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Suppose marathon finish times are right skewed with mean 4.75 hours and standard deviation of 36...

Suppose marathon finish times are right skewed with mean 4.75 hours and standard deviation of 36 minutes. Based on this information, what distribution will the means of finish times for twenty-five (25) randomly selected people running in the 2012 Niagara Falls International Marathon most likely follow?

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Answer #1

Here we have

μ = 4.75, σ 36

Sample size: n=25

Since sample size is small so we cannot apply CLT. The sampling distribution of sample mean will be skewed to right but less than population distribution with mean

tli = μ = 4.75

and standard deviation is

sigma_{ar{x}}=rac{sigma}{sqrt{n}}=rac{36}{sqrt{25}}=7.2

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