Statistics

# Solutions

StatisticsSolutions
1.
1. P(L′) = P(S)
2. P(M OR S)
3. P(F AND L)
4. P(M|L)
5. P(L|M)
6. P(S|F)
7. P(F|L)
8. P(F OR L)
9. P(M AND S)
10. P(F)
3.

P(N) = $15 42 15 42$ = $5 14 5 14$ = .36

5.

P(C) = $5 42 5 42$ = .12

7.

P(G) = $20 150 20 150$ = $2 15 2 15$ = .13

9.

P(R) = $22 150 22 150$ = $11 75 11 75$ = .15

11.

P(O) = $150−22−38−20−28−26 150 150−22−38−20−28−26 150$ = $16 150 16 150$ = $8 75 8 75$ = .11

13.

P(E) = $47 194 47 194$ = .24

15.

P(N) = $23 194 23 194$ = .12

17.

P(S) = $12 194 12 194$ = $6 97 6 97$ = .06

19.

$13 52 13 52$ = $1 4 1 4$ = .25

21.

$3 6 3 6$ = $1 2 1 2$ = .5

23.

$P(R)= 4 8 =.5 P(R)= 4 8 =.5$

25.

P(O OR H)

27.

P(H|I)

29.

P(N|O)

31.

P(I OR N)

33.

P(I)

35.

The likelihood that an event will occur given that another event has already occurred.

37.

1

39.

the probability of landing on an even number or a multiple of three

41.

P(J) = .3

43.

P(Q AND R) = P(Q)P(R)

.1 = (.4)P(R)

P(R) = .25

45.

0.376

47.

C|L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.

49.

L AND C is the event that the person chosen is a voter of the ethnicity in question who prefers life without parole over the death penalty for a person convicted of first degree murder.

51.

.6492

53.

No, because P(L AND C) does not equal 0.

55.

P(musician is a male AND had private instruction) = $15 130 15 130$ = $3 26 3 26$ = .12

57.

P(being a female musician AND learning music in school) = $38 130 38 130$ = $19 65 19 65$ = .29

P(being a female musician)P(learning music in school) = $( 72 130 )( 62 130 ) ( 72 130 )( 62 130 )$ = $4,464 16,900 4,464 16,900$ = $1,116 4,225 1,116 4,225$ = .26

No, they are not independent because P(being a female musician AND learning music in school) is not equal to P(being a female musician)P(learning music in school).

58.
Figure 3.23
60.

$35,065100,45035,065100,450$

62.

To pick one person from the study who is Japanese American AND uses the product 21 to 30 times a day means that the person has to meet both criteria: both Japanese American and uses the product 21 to 30 times a day. The sample space should include everyone in the study. The probability is $4,715100,4504,715100,450$.

64.

To pick one person from the study who is Japanese American given that person uses the product 21 to 30 times a day, means that the person must fulfill both criteria and the sample space is reduced to those who uses the product 21 to 30 times a day. The probability is $471515,273471515,273$.

67.
1. You can't calculate the joint probability knowing the probability of both events occurring, which is not in the information given; the probabilities should be multiplied, not added; and probability is never greater than 100 percent
2. A home run by definition is a successful hit, so he has to have at least as many successful hits as home runs.
69.

0

71.

.3571

73.

.2142

75.

Physician (83.7)

77.

83.7 − 79.6 = 4.1

79.

P(Occupation < 81.3) = .5

81.
1. The Forum Research surveyed 1,046 Torontonians.
2. 58 percent
3. 42 percent of 1,046 = 439 (rounding to the nearest integer)
4. .57
5. .60.
82.
1. yes; P(getting a pork chop) = P(not getting a chicken breast)
2. getting a pork chop and getting a chicken breast
3. no
83.
1. 20/40 = 1/2
2. 5/40 = 1/8
3. 39/40
4. 4/40 = 1/10
5. 33/40
6. 15/40 = 3/8
7. 0/40 = 0
84.

Compute the probabilities.

1. 20/40 = 1/2
2. 8/40 = 1/5
3. 40/40 = 1
4. 16/40 = 2/5
5. 18/40 = 9/20
6. 40/40 = 1
85.
1. {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2. $5 8 5 8$
3. $2 3 2 3$
4. $2 8 2 8$
5. $6 8 6 8$
6. No, because P(G AND E) does not equal 0.
87.

### NOTE

The coin toss is independent of the card picked first.

