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Math 542 Practice Test 1
Section 1.2
1. You roll a pair of fair four-sided dice (with faces labeled 1, 2, 3, and 4) and add the numbers
on the sides that land face down.
A.
B.
C.
D.
Give the elements of the sample space.
Describe the distribution of the sum.
Find the probability that the sum of the dice is greater than 5.
Find the probability that the sum of the dice is odd.
Ans:
A. {2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8}
B.
x
2
3
4
5
6
7
8
P(x) 1/16 2/16 3/16 4/16 3/16 2/16 1/16
C. 3/8
D. 1/2
2. You roll a pair of fair four-sided dice (with faces labeled 1, 2, 3, and 4) and multiply the
numbers on the sides that land face down.
A. Give the elements of the sample space in the form of a table.
B. Describe the distribution of the product.
C. Calculate the probability the product is greater than 4 or the first die is odd.
Ans:
A.
First Die
Second Die
1
2
3
4
1
1
2
3
4
2 3 4
2 3 4
4 6 8
6 9 12
8 12 16
B.
x
1
2
3
4
6
8
9
12
16
P(x) 1/16 2/16 2/16 3/16 2/16 2/16 1/16 2/16 1/16
C.13/16
3. A number, call it X, is randomly chosen from the interval (5, 6) . Calculate the following
probabilities:
A. P (0  X  5)
B. P (4  X  8)
C. Is X a discrete or continuous random variable? Explain.
Ans:
A. 5/11
B. 2/11
C. X is continuous because it can take any value between -5 and 6
4. Bob is going to ask several different girls for a date. He figures that since each girl can either
say yes or no, the probability that each girl says yes is 0.5 so about half of the girls should say
yes. Explain what is wrong with this reasoning.
Ans: There are only 2 outcomes, but they are not equally likely.
5. An experiment consists of selecting 2 cubes with replacement from a bag consisting of 1 red,
1 blue, and 1 green cube.
A. List the elements of the sample space.
B. Assuming all outcomes are equally likely, find the probability of getting 2 cubes of different
colors.
C. Let the random variable X denote the number of green cubes in the 2 cubes selected. Is X a
discrete or continuous random variable? Explain.
Ans:
A.
2nd Cube
R
B
G
1st Cube
R
B
G
RR RB RG
BR BB BG
GR GB GG
B. 6/9
C. X is discrete because it can take values of only 0, 1, and 2.
6. A point is randomly selected from inside a circle with radius r. Find the probability the point
is closer to the center than to the perimeter.
Ans: The point is closer to the center than to the perimeter if the point is chosen from a circle of
radius r / 2 centered at the center of the larger circle. The probability of this occurring is
 (r / 2) 2 1

 r2
4
7. Suppose that a gambler simulates 20 plays of a game, wins 1 time, and estimates P(winning)
= 1/20 = 0.05. He then simulates 5,000 plays of the game, wins 450 times, and estimates
P(winning) = 450/5,000 = 0.09. Which estimate, 0.05 or 0.09 is probably closer to the true
theoretical probability and why?
Ans: The number 0.09 is probably closer because it is based on more trials. The Law of Large
Numbers says that the more trials, the closer the relative frequency is to the theoretical
probability.
8. A gambler uses theory to calculate the probability of winning a card game and gets
P(winning) = 0.10. Which of these options best describes the meaning of this probability?
A.
B.
C.
D.
He is guaranteed to win exactly 10% of the time.
In the long run he will win approximately 10% of the time
He will win 10 times.
He will win on every 10th play.
Ans: B
9. Suppose you guess an answer to a math problem (not a multiple choice question). Is the
probability that you guess correctly equal to 1/2? Why?
A. Yes, because you have a 50-50 chance of getting it right.
B. No, because this random experiment does not have a sample space.
C. No, because the two outcomes of right and wrong are not equally likely.
D. Yes, because the answer is either right or wrong.
Ans: C
Section 1.3
1. If 8 boats enter a race, how many possibilities are there for the first through third place
finishers?
Ans: 8  7  6  336
2. Five soccer players, three football players, and two baseball players are going to sit in a row
of chairs. In how many ways can these athletes be arranged?
Ans:
10!
 2520
5! 3! 2!
3. Cindy must write evaluation reports on three hospitals and two health clinics as part of her
degree program in public health. If there are six hospitals and five clinics in her area, how many
different reports could she write?
6  5
Ans:       200
 3  2
4. The manager of a baseball team has filled the first, fifth, and seventh slots in the nine-player
starting lineup. If he has 12 players left to choose from, how many different starting lineups are
possible?
Ans:
12
P6  665, 280
5. Find the coefficient of the x 4 y 6 term in the expansion of ( x  y)10
10 
Ans:    210
6
Section 1.4
1. Consider a random experiment with the sample space S  {1, 2,3, 4} and let A and B be
events. State which of the axiom(s) of probability justifies each statement below:
A. P({1}  {2})  P({1})  P({2})
B. P({1}  {2}  {3}  {4})  1
P( A)
C. If P( B)  0 , then
0
P( B)
D. P( A  A)  P( A)  P( A)
Ans: 3, 2, 1, 3
2. Determine whether or not it is reasonable to assume that each pair of events is disjoint.
Explain your reasoning.
A. A student goes to the store to purchase a single calculator. Let A be the event she purchases a
Texas Instruments and B be the event she purchases a Hewlett Packard.
B. A hiker hikes through a forest. Let A be the event he sees a bird and B be the event he crosses
a stream.
Ans:
A. These events are disjoint because she can’t purchase both a Texas Instruments and a Hewlett
Packard.
B. These are not disjoint because he can see a bird and cross a stream only his hike.
3. A parking lot contains a total of 45 vehicles each of which is either a pickup or blue in color.
There are 30 pickups and 25 blue vehicles. Consider the random experiment of choosing a
vehicle from this lot. Find the probability of choosing a blue pickup.
Ans: Let A denote the event of selecting a pickup and B denote the event of selecting a blue
vehicle. Then P ( A  B )  1 , P( A)  30 / 45 , and P( B)  25 / 45 . Thus by the addition rule,
1
30 25
2
  P( A  B)  P( A  B) 
45 45
9
4. A game is played by flipping a fair coin until the first tail appears. Find the probability the
game lasts more than 3 flips.
Ans: We use complements and disjoint events:
P  more than 3  1– P 1 or 2   1–  P 1  P  2  
1 1 1 1
 1     
2 2 2 4
Section 1.5
1. 100 people were given either a 10-mg pill of a new experimental drug or a placebo. They
recorded whether or not they got a headache. The following table summarizes the results.
Received drug Received placebo Totals
15
65
80
Headache
17
3
20
No headache
32
68
100
Totals
A. Find the probability of selecting someone who received the drug or got a headache.
B. If 3 different subjects are randomly selected without replacement, find the probability that at
least 1 received a placebo.
C. If one subject is randomly selected, find P(Headache | Received drug).
D. Are the events of getting a headache and receiving a placebo disjoint? Why?
Ans:
32 80 15
97



 0.97
A. 100 100 100 100
B.
P(at least one placebo)  1  P(no placebo)
32 31 30
 1
   0.969
100 99 98
C.
P ( H | D) 
P( H  D) 15 /100 15


 0.469
P ( D)
32 /100 32
D. No they are not disjoint because 65 people received a placebo and got a headache.
2. Suppose the student body at a certain university contains 1200 students and is 55% female
and 45% male. Consider the experiment of randomly choosing 5 students and recording their
genders.
A. Treating the selections as dependent, find the probability that the first and fourth students are
female.
B. Repeat part a, but treat the selections as independent.
C. Treating the selections as independent, find the probability that we choose exactly two
females.
D. If we were to select 200 students, would it be reasonable to treat the selections as
independent? Why or why not?
Ans:
A. There are a total of 660 females and 540 males at the university so that
660 540 539 659 538
P(FMMFM) 




