CHAPTER 3 - Discrete Random Variables and Probability Distributions
SHORT
ANSWER
1. Three automobiles are selected at random, and each is categorized as having a diesel (S) or nondiesel (F) engine
(so outcomes are SSS, SSF, etc.). If X = the number of cars among the three with diesel engines, list each
outcome in S and its associated X value.
A
NS
:
Outcome FFF SFF FSF FFS FSS SFS SSF SSS
X 3 2 2 2 1 1 1 0
PTS: 1
2. Let X = the number of nonzero digits in a randomly selected zip code. What are the possible values of X? Give three
possible outcomes and their associated X values.
ANS:
In my perusal of a zip code directory, I found no 00000, nor did I find any zip codes with four zeros, a fact which
was not obvious. Thus possible X values are 2, 3, 4, 5 (and not 0 or 1). X = % for the outcome 15213, X=4 for the
outcome 44074, and X=3 for 94322.
PTS: 1
3. If the sample space is an infinite set, does this necessarily imply that any random variable X defined from S will
have an infinite set of possible values? If yes, say why. If no, give an example.
ANS:
No. In the experiment in which a coin is tossed repeatedly until an H results, let Y = 1 if the experiment terminates
with at most 5 tosses and Y = 0 otherwise. The sample space is infinite, yet Y has only two possible values.
PTS: 1
4. An automobile service facility specializing in engine tune-ups knows that 50% of all tune-ups are done on four-
cylinder automobiles, 40% on six-cylinder automobiles, and 10% on eight-cylinder automobiles. Let X = the
number of cylinders on the next car to be tuned. What is the pmf of X?
ANS:
x 4 6 8
P(x) .50 .40 .10
PTS: 1
5. The n candidates for a store manager have been ranked 1,2,3,…,n. Let X = the rank of a randomly selected
candidate, so that X has pmf
,(this is called the discrete uniform distribution). Compute E(X) and V(X) using the shortcut formula. [Hint: The
sum of the first n positive integers is n(n + 1)/2, whereas the sum of their squares is n(n + 1)(2n + 1)/6.]
ANS:
, PTS: 1
6. A chemical supply company currently has in stock 100lb of a certain chemical, which it sells to customers in 5-lb
lots. Let X = the number of lots ordered by a randomly chosen customer, and suppose that X has pmf
x 1 2 3 4
P(x) .2 .3 .3 .2
Compute E(X) and V(X). Then compute the expected number of pounds left after the next customer’s order is
shipped, and the variance of the number of pounds left. (Hint: The number of pounds left is a linear function of X.)
ANS:
Each lot weighs 5 lbs, so weight left = 100 – 5x. Thus the expected weight left is 100 – 5E(X) = 87.5, and the
variance of the weight left is V(100 – 5X) = V(-5X) = 25V(x) = 26.25.
PTS: 1
7. Twenty-five percent of all telephones of a certain type are submitted for service while under warranty. Of these,
60% can be repaired whereas the other 40% must be replaced with new units. If a company purchases ten of these
telephones, what is the probability that exactly two will end up being replaced under warranty?
ANS:
Let S represent a telephone that is submitted for service while under warranty and must be replaced. Then p = P(S) =
P(replaced | submitted) P(submitted) = (.40)(.25) = .10. Thus, X, the number among the company’s 10 phones that
must be replaced, has a binomial distribution with
PTS: 1
8. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a
laboratory assistant to randomly select 15 of the specimens for analysis.
a. What is the pmf of the number of granite specimens selected for analysis?
b. What is the probability that all specimens of one of the two types of rock are selected for analysis?
c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation
of its mean value?
ANS:
a. Possible values of X are 5,6,7,8,9,10. (In order to have less than 5 of the granite, there would have to be
more than 10 of the basaltic).