1. {(G,H) (G,T) (B,H) (B,T) (R,H) (R,T)}
2. P(A) = P(blue)P(head) = $( 3 10 ) ( 3 10 )$$( 1 2 ) ( 1 2 )$ = $3 20 3 20$
3. Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P(A AND B) = 0.
4. No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A; if the card chosen is blue it is also (red or blue). P(A AND C) = P(A) = $3 20 . 3 20 .$
89.
1. S = {(HHH), (HHT), (HTH), (HTT), (THH), (THT), (TTH), (TTT)}
2. $4 8 4 8$
3. Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P(A AND B) = 0.
91.
1. If Y and Z are independent, then P(Y AND Z) = P(Y)P(Z), so P(Y OR Z) = P(Y) + P(Z) – P(Y)P(Z).
2. .5
93.

iii i iv ii

95.
1. P(R) = .44
2. P(R|E) = .56
3. P(R|O) = .31
4. No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P(R|E) ≠ P(R).
5. No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P(R|E) > P(R).
97.
1. P(type O OR Rh–) = P(type O) + P(Rh–) – P(type O AND Rh–)

0.52 = 0.43 + 0.15 – P(type O AND Rh–); solve to find P(type O AND Rh–) = .06

6 percent of people have type O, Rh– blood

2. P(NOT(type O AND Rh–)) = 1 – P(type O AND Rh–) = 1 – .06 = .94

94 percent of people do not have type O, Rh– blood

99.
1. Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
2. P(C OR N) = P(C) + P(N) – P(C AND N) = .36 + .12 – .08 = .40
3. P(NEITHER chocolate NOR nuts) = 1 – P(C OR N) = 1 – .40 = .60
101.

0

103.

$10 67 10 67$

105.

$10 34 10 34$

107.

d

110.

b

112.
1. $26 106 26 106$
2. $33 106 33 106$
3. $21 106 21 106$
4. $( 26 106 ) ( 26 106 )$ + $( 33 106 ) ( 33 106 )$$( 21 106 ) ( 21 106 )$ = $( 38 106 ) ( 38 106 )$
5. $21 33 21 33$
114.

a

117.
1. P(C) = .4567
2. not enough information
3. not enough information
4. no, because over half (0.51) of men have at least one false-positive text
119.
1. P(J OR K) = P(J) + P(K) − P(J AND K); .45 = .18 + .37 – P(J AND K); solve to find P(J AND K) = .10
2. P(NOT (J AND K)) = 1 – P(J AND K) = 1 – 010 = .90
3. P(NOT (J OR K)) = 1 – P(J OR K) = 1 – .45 = .55
120.
1. Figure 3.24
2. P(GG) = $( 5 8 )( 5 8 ) ( 5 8 )( 5 8 )$ = $25 64 25 64$
3. P(at least one green) = P(GG) + P(GY) + P(YG) = $25 64 25 64$ + $15 64 15 64$ + $15 64 15 64$ = $55 64 55 64$
4. P(G|G) = $5 8 5 8$
5. Yes, they are independent because the first card is placed back in the bag before the second card is drawn. The composition of cards in the bag remains the same from draw one to draw two.
122.
1. <2020–64>64Totals
Female .0244 .3954 .0661 .486
Male .0259 .4186 .0695 .514
Totals .0503 .8140 .1356 1
Table 3.27
2. P(F) = .486
3. P(>64|F) = .1361
4. P(>64 and F) = P(F) P(>64|F) = (.486)(.1361) = .0661
5. P(>64|F) is the percentage of female drivers who are 65 or older and P(>64 and F) is the percentage of drivers who are female and 65 or older.
6. P(>64) = P(>64 and F) + P(>64 and M) = .1356
7. No, being female and 65 or older are not mutually exclusive because they can occur at the same time P(>64 and F) = .0661.
124.
1. Car, Truck or Van Walk Public Transportation Other Totals
Alone .7318
Not Alone .1332
Totals .8650 .0390 .0530 .0430 1
Table 3.28
2. If we assume that all walkers are alone and that none from the other two groups travel alone (which is a big assumption) we have: P(Alone) = .7318 + .0390 = .7708.
3. Make the same assumptions as in (b) we have: (.7708)(1,000) = 771
4. (.1332)(1,000) = 133
126.

The completed contingency table is as follows:

Method A Method B Method C Other Totals
Female 0 70 136 49 255
Male 2,146 463 60 135 2,804
Totals 2,146 533 196 184 3,059
Table 3.29
1. $255 3059 255 3059$
2. $196 3059 196 3059$
3. $718 3059 718 3059$
4. 0
5. $463 3059 463 3059$
6. $136 196 136 196$
7. Figure 3.25
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