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# PPS7_Solution_.doc

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York University

Mathematics and Statistics

MATH 2565

Jagmohan Chawla

Winter

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Chapter 17 Probability Models 339
PPS7 (Solution)
Problem 17.27: Tennis, anyone? The first serves can be considered Bernoulli
trials. There are only two possible outcomes, successful and unsuccessful. The
probability of any first serve being good is given as p = 0.70. Finally, we are
assuming that each serve is independent of the others. Since she is serving 6
times, use Binom(6,0.70).
Let X the number of successful serves in n = 6 first serves.
a)
P(all six servesin ) P(X 6)
(0.70) (0.30) 0
6
0.118
b)
P(exactlyfour servesin) P(X 4)
(0.70) (0.30) 2
4
0.324
c)
P(at least four servesin) P(X 4) P(X 5) P(X 6)
4 2 5 1 6 0
40.70) (0.30) 5 .70) (0.30) 6 .70) (0.30)
0.744
d)
P(no more thanfourservesin) P(X 0) P(X 1) P(X 2) P(X 3) P(X 4)
(0.70) (0.30) 6 (0.70) (0.30) 5 (0.70) (0.30) 4
0 1 2
(0.70) (0.30) 3 (0.70) (0.30) 2
3 4
0.580
Problem 17.29: And more tennis. The first serves can be considered Bernoulli
trials. There are only two possible outcomes, successful and unsuccessful. The
probability of any first serve being good is given as p = 0.70. Finally, we are
assuming that each serve is independent of the others. Since she is serving 80
times, use Binom(80,0.70).
Let X the number of successful serves in n = 80 first serves.
a) E(X) np 80(0.70) 56 first serves in. 340 Part 4 Randomness and Probability
SD(X) npq 80(0.70)(0.30) 4.10 first serves in.
b) Since np = 56 and nq = 24 are both greater than 10,
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.
d) Using Binom(80, 0.70):
P(at least 65 firstserves) P(X 65)
P(X 65) P(X 66)
(0.70) (0.30)
65
0.0161
According to the Binomial model, the probability that she makes at least 65
first serves out of 80 is approximately 0.0161.
Using N(56, 4.10):
x
z
6556
z
4.10
z 2.195
P(X 65) P(z 2.195) 0.0141
Problem17.31: Apples.
a) A binomial model and a normal model are both appropriate for modeling the
number of cider apples that may come from the tree.
Let X = the number of cider apples found in the n = 300 apples from the tree.
The quality of the apples may be considered Bernoulli trials. There are only
two possible outcomes, cider apple or not a cider apple. The probability that
an apple must be used for a cider apple is constant, given as p = 0.06. The
trials are not independent, since the population of apples is finite, but the
apples on the tree are undoubtedly less than 10% of all the apples that the
farmer has ever produced, so model with Binom(300, 0.06). E(X) np 300(0.06) 18 cider apples.
SD(X) npq 300(0.06)(0.94) 4.11 cider apples.
Since np = 18 and nq = 282 are both greater than 10,
approximated by the Normal model, N(18, 4.11).
b) Using Binom(300, 0.06):
P(at most 12 cider apples) P(X 12)
P(X 0)
30 0
00.06) (0.94)
0.085
According to the Binomial model, the probability that no more than 12 cider
apples come from the tree is approximately 0.085.
Using N(18, 4.11):
x

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