Question

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 139 millimeters, and a standard deviation of 6 millimeters. If a random sample of 40 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 0.8 millimeters? Round your answer to four decimal places.

Answer 1

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters

Explanation:

To solve this question, we need to understan the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 139, \sigma = 6, n = 40, s = (6)/(√(40)) = 0.9487`

Either the sample mean differs by 0.8 mm or less from the population mean, or it differs by more. The sum of these probabilities is decimal 1.

Probability it differs by less than 0.8mm

pvalue of Z when X = 139 + 0.8 = 139.8mm subtracted by the pvalue of Z when X = 139 - 0.8 = 138.2 mm

X = 139.8

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

By the Central Limit Theorem

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

`Z = (139.8 - 139)/(0.9487)`

`Z = 0.84`

`Z = 0.84` has a pvalue of 0.7995

X = 138.2

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

`Z = (138.2 - 139)/(0.9487)`

`Z = -0.84`

`Z = -0.84` has a pvalue of 0.2005

0.7995 - 0.2005 = 0.5990

What is the probability that the sample mean would differ from the population mean by more than 0.8 millimeters?

p + 0.5990 = 0.4010

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters