Question

the operation manager at a tire manufacturing company believes that the mean mileage of a tire is 20,138 miles, with a variance of 10,304,100. what is the probability that the sample mean would be less than 19,448 miles in a sample of 171 tires if the manager is correct? round your answer to four decimal places.

Answer 1

The probability that the sample mean would be less than 19,448 miles is 0.9972

Given,

The mean mileage of a tire, μ = 20,138

The standard deviation, σ = $\sqrt{10,304,100}$ = 3210

Sample size, n = 171

We have to find the probability that the sample mean would be less than 19,448 miles.

That is,

P(X < 19448)

Since the sample size in this instance is greater than 30, we can utilise the central limit theorem and the z score formula provided by:

z = (X - μ) / (σ/√n)

The following would be the sample mean's distribution:

X ≈ N (μ, σ/√n)

In this situation, the z score was found to be:

z = (19448 - 20138) / (3210/√171)

z = -690 / 245.48

z = 2.81

Using the conventional table, we get the following:

P(z < 2.81) = 0.9972

That is,

The probability that the sample mean would be less than 19,448 miles is 0.9972

Learn more about probability here;

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