1
Chapter 5.1 # 7, 9, 11, 13, 15, 17, 19, 25, 27
For exercises 6 through 11, determine whether the distribution represents a probability
distribution. If it does not, state why.
7.
X
3
6
8
12
P(x) 0.3 0.5 0.7 -0.8
No, we cannot have negative numbers, and P(x) does not equal 1.
9.
=
X
P(x)
1
10
= 1, yes 1.
10
11.
X
P(x)
2
3
4
5
Each P(x) is between 0 and 1.
5
1.2
10
0.3
 2  1 , No, not equal to 1.
 p ( x) 
10
 1.
10
15
0.5
1. P(x) = 1.2 is not between 0 and 1, 2.
 p ( x)  2  1
For Exercises 12 through 18, state whether the variable is discrete or continuous.
13. The number of cheeseburgers a fast-food restaurant serves each day.
Discrete
15. The weight of a Siberian tiger.
Continuous
17. The number of mathematics majors in your school.
Discrete
For Exercises 19 through 26, construct a probability distribution for the data and draw a graph for
the distribution.
19. Medical tests the probabilities that a patient will have 0, 1, 2, or 3 medical tests performed on
entering a hospital are
6 5 3
1
, , , and
respectively.
15 15 15
15
2
X
0
1
2
3
P(X)
6/15
5/15
3/15
1/15
25. Student Classes The probabilities that a student is registered for 2, 3, 4, or 5 classes are 0.01, 0.34,
0.62, and 0.03, respectively.
27. Selecting a Monetary Bill A box contains three $1 bills, two $5 bills, five $10 bills, and one $20
bill. Construct a probability distribution for the data if x represents the value of a single bill drawn at
random and then replaced.
Chapter 5.2 #3, 7, 11, 12, 13, 15, 18
3. Number of Credit Cards A bank vice president feels that each savings account customer has, on
average, three credit cards. The following distribution represents the number of credit cards
people won. Find the mean, variance, and standard deviation. Is the vice president correct?
3
Number of
Credit
Cards X
Probability
P(X)
Mean:  
0
1
2
3
4
0.18
0.44
0.27
0.08
0.03
 X P( X )  0  0.18 1 0.44  2  0.27  3  0.08  4  0.03  1.34
Variance  2  
X
2
 P( X )    2 
02  0.18  12  0.44  22  0.27  32  0.08  42  0.03  1.342  2.72  1.7956  0.9
Standard deviation
  0.9  1
No, the vice president is not correct. On average, each person has about one credit card.
7. Commercials during children’s TV programs a concerned parents group determined the number of
commercials shown in each of five children’s programs over a period of time. Find the mean, variance,
and standard deviation shown.
Number of commercials X 5
6
7
8
9
Probability P(x)
0.2 0.25 0.38 0.10 0.07
x
x2
5 25
6 36
7 49
8 72
9 81
Sum =
P(x) X P(x)
x 2 P (x)
0.2
0.25
0.38
0.10
0.07
5
9
18.62
6.4
5.67
44.69
1
1.5
2.66
0.8
0.63
6.59
   X P( X )  5(0.2) + 6(0.25) +7(0.38) + 8(0.10) + 9(0.07)
= 6.59 or 6.6
  44.69  (6.59)2  44.69  43.43  1.26
2
  1.26  1.123 1.12
11. Insurance an insurance company insures a person’s antique coin collection worth $20,000 for an
annual premium of $300. If the company figures that the probability of the collection being stolen is
0.002, what will be the company’s expected profit?
P (of being stolen) = 0.002
Worth=20,000
P (of not being stolen) = 0.998
premium=300=20,000-300=19,700
x
P(x)
x p(x)
4
Profit 300
0.998 299.4
Loss -19,700 0.002 -39.4
Sum =
260
Expected profit is $260.0.
12. Job Bids a landscape contractor bids on jobs where he can make $3000 profit. The probabilities of
getting one, two, three, or four jobs per month are shown.
Number of jobs X 1
2
3
4
Probability P(x)
0.2 0.3 0.4 0.1
Find the contractor’s expected profit per month.
Mean:  
 X P( X )  1(0.2)  2(0.3)  3(0.4)  4(0.1)  2.4
Per month the contractor’s expected profit is 2.4($3000) = $7200
13. Rolling Dice if a person rolls doubles when he tosses two dice, he wins $5. For the game to be fair,
how much should the person pay to play the game?
Win
Gain x
$5
Probability P(x)
1/6
   X P( X )  5(1/ 6)  0.83 Therefore, he has to pay $0.83 for the game to be fair.
15. Lottery prizes a lottery offers one $1000 prize, one $500 prize, and five $100 prizes. One thousand
tickets are sold at $3 each. Find the expectation if a person buys one ticket.
win
win
win
loss
Sum
x
1000-3
500-3
100-3
-3
=
P(x)
1/1000
1/1000
5/1000
993/1000
x p(x)
997/1000
497/1000
485/1000
-2979/1000
-1000/1000=-1
1
1
5
1000 (
)  500 (
)  100 (
)  3  1.00
1000
1000
1000
Expectation of buying one ticket is losing one dollar
Or -$1.00.
18. Life Insurance a 35-year-old woman purchases a $100,000 term life insurance policy for an annual
payment of $360. Based on a period life table for the U.S. government, the probability that she will
survive the year is 0.999057. Find the expected value of the policy for the insurance company.
Win
Lose
5
Gain X
Probability P(X)
$360
0.999057
-99,640
0.000943
E ( x)     x p( x)  360(.999057)  99640(.000943)  265.70
The expected value of the policy for the insurance company is $265.70
Section 5-3 #’s 1, 3ae, 5, 7, 11, 14bgh, 15, 17 ,19, 21 ,25
1) Which of the following are binomial experiments or can be reduced to binomial experiments?
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
Surveying 100 people to determine if they like Sudsy Soap Yes
Tossing a coin 100 times to see how many heads occur Yes
Drawing a card with replacement from a deck and getting a heart Yes
Asking 1000 people which brand of cigarettes they smoke No
Testing four different brands of aspirin to see which brands are effective No
Testing one brand of aspirin by using 10 people to determine whether it is effective Yes
Asking 100 people if they smoke Yes
Checking 1000 applicants to see whether they were admitted to White Oak College Yes
Surveying 300 prisoners to see how many different crimes they were convicted of No
Surveying 300 prisoners to see whether this is their first offense Yes
3) Compute the probability of X successes, using the binomial formula.
P( X ) 
n!
p X q n X 
( n  X )! X !
a) n=6, X=3, p=0.03
n
CX
p X q n X
then q= 1 -0.03 = 0.97
P ( X )  6 C3 0.03 0.97 6 3  20(0.000027)(0.912673)  0.0005
3
e) n=10, X=5, p=0.37
P( X ) 
then q= 1 – 0.37 = 0.63
C5 0.37 0.6310 5  252(0.37)5 (0.63) 5  0.173
5
10
5) True/False Exam a student takes a 20-question, true/false exam and guesses on each question.
Find the probability of passing if the lowest passing grade is 15 correct out of 20. Would you
consider this event likely to occur? Explain your answer.
n=20, X  15, p= 1 , q= 1
2
2
6
1
1
1
1
P( X )  20 C15 ( )15 ( ) 2015  15,504( )15 ( )5  0.015
2
2
2
2
1
1
1
1
P( X )  20 C16 ( )16 ( ) 2016  4,845( )16 ( ) 4  0.005
2
2
2
2
1 17 1 2017
1 17 1 3
P( X )  20 C17 ( ) ( )
 1,140( ) ( )  0.001
2
2
2
2
1 18 1 2018
1 18 1 2
P( X )  20 C18 ( ) ( )
 190( ) ( )  0.0002
2
2
2
2
1 19 1 2019
1 19 1 1
P( X )  20 C19 ( ) ( )
 20( ) ( )  0.00002
2
2
2
2
1 20 1 20 20
1 20 1 0
P( X )  20 C20 ( ) ( )
 1( ) ( )  0.000000954
2
2
2
2
so... P( X )  0.015  0.005  0.001  0.0002  0.00002  0.000000954  0.021
No, this event would not likely occur because it’s only about a 2% chance.
