+82 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)?
+23246 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
+23246 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
