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Batches that consist of 50 coil springs from a production process are checked for conformance to customer requirements. The mean number of n

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Batches that consist of 50 coil springs from a production process are checked for conformance to customer requirements. The mean number of nonconforming coil springs in a batch is 5. Assume that the number of nonconforming springs in a batch, denoted as X, is a binomial random variable.

a. What are n and p?

b. What is P(X ≤ 2)?

c. What is P(X ≥ 49)?

Nov 21, 2014

#1
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a)  n = number in the batch, so n = 50

p = probability of non-conformance = 5/50  =  1/10

b)  P(X≤2)  =  P(X=0) + P(X=1) + P(X=2)

=  50nCr0·(1/10)^0·(9/10)·50 + 50nCr1·(1/10)^1·(9/10)·49 + 50nCr2·(1/10)^2·(9/10)·48

=  0.112

c)  P(X≥49)  =   P(X=49) + P(X=50)

=  50nCr49·(1/10)^49·(9/10)·1 + 50nCr50·(1/10)^50·(9/10)·0

=  \$.51 x 10^-48

Nov 22, 2014

#1
+5

a)  n = number in the batch, so n = 50

p = probability of non-conformance = 5/50  =  1/10

b)  P(X≤2)  =  P(X=0) + P(X=1) + P(X=2)

=  50nCr0·(1/10)^0·(9/10)·50 + 50nCr1·(1/10)^1·(9/10)·49 + 50nCr2·(1/10)^2·(9/10)·48

=  0.112

c)  P(X≥49)  =   P(X=49) + P(X=50)

=  50nCr49·(1/10)^49·(9/10)·1 + 50nCr50·(1/10)^50·(9/10)·0

=  \$.51 x 10^-48

geno3141 Nov 22, 2014