Following the same pattern for the other values, we arrive at the pmf, in table form below.
x 5 6 7 8 9 10
P(x) .0163 .1354 .3483 .3483 .1354 .0163
SHORT
ANSWER
1. Three automobiles are selected at random, and each is categorized as having a diesel (S) or nondiesel (F) engine
(so outcomes are SSS, SSF, etc.). If X = the number of cars among the three with diesel engines, list each
outcome in S and its associated X value.
A
NS
:
Outcome FFF SFF FSF FFS FSS SFS SSF SSS
X 3 2 2 2 1 1 1 0
PTS: 1
2. Let X = the number of nonzero digits in a randomly selected zip code. What are the possible values of X? Give three
possible outcomes and their associated X values.
ANS:
In my perusal of a zip code directory, I found no 00000, nor did I find any zip codes with four zeros, a fact which
was not obvious. Thus possible X values are 2, 3, 4, 5 (and not 0 or 1). X = % for the outcome 15213, X=4 for the
outcome 44074, and X=3 for 94322.
PTS: 1
3. If the sample space is an infinite set, does this necessarily imply that any random variable X defined from S will
have an infinite set of possible values? If yes, say why. If no, give an example.
ANS:
No. In the experiment in which a coin is tossed repeatedly until an H results, let Y = 1 if the experiment terminates
with at most 5 tosses and Y = 0 otherwise. The sample space is infinite, yet Y has only two possible values.
PTS: 1
4. An automobile service facility specializing in engine tune-ups knows that 50% of all tune-ups are done on four-
cylinder automobiles, 40% on six-cylinder automobiles, and 10% on eight-cylinder automobiles. Let X = the
number of cylinders on the next car to be tuned. What is the pmf of X?
ANS:
x 4 6 8
P(x) .50 .40 .10
PTS: 1
5. The n candidates for a store manager have been ranked 1,2,3,…,n. Let X = the rank of a randomly selected
candidate, so that X has pmf
,(this is called the discrete uniform distribution). Compute E(X) and V(X) using the shortcut formula. [Hint: The
sum of the first n positive integers is n(n + 1)/2, whereas the sum of their squares is n(n + 1)(2n + 1)/6.]
ANS:
, PTS: 1
6. A chemical supply company currently has in stock 100lb of a certain chemical, which it sells to customers in 5-lb
lots. Let X = the number of lots ordered by a randomly chosen customer, and suppose that X has pmf
x 1 2 3 4
P(x) .2 .3 .3 .2
Compute E(X) and V(X). Then compute the expected number of pounds left after the next customer’s order is
shipped, and the variance of the number of pounds left. (Hint: The number of pounds left is a linear function of X.)
ANS:
Each lot weighs 5 lbs, so weight left = 100 – 5x. Thus the expected weight left is 100 – 5E(X) = 87.5, and the
variance of the weight left is V(100 – 5X) = V(-5X) = 25V(x) = 26.25.
PTS: 1
7. Twenty-five percent of all telephones of a certain type are submitted for service while under warranty. Of these,
60% can be repaired whereas the other 40% must be replaced with new units. If a company purchases ten of these
telephones, what is the probability that exactly two will end up being replaced under warranty?
ANS:
Let S represent a telephone that is submitted for service while under warranty and must be replaced. Then p = P(S) =
P(replaced | submitted) P(submitted) = (.40)(.25) = .10. Thus, X, the number among the company’s 10 phones that
must be replaced, has a binomial distribution with
PTS: 1
8. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a
laboratory assistant to randomly select 15 of the specimens for analysis.
a. What is the pmf of the number of granite specimens selected for analysis?
b. What is the probability that all specimens of one of the two types of rock are selected for analysis?
c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation
of its mean value?
ANS:
a. Possible values of X are 5,6,7,8,9,10. (In order to have less than 5 of the granite, there would have to be
more than 10 of the basaltic).
Following the same pattern for the other values, we arrive at the pmf, in table form below.
x 5 6 7 8 9 10
P(x) .0163 .1354 .3483 .3483 .1354 .0163
