Question

The diameter of a mason jar is 3 inches but can be as large as 3.03 inches and as small as 2.97 inches. Twenty-five samples of mason jars are taken and it is discovered that these components have a grand mean of 3.01 inches and a standard deviation of 0.02 inches. What is the probability of producing a bad product? (4pts)

Answer 1

18.15% probability of producing a bad product

Explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

`\mu = 3.01, \sigma = 0.02`

What is the probability of producing a bad product?

Less than 2.97 or more than 3.03.

Less than 2.97

pvalue of Z when X = 2.97. So

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

`Z = (2.97 - 3.01)/(0.02)`

`Z = -2`

`Z = -2` has a pvalue of 0.0228

More than 3.03

1 subtracted by the pvalue of Z when X = 3.03. So

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

`Z = (3.03 - 3.01)/(0.02)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

1 - 0.8413 = 0.1587

Then

0.0228 + 0.1587 = 0.1815

18.15% probability of producing a bad product