Question

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?

Answer 1

The question involves calculating probabilities and expected numbers using the binomial distribution. We assess the chance a shipment passes with a 10% defect rate, determine the expected number of defects in a sample of 5, and find the probability of 4 defects with a 20% defect rate.

Step-by-step explanation:

The question posed relates to the binomial probability distribution, which calculates the chance of a certain number of successes in a series of independent trials. In this case, 'success' refers to encountering a defective item, and the trials are the inspection of the items in the shipment.

a. Probability that the inspection procedure will pass the shipment:

To pass the inspection, a maximum of one defective item is allowed in the sample of 5, with a defect rate of 10%. Using the binomial distribution, the probability P of passing is P(X ≤ 1), where X is the number of defective items detected. This includes the probability of finding exactly 0 and exactly 1 defective item, which are calculated as P(X = 0) + P(X = 1).

b. Expected number of defectives:

The expected number of defectives in the sample of 5 is calculated using the formula for expected value E(X) = n * p, where n is the sample size and p is the probability of a defective item.

c. Probability with a 20% defective rate:

With the defect rate at 20%, we again use the binomial formula to find the probability of exactly 4 defectives P(X = 4) in a sample of 5.

Answer 2

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.
`P(X\leq 1)`

`P(X\leq 1)=P(0\leq X\leq 1)=P(X=0,1)`

So,
`P(X\leq 1)=P(X=0)+P(X=1)`

`P(X\leq 1)=^5C_0(0.1)^0(1-0.1)^5+^5C_1(0.1)^1(1-0.1)^4`

`P(X\leq 1)=1(1)(0.9)^5+5(0.1)(0.9)^4`

`P(X\leq 1)=0.59049+0.32805`

`P(X\leq 1)=0.918540`

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

`E(X)=n* p`

`E(X)=5* 0.1`

`E(X)=0.5`

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

`P(X=4)=^5C_4(0.2)^4(1-0.2)^1`

`P(X=4)=5(0.2)^4(0.8)^1`

`P(X=4)=0.0064`

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