 0.027600
1200 1199 1198 1197 1196
2
3
B. P(FMMFM)  (0.55) (0.45)  0.027565
5
C. There are    10 ways of choosing exactly two females, each of which has probability
 2
0.027565. Thus the total probability is 10(0.027565)  0.27565 .
D. Since 200 is more than 5% of the population, it would not be reasonable to treat the
selections as independent.
3. Suppose a bag contains 2 red cubes and 3 blue cubes. You pick 2 cubes without replacement.
Find the probability that the first is blue and the second is red.
2 3 3
Ans: P( B  R)   
5 4 10
4. If 5 people are randomly selected with replacement from a group of 15 people of which Joe is
a member, find the probability that Joe is selected.
5
 14 
Ans: P(Joe is selected)  1  P(Joe is not selected)  1     0.292
 15 
Section 1.6
1. Health officials estimate that about 0.5% of the population has a certain disease. There is a
test that when given to people who have the disease would correctly identify it 95% of the time.
This test also gives a false positive about 3% of the time. A person is selected at random and gets
a positive result when given this test. What is the conditional probability that the person actually
has the disease?
Ans: P(have the disease | test positive) 
0.005(0.95)
 0.137
0.005(0.95)  0.995(0.03)
2. Three toy boxes, call them boxes A, B, and C, contain 20, 50, and 30 stuffed animal,
respectively. Of the animals in box A, 45% are bears; in box B, 35% are bears; and in box C,
80% are bears. A girl randomly chooses a stuffed animal from one of the boxes and gets a bear.
Find the conditional probability the bear was chosen from box A.
Ans: P( A | animal is a bear) 
0.2(0.45)
 0.178
0.2(0.45)  0.5(0.35)  0.3(0.8)
3. If events A1 and A2 form a partition of the sample space, B is an event, P  A1   0.75 ,
P  A2   0.25 , P  A1 | B   0.85 , and P  B | A1   0.64 , find P  B | A2  .
Ans: By Bayes’ Theorem, 0.85 
P  B | A2   0.339
0.75(0.64)
. Solving this for P  B | A2  yields
0.75(0.64)  0.25P  B | A2 
Section 1.7
1. A fair six-sided die is rolled. Let A be the event the resulting number is even and B be the
event the number is greater than 4. List the elements of both events and use the definition of
independence to show that A and B are independent.
1
1
A  {2, 4, 6}, P( A)  , B  {5, 6}, P( B) 
2
3
Ans:
1 1 1
P( A  B)  P({6})     P ( A) P ( B )
6 2 3
2. A math professor has 10 algebra and 5 calculus books on her bookshelf. If she randomly
selects 3 books, find the probability that
A. they are all algebra books,
B. the third book is the first calculus book selected, and
C. at least 1 book is an algebra book.
Ans:
10 9 8 24
  
 0.264
A. 15 14 13 91
10 9 5 15
    0.165
B. 15 14 13 91
C.
P(at least 1 algebra book)  1- P(no algebra books)
5 4 3 89
 1   
 0.978
15 14 13 91
3. Determine whether or not it is reasonable to assume that each pair of events is independent.
Explain your reasoning.
A. You go to the beach. Let A be the event you swim all day. Let B be the event you get a
sunburn.
B. A car is randomly chosen from a parking lot. Let A bet the event the car is gray and B be the
event the car has an out-of-state license plate.
Ans:
A. These events are not independent. If you swim all day, you will be in the sun the entire day,
so you increase the probability of getting a sunburn.
B. It is reasonable to assume these are independent.
4. You roll a pair of fair five-sided dice (with faces 1, 2, 3, 4, and 5). Find the probability that
the second die is greater than 2 or the first die is even.
Ans:
P(2nd > 2  1st is even)  P(2nd > 2)  P(1st is even)  P(2nd > 2  1st is even)
3 2 3 2 19
    
5 5 5 5 25
1. Which one of these options best defines a random variable?
A. A random variable is measure of probability.
B. Random variables are data.
C. Random variables are functions that assign probabilities to outcomes of a random
experiment.
D. A random variable is a function that takes in an outcome from a random experiment and
returns a real number.
Ans: D
2. Consider the random variable X with the pmf described below
2
3
4
x 1
f(x) 0.4 0.3 0.2 0.1
A. Sketch the probability histogram of X.
B. Calculate the median and mode of X.
C. Calculate P  X {1,3, 4}
D. If F ( x ) is the c.d.f. of X, find the value of F (3.5)
Ans:
A.
0.5
f(x)
0.4
0.3
0.2
0.1
0
1
2
3
4
x
B. The mode is 1 since f ( x ) has its maximum value at x  1 . The median is 2 since
P( X  2)  0.7 and P( X  2)  0.6 .
C. 0.4  0.2  0.1  0.7
D. F (3.5)  P( X  3.5)  P( X  3)  0.9
3. Let X be a uniformly distributed discrete random variable with range {1, 2,
the following probabilities:
,15} . Calculate
A. P( X  4)
B. P(12  X  18)
C. P ( X  9| X  6)
Ans:
4
 0.267
A.
15
4
 0.267
B.
15
P( X  9  X  6) P( X  9) 7 /15
7


  0.7
P( X  6)
P( X  6) 10 /15 10
C.
4. A fair six-sided die is rolled repeatedly until the first 4 appears. Let the random variable X
denote the number of rolls until the first 4 appears. Find a formula for the pmf of X.
Ans: The only way that X can equal x is if the first x  1 rolls are not 4 and the last roll is a 4.
5
Thus f ( x)  P( X  x)   
6
x 1
1
 
6
5. Could the following table describe the pmf of a random variable? Explain why or why not.

