Mann - Introductory Statistics, Fifth Edition, Solutions Manual 141

5.1 Random variable: A variable whose value is determined by the outcome of a random experiment is called a random variable. An example of this is the income of a randomly selected family.

Discrete random variable: A random variable whose values are countable is called a discrete random variable. An example of this is the number of cars in a parking lot at any particular time.

Continuous random variable: A random variable that can assume any value in one or more intervals is called a continuous random variable. An example of this is the time taken by a person to travel by car from New York City to Boston.

5.2 a. a continuous random variable b. a discrete random variable

c. a discrete random variable d. a continuous random variable

e. a discrete random variable f. a continuous random variable

5.3 a. a discrete random variable b. a continuous random variable

c. a continuous random variable d. a discrete random variable

e. a discrete random variable f. a continuous random variable

5.4 The number of households x watching ABC is a discrete random variable because the values of x are countable: 0, 1, 2, 3, 4 and 5.

5.5 The number of cars x that stop at the Texaco station is a discrete random variable because the values of x are countable: 0, 1, 2, 3, 4, 5 and 6.

5.6 The probability distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities. As an example, the following table lists the probability distribution of x where x is the number of cars owned by a family living in a city.

x / P(x)
0 / 0.04
1 / 0.32
2 / 0.35
3 / 0.18
4 / 0.11

The probability distribution of a discrete random variable can be presented in the form of a mathematical formula, a table, or graph.

5.7 The two characteristics of the probability distribution of a discrete random variable x are:

1. The probability that x assumes any single value lies in the range 0 to 1, that is, 0.

2. The sum of the probabilities of all values of x for an experiment is equal to 1, that is: .

5.8 a. This table does represent a valid probability distribution of x because it satisfies both conditions required for a valid probability distribution.

b. This table does not represent a valid probability distribution of x because the sum of the probabilities of all outcomes listed in the table is not 1, which violates the second condition of a probability distribution.

c. This table does not represent a valid probability distribution of x because the probability of x = 7 is negative, which violates the first condition of a probability distribution.

5.9 a. This table does not satisfy the first condition of a probability distribution because the probability of x = 5 is negative. Hence, it does not represent a valid probability distribution of x.

b. This table represents a valid probability distribution of x because it satisfies both conditions required for a valid probability distribution.

c. This table does not represent a valid probability distribution of x because the sum of the probabilities of all outcomes listed in the table is not 1, which violates the second condition of a probability distribution.

5.10 a. P(x = 3) = .15

b. P(x 2) = P(0) + P(1) + P(2) = .11 + .19 + .28 = .58

c. P(x 4) = P(4) + P(5) + P(6) = .12 + .09 + .06 = .27

d. P(1 x 4) = P(1) + P(2) + P(3) + P(4) = .19 + .28 + .15 + .12 = .74

e. P(x 4) = P(0) + P(1) + P(2) + P(3) = .11 + .19 + .28 + .15 = .73

f. P(x 2) = P(3) + P(4) + P(5) + P(6) = .15 + .12 + .09 + .06 = .42

g. P(2 x 5) = P(2) + P(3) + P(4) + P(5) = .28 + .15 + .12 + .09 = .64

5.11 a. P(x = 1) = .17

b. P(x 1) = P(0) + P(1) = .03 + .17 + = .20

c. P(x 3) = P(3) + P(4) + P(5) = .31 + .15 + .12 = .58

d. P(0 x 2) = P(0) + P(1) + P(2) = .03 + .17 + .22 = .42

e. P(x 3) = P(0) + P(1) + P(2) = .03 + .17 + .22 = .42

f. P(x 3) = P(4) + P(5) = .15 + .12 = .27

g. P(2 x 4) = P(2) + P(3) + P(4) = .22 + .31 + .15 = .68

5.12 Let x = number of exercise machines sold at Elmo’s on a given day.

b. i. P(exactly 6 ) = P(6) = .14

ii. P(more than 8) = P(x > 8) = P(9) + P(10) = .16 + .12 = .28

iii. P(5 to 8) =P (5 8) = P(5) + P(6) + P(7) + P(8) = .11 + .14 + .19 + .20 = .64

iv. P(at most 6) = P(x 6) = P(4) + P(5) + P(6) = .08 + .11 + .14 = .33

5.13 a.

b. i. P(exactly 3) = P(3) = .25

ii. P(at least 4) = P(x 4) = P(4) + P(5) + P(6) = .14 + .07 + .03 = .24

iii. P(less than 3) = P(x 3) = P(0) + P(1) + P(2) = .10 + .18 + .23 = .51

iv. P(2 to 5) = P(2) + P(3) + P(4) + P(5) = .23 + .25 + .14 + .07 = .69

5.14 a. Let x denote the number of TV sets owned by a family. The following table gives the probability distribution of x.

x / P(x)
0 / 120 / 2500 = .048
1 / 970 / 2500 = .388
2 / 730 / 2500 = .292
3 / 410 / 2500 = .164
4 / 270 / 2500 = .108

b. The probabilities listed in the table of part a are exact because they are based on data from the entire population.

c. i. P(x = 1) = .388

ii. P(x > 2) = + P(3) + P(4) = .164 + .108 = . 272

iii. P(x 1) = P(0) + P(1) = .048 + .338 = .436

iv. P(1 3) = P(1) + P(2) + P(3) = .388 + .292 + .164 = .844

5.15 a.

x / P(x)
1 / 8 / 80 = .10
2 / 20 / 80 = .25
3 / 24 / 80 = .30
4 / 16 / 80 = .20
5 / 12 / 80 = .15

b. The probabilities listed in the table of part a are approximate because they are obtained from a sample of 80 days.

c. i. P(x = 3) = .30

ii. P(x 3) = P(3) + P(4) + P(5) = .30 + .20 + .15 = .65

iii. P(2 4) = P(2) + P(3) + P(4) = .25 + .30 + .20 = .75

iv. P(x ≤4) = P(2) + P(3) + P(4) = .10 + .25 + .30 = .65

5.16 Let L = car selected is a lemon and G = car selected is good

Then, P(L) = .05 and P(G) = 1 – .05 = .95

Let x be the number of lemons in two selected cars. The following table lists the probability distribution of x. Note that x = 0 if both cars are good, x = 1 if one car is good and the other car is a lemon, and x = 2 if both cars are lemons. The probabilities are written in the table using the above tree diagram. The probability of x = 1 is obtained by adding the probabilities of LG and GL.

Outcomes / x / P(x)
GG / 0 / .9025
LG or GL / 1 / .0950
LL / 2 / .0025

5.17 Let Y = owns a cell phone and N = does not own a cell phone.

Then P(Y) = .64 and P(N) = 1 – .64 = .36

Let x be the number of adults in a sample of two who own a cell phone. The following table lists the probability distribution of x. Note that x = 0 if neither adult owns a cell phone, x = 1 if one adult owns a cell phone and the other does not, and x = 2 if both adults own a cell phone. The probabilities are written in the table using the tree diagram above. The probability that x = 1 is obtained by adding the probabilities of YN and NY

Outcomes / x / P(x)
NN / 0 / .1296
YN or NY / 1 / .4608
YY / 2 / .4096

5.18 Let: A = adult selected is against using animals for research

N = adult selected is not against using animals for research

Then, P(A) = .30 and P(N) = 1 – .30 = .70

Let x be the number of adults in a sample of two adults who are against using animals for research. The following table lists the probability distribution of x.

Outcomes / x / P(x)
NN / 0 / .49
AN or NA / 1 / .42
AA / 2 / .09

5.19 Let: Y = teen said teachers were “totally clueless” about using the internet for teaching and learning.

N = teen said teachers were not “totally clueless” about using the internet for teaching and learning.

Then P(Y) = .78 and P(N) = 1 – .78 = .22

Let x denote the number of teens in a sample of two teens who believe that teachers are “totally clueless” about using the internet for teaching and learning. The following table lists the probability distribution of x.

Outcomes / x / P(x)
NN / 0 / .0484
AN or NA / 1 / .3432
AA / 2 / .6084

5.20 Let: A = first person selected is lefthanded C = second person selected is left handed

B = first person selected is righthanded D = second person selected is righthanded

Outcomes / x / P(x)
BD / 0 / .5455
AD or BC / 1 / .4090
AC / 2 / .0455

5.21 Let A = first athlete selected used drugs C = second athlete selected used drugs

B = first athlete selected did not use drugs D = second athlete selected did not use drugs

Let x be the number of athletes who used illegal drugs in a sample of two athletes. The following table lists the probability distribution of x.

Outcomes / x / P(x)
BD / 0 / .4789
AD or BC / 1 / .4422
AC / 2 / .0789

5.22 The mean of discrete random variable x is the value that is expected to occur per repetition, on average, if an experiment is repeated a large number of times. It is denoted by and calculated as

The standard deviation of a discrete random variable x measures the spread of its probability distribution. It is denoted by and is calculated as

5.23 a.

x / P(x) / xP(x) / x2P(x)
0 / .16 / 0 / 0
1 / .27 / .27 / .27
2 / .39 / .78 / 1.56
3 / .18 / .54 / 1.62
P(x)=3.45

x / P(x) / xP(x) / x2P(x)
6 / .40 / 2.40 / 14.40
7 / .26 / 1.82 / 12.74
8 / .21 / 1.68 / 13.44
9 / .13 / 1.17 / 10.53

5.24 a.

x / P(x) / xP(x) / x2P(x)
3 / .09 / .27 / .81
4 / .21 / .84 / 3.36
5 / .34 / 1.70 / 8.50
6 / .23 / 1.38 / 8.28
7 / .13 / .91 / 6.37

x / P(x) / xP(x) / x2P(x)
0 / .43 / 0 / 0
1 / .31 / .31 / .31
2 / .17 / .34 / .36
3 / .09 / .27 / .81

5.25

x / P(x) / xP(x) / x2 P(x)
0 / .73 / 0 / .00
1 / .16 / .16 / .16
2 / .06 / .12 / .24
3 / .04 / .12 / .36
4 / .01 / .04 / .16

error

5.26

x / P(x) / xP(x) / x2 P(x)
4 / .17 / .68 / 2.27
5 / .26 / 1.30 / 6.50
6 / .19 / 1.14 / 6.84
7 / .13 / .91 / 6.37
8 / .11 / .88 / 7.04
9 / .10 / .90 / 8.10
10 / .04 / .40 / 4.00

pigs

5.27

x / P(x) / xP(x) / x2 P(x)
0 / .05 / 0 / 0
1 / .12 / .12 / .12
2 / .19 / .38 / .76
3 / .30 / .90 / 2.70
4 / .20 / .80 / 3.20
5 / .10 / .50 / 2.50
6 / .04 / .24 / 1.44

camcorders

On average, 2.94 camcorders are sold per day at this store.

5.28 Let x be the number of exercise machines sold on a given day at Elmo’s.

x / P(x) / xP(x) / x2 P(x)
4 / .08 / .32 / 1.28
5 / .11 / .55 / 2.75
6 / .14 / .84 / 5.04
7 / .19 / 1.33 / 9.31
8 / .20 / 1.60 / 12.80
9 / .16 / 1.44 / 12.96
10 / .12 / 1.20 / 12.00

machines

machines.

The value of the mean,=7.28, indicates that Elmo’s sells an average of 7.28 exercise machines per day.

5.29

x / P(x) / xP(x) / x2 P(x)
0 / .25 / 0 / 0
1 / .50 / .50 / .50
2 / .25 / .50 / 1.00

head

heads

The value of the mean, = 1.00, indicates that, on average, we will expect to obtain 1 head in every two tosses of the coin.

5.30

x / P(x) / xP(x) / x2 P(x)
0 / .14 / .00 / .00
1 / .28 / .28 / .28
2 / .22 / .44 / .88
3 / .18 / .54 / 1.62
4 / .12 / .48 / 1.92
5 / .06 / .30 / 1.50
6.20

2.04 potential weapons

potential weapons

The value of the mean, = 2.04, indicates, on average, 2.04 potential weapons are found per day.

5.31

x / P(x) / xP(x) / x2 P(x)
0 / .048 / 0 / 0
1 / .388 / .388 / .388
2 / .292 / .584 / 1.168
3 / .164 / .492 / 1.476
4 / .108 / .432 / 1.728

TV sets

Thus, there is an average of 1.90 TV sets per family in this town, with a standard deviation of 1.079 sets.

5.32

x / P(x) / xP(x) / x2 P(x)
1 / .10 / .10 / .00
2 / .25 / .50 / 1.00
3 / .30 / .90 / 2.70
4 / .20 / .80 / 3.20
5 / .15 / .75 / 3.75

3.05 installations

1.203 installations

Thus, the average number of installations is 3.05 per day with a standard deviation of 1.203 installations.

5.33

x / P(x) / xP(x) / x2 P(x)
0 / .9025 / 0 / 0
1 / .0950 / .0950 / .0950
2 / .0025 / .0050 / .0100

car

5.34

x / P(x) / xP(x) / x2 P(x)
0 / .1296 / 0 / 0
1 / .4608 / .4608 / .4608
2 / .4096 / .8192 / 1.6384

adults

adult

5.35

x / P(x) / xP(x) / x2 P(x)
10 / .15 / 1.50 / 15.00
5 / .30 / 1.50 / 7.50
2 / .45 / 0.90 / 1.80
0 / .10 / 0 / 0
24.30

million

million.

Thus, the contractor is expected to make $3.9 million profit with a standard deviation of $3.015 million.

5.36 Note that the price of the ticket ($2) must be deducted from the amount won. For example, the $5 prize results in a net gain of $5 – $2 = $3.

x / P(x) / xP(x) / x2 P(x)
3 / 1000/10,000=.10 / .30 / 0.9
8 / 100/10,000=.01 / .08 / 0.64
998 / 5/10,000=.0005 / .499 / 498.002
4998 / 1/10,000=.0001 / .4998 / 2498
2 / 8894/10,000=.8894 / 1.7788 / 3.5576

Thus, on average, the players who play this game are expected to lose $.40 per ticket with a standard deviation of $54.78.

5.37

x / P(x) / xP(x) / x2 P(x)
0 / .5455 / 0 / 0
1 / .4090 / .4090 / .4090
2 / .0455 / .0910 / .1820

person

5.38

x / P(x) / xP(x) / x2 P(x)
0 / .4789 / .0000 / .0000
1 / .4422 / .4422 / .4422
2 / .0789 / .1578 / .3156

athlete

5.39 3! = 3 2 1 = 6

(9 3)! = 6! = 654321 = 720

9! = 987654321 = 362,880

(14 – 12)! = 2! = 21 = 2

= == = 10

= == = 35

= == = 84

= == = 1

5.40 6! = 6 54321 = 720