Problems Probability Models Worksheets Page 8
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12
0
0
12
10
2
b)
1
0 87
0 13
0 812
c)
0 87
0 13
0 87
0 13
0 475
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12
0
10
12
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12
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d)
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e)
0 87
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0 87
0 13
0 998
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6
7
12
18. a) E(Y) = np = 10(0 .80) = 8 bull’s-eyes hit.. SD(Y) =
10 0 8 0 2
1 26
bull’s-eyes hit.
npq
.
.
.
10
10
10
10
0
0
10
8
2
b)
0 8
0 2
0 107
c)
0 8
0 2
0 8
0 2
0 624
.
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...
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10
0
8
10
10
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2
6
4
10
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d)
0 8
0 2
0 302
e)
0 8
0 2
0 8
0 2
0 967
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.
...
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8
6
10
6
6
6
0
4
2
19. a)
0 7
0 3
0 118
b)
0 7
0 3
0 324
.
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6
4
6
6
6
6
4
2
6
0
0
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4
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c)
0 7
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0 744
d)
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0 580
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4
6
0
4
12
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12
20. a)
0 125
0 875
0 201
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0
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12
2
10
12
0
b)
0 125
0 875
0 125
0 875
0 453
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...
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2
12
12
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3
9
4
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c)
0 125
0 875
0 125
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0 171
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3
4
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d)
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4
21. a) E(X) = np = 80(0 .70) = 56 first serves in.. SD(X) =
80 0 7 0 3
4 10
first serves in.
npq
.
.
.
b) Since np = 56 and nq = 24 are both greater than 10, Binom(80,0 .70) may be approximated by the Normal model, N(56, 4.10).
c) According to the Normal model, in matches with 80 serves, she is expected to make between
51.9 and 60.1 first serves approximately 68% of the time, between 47.8 and 64.2 first serves
approximately 95% of the time, and between 43.7 and 68.3 first serves approximately 99.7% of
the time.
80
80
65
15
80
0
d) Using Binom(80, 0.70):
0 7
0 3
0 7
0 3
0 0161
. According to
.
.
...
.
.
.
65
80
the Binomial model, the probability that she makes at least 65 first serves out of 80 is
approximately 0.0161.
65 56
Using N(56, 4.10): P(X ≥ 65) ≈
= P(z > 2.195) ≈ 0.0 141. According to the
P z
4 1
.
Normal model, the probability that she makes at least 65 first serves out of 80 is
approximately 0.0141.
22. a) E(Y) = np = 200(0.80) = 160 bull’s-eyes. SD(Y) =
200 0 8 0 2
5 66
bull’s-eyes.
npq
.
.
.
b) Since np = 160 and nq = 40 are both greater than 10, Binom(200,0.80) may be approximated by the Normal model, N(160,
5.66).
c) According to the Normal model, in matches with 200 arrows, she is expected to get
between 154.34 and 165.66 bull’s-eyes approximately 68% of the time, between
148.68 and 171.32 bull’s-eyes approximately 95% of the time, and between 143.02
and 176.98 bull’s-eyes approximately 99.7% of the time.
d) Using Binom(200, 0.80):
200
200
0
200
140
60
0 8
0 2
0 8
0 2
0 0005
. According to the
.
.
...
.
.
.
0
140
Binomial model, the probability that she makes at most 140 bull’s-eyes out of 200 is
approximately 0.0005.
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