2
x -3 -0.6 1
f(x) 0.2 0.4 0.1 0.05 0.3
Ans: The values of f do not add up to 1, so this table cannot describe a pmf
Section 2.3
1. A box of candy contains 25 candy bars, of which 10 contain peanuts and 15 do not contain. If
you select five bars at random, find the probability that exactly three of the bars contain peanuts.
Ans: Let X denote the number of selected bars that contain peanuts. Then X has a hypergeometric
10  25 
 

3  5  3 

distribution and P( X  3) 
 0.1109
 35 
 
5
2. A certain chemical found in polluted water has been found to be lethal to 20% of the fish that
are exposed to it for 24 hours. Thirty fish are placed in a tank containing the chemical. Find the
probability that less than three of the fish die 24 hours later.
Ans: Let X denote the number of fish that die. Then X is b(30, 0.2) and
P( X  3)  P( X  2)  0.0442
3. A manufacturer of candies claims that 30% of its plain chocolate candies are blue. A random
sample of 50 such candies contains 20 blue candies. Use the unusual event principle to test the
claim.
Ans: Let X denote the number of blue candies in a sample of 50. Then assuming the claim is true,
X is b(50, 0.3) . We would expect there to be 50(0.3)  15 blue candies in a sample of 50. Since
20 is greater than expected, we calculate
P( X  20)  1  P( X  19)  0.0848
Since this probability is greater than 0.05, the observed value is not unusually large, so we do not
reject the claim.
Section 2.4
1. Dandelions appear in a lawn at an average of one per 10 ft2 section. What is the probability
that exactly two dandelions will appear in a 40 ft2 section of lawn?
Ans: Let X denote the number of dandelions in a 40 ft2 section of lawn. Then X has a Poisson
distribution with parameter   4 so that
42
P( X  2)  e4  0.147
2!
2. Suppose pine boards contain an average of 3 knots per 8 feet of board. Assuming the Poisson
distribution applies, find the probability that a randomly selected 10-foot length of board
contains exactly 5 knots.
Ans: There are an average of 3/8 knots/ft. In a 10-foot length of board, there are an average of
10(3/8) = 15/4 knots. Let X denote the number of knots in a 10-ft length of board. Then X has a
Poisson distribution with parameter   15 / 4 so that
(15 / 4)5 15/4
P( X  5) 
e
 0.145
5!
3. A random variable X has a Poisson distribution with parameter  and the property
P( X  2)  P( X  1)  e . Find the value of  .
Ans: Using the pmf of the Poisson distribution, we have
2
e


1
e   e 
  2  2  2  0
2!
1!
Solving this quadratic for  yields the solutions 1  3 . Since  must be positive, the only
solution that is feasible is   1  3 .
Section 2.5
1. The random variable X has the pmf given in the table below.
-2 0
4
x
f(x) 0.2 0.5 0.3
A. Find E ( X )
B. Find Var ( X )
Ans:
A. 0.8
B. 4.96
2. The random variable X has the pmf given in the table below.
0
1
2
x
f(x) 1/4 1/2 1/4
A. Find E ( X )
B. Find Var ( X )
Ans:
A. 1
B. 1/2
3. In a large city, houses have between 1 and 10 trees in their yards and these values are equally
likely. Find the mean and variance of the number of trees in yards in this city.
Ans: Let X denote the number of trees in a randomly selected yard. Based on the given
information, X has a uniform distribution with range {1, ,10} so that its mean and variance are

