What is the probability that a random sample size (__) will have a mean within 11 days of the mean

Question

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with the mean  μ=247 and standard deviation σ=18.

What is the probability that a random sample size of 17 will have a mean gestation period within 11 days of the mean?

Please also explain how I can find the answer

Answer 1

What we're given is the population mean and standard deviation and we have to draw conclusions about a sample from that population. We'll use the z value to find the probability but we have to base that z value on the size of the sample.

For a sample, the Z value can be found with the equation

Z = (X - µ)/s

We use σ if we're talking about the population, but we're interested in the sample so we have to use the sample standard deviation s. We estimate s using the equation

s = σ/√n

That is, the estimated sample standard deviation will depend on the population standard deviation and the sample size. For this problem

s = 18/√17 = 4.366

Now we need to find the Z scores to find the probability of the sample average being within 11 days of the population mean. Eleven days within the mean is 5.5 days above and 5.5 days below the mean. This means we have to find the Z scores when (X-µ) = 5.5 and (X-µ) = -5.5 (5.5 days above and below the population mean).

For the upper value

Z = (x-µ)/s = 5.5/4.366 = 1.26

For the lower value

Z = (x-µ)/s = -5.5/4.366 = -1.26

Using a calculator or z table we can find that the probability for Z = 1.26 is 0.8962.

For Z = -1.26, the probability is .1038

The probability that the sample is between these two is 0.8962 - .1038 = 0.7924, or 79.24%

So the probability that the sample mean is within 11 days (± 5.5 days) of the population mean is 79.24%

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