Question

A distributor receives a large shipment of components. Thedistributor would like to accept the shipment if 10% or fewer ofthe components are defective and to return it if more than 10% ofthe components are defective. She decides to sample 10 components, and to return the shipment if more than 1 of the 10 is defective.
a. If the proportion of defectives inthe batch is in fact 10%, what is the probability thatshe will return the shipment?

b. If the proportion ofdefectives in the batch is 20%, what is the probability that she will return the shipment?

Answer 1

Answer:

a. 0.2639 or 26.39%

b. 0.6242 or 62.42%

Step-by-step explanation:

This problem can be treated as a binomial model with probability of success p = proportion of defectives in the batch.

Number of trials (n) = 10.

For any number of successes 'k', the probability is:

$P(X=k) = \frac{n!}{(n-k)!k!}*p^k*(1-p)^{n-k}$

a. If p =0.10, what is P(x >1):

$P(X>1) = 1- P(X=0)-P(X=1)\\P(X>1) =1- \frac{10!}{(10-0)!0!}*0.10^0*(1-0.10)^{10-0}- \frac{10!}{(10-1)!1!}*0.10^1*(1-0.10)^{10-1}\\P(X>1) = 1-0.348678-0.287420\\P(X>1) =0.2639$

The probability is 0.2639 or 26.39%.

b. If p =0.20, what is P(x >1):

$P(X>1) = 1- P(X=0)-P(X=1)\\P(X>1) =1- \frac{10!}{(10-0)!0!}*0.20^0*(1-0.20)^{10-0}- \frac{10!}{(10-1)!1!}*0.20^1*(1-0.20)^{10-1}\\P(X>1) = 1-0.107374-0.268435\\P(X>1) =0.6242$

The probability is 0.6242 or 62.42%.