10  1 11
102  1 99
  5.5 and  2 

 8.25
2
2
12
12
4. If you bet $5 on the number 7 in roulette, you can either lose $5 or profit $175. The expected
value is $-0.26. Which of these choices best explains the meaning of this expected value?
A. You should bet $0.26 instead of $5.
B. You will lose exactly $0.26 each time you play.
C. You should expect to win $0.26 each time you play.
Total Profit
D. If you play the game many times, the fraction
will be approximately -0.26.
# Times Played
Ans: D
Section 2.6
1. Let X be a random variable with pmf as shown in the table. Find
A. E ( X ) ,
B. E  X 2  ,
D. E 3 X 2  2 X  3 , and
D. Var ( X ) .
Ans:
A. 11/8 = 1.375
x
0
1
2
3
f(x)
1/4
3/8
1/8
1/4
B. 25/8 = 3.125
C. 77/8 = 9.625
D. 79/64 = 1.234
2. Let X be a random variable such that E ( X ) and E  X 2  both exist. Define the function
g (a )  E ( X  a ) 2  .
Show that g does not have a maximum value.
Ans: Using linearity properties of the expected value, we get
g (a)  E  X 2   2aE ( X )  a 2
But since E ( X ) and E  X 2  are fixed values, lim g (a)  lim g (a)   , showing that g does not
a 
a 
have a maximum value.
3. Suppose X is a random variable such that E  X 2   24 . If a is a number such that
E  X  a  2 and E ( X  a ) 2   20 , find possible value(s) of a.
Ans: We have
E  X  a  2  E( X )  2  a
E ( X  a) 2   20   2 E ( X )  a 2  20  E  X 2 
Combining these two equations with the fact that E  X 2   24 yields
a ( a  2)  0
so that a  0 or 2 .
Section 2.7
1. List at least two important uses of the moment-generating function.
Ans: We can use the first and second derivatives of the m.g.f. to find the mean and variance of
the random variable. If we can show that two random variables have the same m.g.f., then we
can conclude that they have the same distribution.
1
2
2
2. Could M (t )  e 2t  e  t  be the m.g.f. of a discrete random variable? Explain why or
2
5
5
why not.
2
Ans: If this were a m.g.f., it would tell us that P( X  0)   . But probabilities cannot be
5
negative, so this cannot be a m.g.f.
3. If the m.g.f. of a random variable X is M (t )  (1  2t )3/2 . Use this to find E ( X ) and Var ( X ) .
Ans: Note that M '(t )  3(1  2t )5/2 and M ''(t )  15(1  2t ) 7/2 so that
E ( X )  M '(0)  3 and Var ( X )  M ''(0)   M '(0)   15  32  6
2
Section 3.2
1. Find the value of c so that the function f ( x )  c / x , 0  x  1 is the pdf of a random variable
X.
Ans: 1/2
2. Consider the pdf f ( x)  (3/16) x 2 ,  2  x  2 . Find the c.d.f, F ( x ) and sketch its graph over
the interval  3,3 .
1
F(x)
0.75
0.5
0.25
0
0
1

Ans: The c.d.f. is F ( x)    x3  8 
16
1
x  2
2  x  2
x2
3. A random variable X has the c.d.f. F ( x)  1  e x 1,  1  x   . Find the median,  0.5 , for
this random variable.
Ans: F ( x)  1  e x1  0.05  x  (ln(0.5)  1)  0.307
4. The random variable X has the p.d.f f ( x ) 
3x 2
, 0  x  2 . Find the mean, variance, and
8
standard deviation of X.
2
Ans:    x
0
3x 2
dx  1.5
8
2
 2   ( x  1.5)2
0
5. A random variable X has the pdf f ( x) 
3x 2
dx  0.15
8
  0.15  0.387
5
1  x 4  for 1  x  1 .
8
A. Find E ( X ) .
B. Find P  0.5  X  1.5 .
Ans:
1
A. E ( X )   x
1
5
1  x 4  dx  0

8
B. P  0.5  X  1.5   P  0.5  X  1  
1
0.5
5
1  x 4  dx  0.8086

8
6. When a certain first-grade boy plays on the playground, he usually spends a lot of time on the
swings. However, sometimes he does not feel like swinging and swings for only a very short
amount of time. Let the random variable X denote the amount of time the boy spends on the
swing on a randomly chosen trip to the playground. Sketch an example of what the graph of the
pdf of X might look like. Briefly explain why your graph looks the way it does. Specifically
focus on explaining the height of the graph.
Ans: The value of f ( x ) would be small for x close to 0 to model the fact that the boy only
occasionally spends a little amount of time swinging. For x further from 0, the value of f ( x )
would be larger to model the fact that the boy usually spends a lot of time swinging.
Section 3.3
1. Calls arrive at a homework help line according to a Poisson distribution an average of 3 every
10 minutes. Let W denote the waiting time (in minutes) between calls. Find P(W  10) .
Ans: W has an exponential distribution with parameter   3 /10 so that
P( X  10)  e3/10(10)  0.0498
2. If X, the number of minutes a student spends studying for a statistics test, is U (15, 60) , find
A. E(X) and Var(X)
B. P(50 < X < 75)
Ans:
60  15 75
(60  15) 2

 37.5 Var ( X ) 
 168.75
2
2
12
 1  2
B. P(50  X  75)  P(50  X  60)  10 
   0.222
 60  15  9
A. E ( X ) 
3. The life span of a certain type of fan is described by an exponential distribution. Suppose the
mean life span is 25,000 hours. Find the probability that
A. A randomly selected fan will last at least 20,000 hours.
B. A randomly selected fan will last between 20,000 and 30,000 hours.
Ans:
A. Let X denote the life span. Then X has an exponential distribution with parameter
  1/ 25, 000  0.00004 so that
P( X  20,000)  e0.00004(20,000)  0.449
B. Using the same distribution as in part a,
P(20, 000  X  30, 000)  P( X  30, 000)  P( X  20, 000)
 1  e 0.00004(30,000)  1  e 0.00004(20,000)   0.148
4. If X is U  0,10 , find P  X  8 | X  7 .
Ans: P  X  8 | X  7  
P( X  8  X  7) P( X  8) 2 /10 2



P( X  7)
P( X  7) 3 /10 3
5. Suppose X, the gallons of gas purchased by a randomly chosen customer at a gas station, is
U (5, 25)
A. Find P(8  X  28)
B. P  X  12 | X  10
C. If gas costs $3.59 a gallon, find the expected total cost of gas for a randomly selected
customer.
Ans:
17
20
P( X  12  X  10) P( X  12) 13 /15 13
B. P  X  12 | X  10  



P( X  10)
P( X  10) 15 / 20 15
C. Let Y denote the total cost. Then Y  3.59 X so that
 25  5 
E (Y )  3.59 E ( X )  3.59 
  $53.85
 2 
A. P(8  X  28)  P(8  X  25) 
Section 3.4
1. If X is N  4, 22  , find
A. P( X 2  1  12)
B. The 34th percentile
Ans:
A.

P( X 2  1  12)  P( X 2  11)  P  11  X  11

  11  4
11  4 
 P 
Z
  P(3.66  Z  0.34)  0.3668
2
2 

B. We know that P( Z  0.41)  0.34 . Thus the 34th percentile is the number x such that
x4
0.41 
 x  3.18
2
2. The weights of the fish in a certain lake are normally distributed with a mean of 15 lb and a
standard deviation of 6. If one fish is randomly selected, what is the probability that its weight is
between 12.6 and 18.6 lb?
Ans: Let X denote the weight of a randomly selected fish. Then
18.6  15 
 12.6  15
P(12.6  X  18.6)  P 
Z
  P(0.4  Z  0.6)  0.3811
6
6


3. If X is N   ,  2  , find P    ( / 2)  X    ( / 2)
  ( / 2)   
   ( / 2)  
P    ( / 2)  X    ( / 2)   P 
Z




Ans:
1
 1
 P    Z    0.3830
2
 2
Section 3.5
1. Consider a random variable X with pdf f ( x) 
3 2
x , 0  x  4 . Define the variable Y  X 2 .
64
Find the pdf of Y.
Ans: Note that the range of Y is 0  Y  16 and the c.d.f. of Y is


F ( y )  P(Y  y )  P  X 2  y   P X  y  
By the fundamental theorem of Calculus, the pdf of Y is
0
y
3 2
x dx
64
f ( y) 
d
3
F ( y) 
dy
64
 y   12 y
2
1/2
3

y

 128
for 0  y  16
2. Consider a random variable X with pdf f ( x)  e x 1 , 1  x . Define the variable Y  X .
Find the pdf of Y.
Ans: Note that the range of Y is 1  Y and the c.d.f. of Y is
F ( y )  P(Y  y )  P


X  y  P  X  y 2    e x 1 dx
y2
1
By the fundamental theorem of Calculus, the pdf of Y is
f ( y) 
2
d
F ( y)  2 ye y 1 for 1  y
dy
3. Consider a random variable X with pdf f ( x)  e x 1 , 1  x .
A. Find the c.d.f., F ( x ) .
B. Find the inverse c.d.f., F 1 ( x) .
Ans:
x
A. F ( x)   et 1 dt  1  e x1
B. y  1  e
1
 x 1
 x  1  ln(1  y)  F 1 ( y)  1  ln(1  y)
Section 3.6
1. The joint pmf of two variables X and Y is f ( x, y ) 
3x  y
, x  1, 2, y  1, 2
24
A. Find the marginal pmf’s.
B. Are X and Y independent? Why or why not?
Ans:
A. f X ( x) 
3x  1 3x  2 2 x  1


,
24
24
8
fY ( y ) 
3(1)  y 3(2)  y 9  2 y


24
24
24
B. X and Y are not independent because f X ( x) fY ( y)  f ( x, y)
2. The joint pdf of two variables X and Y is f ( x, y)  6 xy 2 , 0  x  1, 0  y  1 . Find
P(Y  X 2 / 2)
Ans: P(Y  X 2 / 2)  
1

x 2 /2
0 0
6 xy 2 dy dx  
1
0
x7
1
dx 
4
32
3. The variables X and Y have the joint pdf f ( x, y ) 
1
for 0  x  1, 0  y  3
3
A. Find P( X  Y )
B. Find the marginal pdf’s and determine if X and Y are independent.
Ans:
A. P( X  Y )  
1
0

