3.25 a company is interested in evaluating its current inspection procedure on large shipments of identical items. te procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. it is known that items are defective at a 10% rate overall. a. what is the probability that the inspection procedure will pass the shipment? b. what is the expected number of defectives in this process of inspecting 5 items? c. if items are defective at a 20% rate overall, what is the probability that you will fnd 4 defectives in a sample of 5?

a) The probability that the inspection procedure will pass the shipment is 0.918540.

b) The expected number of defectives in this process of inspecting 5 items is 0.5.

c) The probability that you will find 4 defectives in a sample of 5 is 0.0064.

Step-by-step explanation:

Given : A company is interested in evaluating its current inspection procedure on large shipments of identical items. Te procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.

To find :

a. What is the probability that the inspection procedure will pass the shipment?

b. What is the expected number of defectives in this process of inspecting 5 items?

c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?

Solution :

Applying binomial distribution,

The probability of success p=10%=0.1

Number of items n=5

a) The probability that the inspection procedure will pass the shipment i.e.

So,

The probability that the inspection procedure will pass the shipment is 0.918540.

b) The expected number of defectives in this process of inspecting 5 items i.e. E(X)

The mean E(X) is defined as

The expected number of defectives in this process of inspecting 5 items is 0.5.

c) The probability that you will find 4 defectives in a sample of 5 i.e P(X=4)

Here, P=20%=0.2

The probability that you will find 4 defectives in a sample of 5 is 0.0064.

two data sets with the same median will not have the same varaibility

step-by-step explanation: