Question

The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation

5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.

(a) what is the probability that a part is defective?

(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in

such a sample? Assume that each part is independent of the others.

Answer 1

a) There is a 15.87% probability that a part is defective.

b) The expected number of parts defective in such a sample is 0.7935.

Explanation:

To solve this question, we use concepts of the normal probability distribution and the binomial probability distribution.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

`E(X) = np`

(a) what is the probability that a part is defective?

The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation

5 pounds. A part is considered defective if the tensile strength is less than 35 pounds.

Here we have
`\mu = 40, \sigma = 5`

This probability is the pvalue of Z when X = 35.

So

`Z = (X - \mu)/(\sigma)`

`Z = (35 - 40)/(5)`

`Z = -1`

`Z = -1` has a pvalue of 0.1587.

There is a 15.87% probability that a part is defective.

(b) If a testing sample consists of 5 parts, what is the expected number of parts defective in such a sample? Assume that each part is independent of the others.

This is the expected value of a binomial distribution when
`n = 5, p = 0.1587`.

So

`E(X) = np = 5*0.1587 = 0.7935`

The expected number of parts defective in such a sample is 0.7935.