3
x
1
5
dy dx 
3
6
1
dy  1,
0 3
B. f X ( x)  
3
fY ( y )  
1
0
1
1
dx  . Since f X ( x) fY ( y)  f ( x, y) , X and Y are
3
3
independent.
Section 3.7
1. A gas station sells two types of gasoline, regular and premium. If the variables X and Y
represent the number of gallons of regular and premium sold in a day, respectively, assume that
X is N  415,152  and that Y is N  250, 202  .
A. If the owner makes a profit of $0.09 for each gallon of regular gas sold and $0.11 for each
gallon of premium gas sold, find the expected total profit in a day.
B. Assuming X and Y are independent, find the probability that the profit is greater than $70
Ans:
A. Let P denote the profit in a day. Then P  0.09 X  0.11Y and
E ( P)  E (0.09 X  0.11Y )  0.09 E ( X )  0.11E (Y )  0.09(415)  0.11( 250)  $64.85
B. Note that P is N  64.85, 0.092 (152 )  0.112 (202 )   N (64.85, 6.6625) so that
70  64.85 

P( P  70)  P  Z 
  P( Z  2.00)  0.0228
6.6625 

2. If X is N  5, 42  , Y is N 16, 2 2  , and X and Y are independent, find P(3 X  Y ) .
Ans: Note that P(3 X  Y )  P(3 X  Y  0) , so let W  3X  Y . Then W is
N  3(5)  16,32 (42 )  (1) 2 (22 )   N (1,148)
and
0  (1) 

P( X  0)  P  Z 
  P( Z  0.08)  0.5319
148 

3
for t  3 , the m.g.f. of Y is M Y (t )  (1  t / 4) 5 for t  3 ,
3t
and X and Y are independent, find the m.g.f. of W  3X  4Y .
3. If the m.g.f. of X is M X (t ) 
Ans: The m.g.f. of W is
M X (3t ) M Y (4t ) 
3
(1  t )5
 (1  (4t ) / 4)5 
 (1  t )6
3  (3t )
1 t
for t  3
Section 3.8
1. Suppose the amount of time it takes a mechanic to rebuild the transmission in a certain type of
car has mean of 8.4 hours with standard deviation of 1.8 hours. If 35 mechanics are randomly
selected, approximate the probability that their mean rebuild time exceeds 9.1 hours.
Ans: Let X 35 denote the mean rebuild time of 35 mechanics. By the central limit theorem, X 35 is


approximately N 8.4,1.82 / 35  N (8.4, 0.548) . Thus
9.1  8.4 

P  X 35  9.1  P  Z 
  P(Z  0.95)  0.1711
0.548 

2. Suppose the number of knots on 10-ft long pine boards is described by a Poisson distribution
with parameter   5.3 . If 42 such boards are randomly chosen, approximate the probability that
the mean number of knots is greater than 8.
Ans: Since the population has a Poisson distribution, the population mean and variance are both
5.3. Let X 42 denote the mean number of knots on 42 boards. By the central limit theorem, X 42 is


approximately N 5.3,5.3 / 42  N (5.3, 0.818) . Thus
8  5.3 

P  X 42  8  P  Z 
  P( Z  2.99)  0.0014
0.181 

3. IQ scores have a mean of   100 and variance of  2  152 If a sample of 35 people are
randomly selected, find the probability their mean IQ is less than 105.
Ans: Let X 35 denote the mean IQ scores of 35 people. By the central limit theorem, X 35 is


approximately N 100,152 / 35  N (100,38.03) . Thus
105  100 

P  X 35  105  P  Z 
  P( Z  0.81)  0.7910
38.03 

4. A particular aircraft strobe light is designed so that the time between flashes is normally
distributed with a mean of 3.00 sec and a standard deviation of 0.40 sec. Find the probability
that for 60 randomly selected pairs of successive flashes, the mean time between flashes is
greater than 3.10 sec.
Ans: Let X 60 denote the mean time between flashes for 60 pairs. Since the population is