7) Driving to Work Alone It is reported that 77% of workers aged 16 and over drive to work alone.
Choose 8 workers at random. Find the probability that
a. All drive to work alone
n  8, p  0.77,X  8
P( X )  8 C8 (0.77)8 (0.23)0  0.124
b. More than one-half drive to work alone
n  8, p  0.77, X  5, 6, 7, or 8
P( X )  8 C5 (0.77)5 (0.23)3  8 C6 (0.77)6 (0.23) 2  8 C7 (0.77) 7 (0.23)1  8 C8 (0.77)8 (0.23)0
 0.184  0.309  0.295  0.124  0.912
c. Exactly 3 drive to work alone
n  8, p  0.77, X  3
P( X )  8 C3 (0.77)3 (0.23)5  0.017
11) College Education and Business World Success R. H. Bruskin Associates Market Research found that
40% of Americans do not think that having a college education is important to succeed in the
business world. If a random sample of five Americans is selected, find these probabilities.
a) Exactly two people will agree with that statement.
n=5, p=0.4, q=0.6, X=2
P( X  2)  5 C2 (0.4) 2 (0.6)3  0.346
b) At most three people will agree with that statement.
n=5, p=0.4, q=0.6, X=2
P( X  3)  P( x  0)  P( x  1)  P( x  2)  P( x  3) 
5
C0 (0.4)0 (0.6)5  5 C1 (0.4)1 (0.6) 4  5 C2 (0.4) 2 (0.6)3 
5
C3 (0.4)3 (0.6) 2  0.913
7
c) At least two people will agree with that statement.
P( x  2)  1  P( x  2)  1  ( P( x  0)  P( x  1))  1  (0.078  0.259)  0.663
d) Fewer than three people will agree with that statement.
P( x  3)  P( x  0)  P( x  1)  P( x  2)  0.078  0.259  0.346  0.683
14) Find the mean, variance, and standard deviation for each of the values of n and p when the
conditions for the binomial distribution are met.
b.
n  300, p  0.3
  300(0.3)  90
 2  300(0.3)(0.7)  63
  63  7.94  7.9
g.
n  50, p 
2
5
2
5
2 3
 2  50( )( )  12
5 5
  50( )  20
  12  3.464  3.5
h.
n  36, p 
1
6
1
6
1 5
2
  36( )( )  5
6 6
  36( )  6
  5  2.236  2.2
15) Social Security Recipients A study found that 1% of social security recipients are too young to vote.
If 800 social security recipients are randomly selected, find the mean, variance, and standard
deviation of the number of recipients who are too young to vote.
  n  p  800(0.01)  8
 2  n  p  q  800(0.01)(0.99)  7.9
  n  p  q  7.9  2.8
8
17) Defective Calculators If 3% of calculators are defective, find the mean, variance, and standard
deviation of a lot of 300 calculators.
n  300, p  0.03
  300(0.03)  9
 2  300(0.03)(0.97)  8.73
  8.73  2.95
19) Watching Fireworks a survey found that 21% of Americans watch fireworks on television on July 4.
Find the mean, variance, and standard deviation of the number of individuals who watch fireworks
on television on July 4 if a random sample of 1000 Americans is selected.
  n  p  1000(0.21)  210
 2  n  p  q  1000(0.21)(0.79)  165.9
  n  p  q  165.9  12.9
21) Survey on Bathing Pets a survey found that 25% of pet owners had their pets bathed
professionally rather than do it themselves. If 18 pet owners are randomly selected, find the
probability that exactly five people have their pets bathed professionally.
n=18, p=0.25, q=0.75, X=5
P( X )  18 C5 (0.25)5 (0.75)13  0.199
25) Survey on Internet Awareness In a survey, 58% of American adults said they had never heard of
the Internet. If 20 American adults are selected at random, find the probability that exactly 12 will
say they have never heard of the Internet.
P( X ) 
20
C12 (0.58)12 (0.42)8  0.177