Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles. What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Explanation:

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.

Central Limit Theorem

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

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that
`\mu = 33208, \sigma = 2503`.

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when
`X = 33208+633 = 33841` subtracted by the pvalue of Z when
`X = 33208 - 633 = 32575`

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

`s = (\sigma)/(√(n)) = (2503)/(√(49)) = 357.57`

So

X = 33841

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

`Z = (33841 - 33208)/(357.57)`

`Z = 1.77`

`Z = 1.77` has a pvalue of 0.9616

X = 32575

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

`Z = (32575- 33208)/(357.57)`

`Z = -1.77`

`Z = -1.77` has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.