normally distributed, X 60 is N 3,0.042 / 60  N (3,0.000206) . Thus
3.1  3 

P  X 60  3.1  P  Z 
  P(Z  6.96)  0.0000
0.000206 

5. Which one of these options best describes the requirements for using the Central Limit
Theorem?
A. The population variances are assumed to be equal.
B. The population has an unknown variance.
C. np0  5
D. The population is normally distributed or n  30 .
Ans: D
6. Answer the following statements true or false. For each false statement, explain why the
statement is false.
A. To use the central limit theorem, the population must be normally distributed.
B. The central limit theorem says that the distribution of the sample mean is the same as the
distribution of the population.
C. To find an observed value of the sample mean X 42 , we need to observed 42 independent
values from the population, add them together, and divide by 42.
D. The central limit theorem says that the variance of the sample mean is greater than the
variance of the population.
Ans:
A. False: If the population is not normally distributed, then the sample size needs to be greater
than 30.
B. False: The CLT says that the sample mean is approximately normal, which may or may not
be the same as the distribution of the population.
C. True
D. False: The CLT says the variance of the sample mean is  2 / n which is less than  2 , the
population variance (when n  2 ).
Section 3.9
1. Suppose the number of calls that arrive at a technical support call center in an hour is
described by a Poisson distribution with a mean of 18 calls per hour. If the call center opens at
8:00 A.M., use software to find the probability that the 9th call of the day arrives before 8:30
A.M.
Ans: Let W denote the time (in hours) until the 9th call arrives. The W has a gamma distribution
is parameters r  9 and   18 so that its pdf is
189
f ( w) 
w91e 18 w  4919625w8e 18 w .
9

1
!
 
The 9th call of the day arrives before 8:30 A.M. if W  0.5 . The probability of this occurring is
0.5
P(W  0.5)   4919625w8e18w dw  0.544
0
2. Use the definition of the Gamma function to show that (t )  (t  1)(t  1) .
Ans: Using integration by parts where u  y t 1 and dv  e y dy , we get

(t )   y t 1 e y dy
0


  y t 1e y   (t  1) y t 2 e  y dy
0
0

 0  (t  1)  y t 2 e y dy
0
 (t  1)(t  1)
3. Minor flaws in the production of metal tubing appear an average of one every two feet.
Assuming the number of flaws in a foot is described by a Poisson distribution, approximate the
probability that 10 flaws appear within 28.41 feet of randomly selected tubing.
Ans: Let W denote the distance from one point in the tubing until the 10th flaw appears. Then W
has a gamma distribution with parameters r  10 and   1/ 2 . But this is the same as a chisquare distribution where n  20 . We want to find P(W  28.41) . From Table C.3, we see that
2
 0.1
(20)  28.41 . This means that P(W  28.41)  1  0.1  0.9 . Thus the desired probability is
0.9.
4. If T has a Student-t distribution with the given number of degrees of freedom, find the
indicated probability:
A.
B.
C.
D.
21 d.f., P (T  1.323)
10 d.f., P(T  2.764)
5 d.f., P(2.015  T  3.365)
27 d.f., P (T  2.473)  (T  1.314)
Ans:
A. 0.9
B. 0.01
C. 0.99  0.95  0.04
D. 0.01  0.9  0.91
Section 3.10
1. When played fairly, the probability of winning a single play of a certain card game is 0.10.
Suppose a gambler plays the game 400 times and wins 59 times.
A. Use the normal distribution to approximate the probability of winning at least 59 times if the
game is played fairly.
B. Based on this result, does it appear the gambler played the game fairly? Explain.
Ans:
A. Let X denote the number of wins in 400 plays of the game. Then X is b(400, 0.10) Let Y be
N (400(0.1), 400(0.1)(0.9))  N (40,36) . Then by the Limit Theorem of De Moivre and Laplace,
58.5  40 

P( X  59)  P(Y  58.5)  P  Z 
  P( Z  3.08)  0.001
36 

B. Since this probability is small, it indicated that the gambler did not play fairly.
2. If X is b(284, 0.65), approximate P(X > 200).
Ans: Let Y be N (284(0.65), 284(0.65)(0.35))  N (184.6, 64.61) . Then by the Limit Theorem of
De Moivre and Laplace,
200.5  184.6 

P( X  200)  P(Y  200.5)  P  Z 
  P( Z  1.98)  0.0239
64.61 

3. If X is b(n, p ) , using the Limit Theorem of De Moivre and Laplace, we can approximate
P( X  a)  P(a  0.5  Y  a  0.5)
where 0  a  n and Y is N (np, np(1  p)) . Give a geometric reason to explain why we add and
subtract 0.5 in this approximation
Ans: If we draw a probability histogram of X, p  P( X  a ) is the area of a rectangle with width
1 centered at a with height p. Drawing the density curve of Y over this histogram, we see that the
area of this rectangle is approximately the area under the curve between a  0.5 and a  0.5 .