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 = = 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
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