following is a probability distribution (with
one probability not given) for females’ rating of Top 40 music.
X=Rating
Probability
1
.04
2
.05
3
.09
4
??
5
.32
6
.26
a. Is X = music rating a discrete variable or a continuous variable? Explain.
Discrete. There are a small number of distinct possible outcomes (the ratings 1-6).
b. What must be the value of the probability for X = 4 (the probability that rating equals 4)?
Explain how you determined this.
P(x=4) is 0.24 since all the probabilities need to add to 1.
c. Determine E(X), the mean rating given by females. {HINT: Find (rating*probability) for each
rating, and then add up those values.}
E(X) = (1)(.04) + 2(.05) + 3(.09) + 4(.24) + 5(.32) + 6(.26) = .04+.10+.27+.96+1.60+1.56 = 4.53
d. Find the probability that the rating given is 3 or less. Note: When we find the probability that a
variable is less than or equal to some value it is called a cumulative probability. To find it, we add
probabilities for all values up to and including that point.
P(X≤3) = P(X=1)+P(X=2)+P(X=3) = 0.04 + 0.05 +0.09 = 0.18
e. For each rating value, determine the cumulative probability.
X = Rating
1
2
3
4
Cumulative
Probability
.04
.09
.18
.42
5
6
.74
1.0
f. Use the cumulative probabilities just found as an aid in finding the probability that the rating of
a randomly selected student would be greater than 4. Show work.
P (rating > 4) = 1 – P(rating ≤ 4) = 1- .42 = .58
3 Suppose somebody randomly guesses at every one of 20 True-False questions.
a. The number of correct guesses is a binomial random variable. What are the values of the
parameters n and p?
n = 20
p = 0.5
b. What is the expected number and standard deviation of correct guesses at the n = 20 answers?
Show how you determined this. Since this is a binomial experiment we can use the following
formulas:
Expected value: np = 20(0.5) = 10
Standard Deviation:
np(1  p) =
20 * 0.5 * (1  0.5) = 2.24
Use Minitab or SPSS to find the probability for parts c through g. Instructions on how to do
this are given in this week’s Lab Activity folder.
c. Assuming random guessing, what is the probability that the number of correct guesses is 13 or
fewer?
0.942341
d. What is the probability that somebody just guessing could get 14 or more correct? Show work
(Hint: This is the complement of the previous part.)
1-0.942341 = 0.057659
e. What is the probability that somebody just guessing would get 18 or more correct? (Hint:
You’ll have to deal with the probability of 17 or less first.
1 − 0.999799 = 0.000201
f. What is the probability that somebody just guessing gets exactly 10 correct? (You’ll have to
click on Probability rather than Cumulative Probability in the Minitab dialog box.)
0.176197
g. What is the probability that the number correct for somebody just guessing is one of 9, 10, or
11 correct? Hint: First find separate probabilities for each of 9, 10, and 11. Show any work.
P(9 right) + P(10 right) + P(11 right) =
0.160179 + 0.176197 + 0.160179 = 0.496555
Probability Distributions - Continuous Solutions
4 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 Minitab of SPSS, 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 = 0.974049, so answer = 1 - .974049 = .025951
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: = 0.974049 – 0.252493 =
.721106
f.
Calculate a z-score for $400.
z = (400 – 360)/90 = 40/90 = 0.44
g. Use the Standard Normal Table of the text (or in this weeks folder) to determine the
cumulative probability for the z-score you calculated in the previous part.
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 Minitab or SPSS 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 g, 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 the Standard Normal Table. Show calculation of the z-score as part of your
solution. (You can verify your answer using Minitab or SPSS if you wish.)
z = (240 – 360)/90 = –120/90 = –1.33, answer = 0.0918
l.
With Minitab or SPSS, find the 75th percentile of amounts spent on textbooks. {This
requires you to find the inverse 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 the Standard Normal Table search for the cumulative probability most near 0.75.
You’re looking “inside” the table where the probabilities are for the value closest to
0.7500. 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