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Chapter 7
7.1 a 0, 1, 2, &
b Yes, we can identify the first value (0), the second (1), and so on.
c It is finite, because the number of cars is finite.
d The variable is discrete because it is countable.
7.2 a any value between 0 and several hundred miles
b No, because we cannot identify the second value or any other value larger than 0.
c No, uncountable means infinite.
d The variable is continuous.
7.3 a The values in cents are 0 ,1 ,2, &
b Yes, because we can identify the first ,second, etc.
c Yes, it is finite because students cannot earn an infinite amount of money.
d Technically, the variable is discrete.
7.4 a 0, 1, 2, &, 100
b Yes.
c Yes, there are 101 values.
d The variable is discrete because it is countable.
7.5 a No the sum of probabilities is not equal to 1.
b Yes, because the probabilities lie between 0 and 1 and sum to 1.
c No, because the probabilities do not sum to 1.
7.6 p(x) = 1/6 for x = 1, 2 ,3 ,4 ,5, and 6
7.7 a x p(x)
0 24,750/165,000 = .15
1 37,950/165,000 = .23
2 59,400/165,000 = .36
3 29,700/165,000 = .18
4 9,900/165,000 = .06
5 3,300/165,000 = .02
b (i) P(X £ 2) = p(0) + p(1) + p(2) = .15 + .23 + .36 = .74
(ii) P(X > 2) = p(3) + p(4) + p(5) = .18 + .06 + .02 = .26
(iii) P(X ³ 4) = p(4) + p(5) = .06 + .02 = .08
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7.8 a P(2 £ X £ 5) = p(2) + p(3) + p(4) + p(5) = .310 + .340 + .220 + .080 = .950
b P(X > 5) = p(6) + p(7) = .019 + .001 = .020
c P(X < 4) = p(0) + p(1) + p(2) + p(3) = .005 + .025 + .310 + .340 = .680
7.9 p(0) = p(1) = p(2) = . . . = p(10) = 1/11 = .091
7.10 a P(X > 0) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8
b P(X ³ 1) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8
c P(X ³ 2) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8
d P(2 £ X £ 5) = p(2) = .3
7.11 P(X ³ 2) = p(2) + p(3) = .4 + .2 = .6
7.12 a P(X < 2) = p(0) + p(1) = .05 + .
.43 = .48
b P(X > 1) = p(2) + p(3) = .31 + .21 = .52
7.13
a P(HH) = .25
b P(HT) = .25
c P(TH) = .25
d P(TT) = .25
7.14 a P(0 heads) = P(TT) = 1/4
b P(1 head) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2
c P(2 heads) = P(HH) = 1/4
d P(at least 1 head) = P(1 head) + P(2 heads) = 1/2 + 1/4 = 3/4
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7.15
7.16 a P(2 heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8
b P(1 heads) = P(HTT) + P(THT) = P(TTH) = 1/8 + 1/8 + 1/8 = 3/8