Let's normalize the given random variable `X : ` subtract its mean `mu ` and divide by its standard deviation `sigma=13.9 `. Then the divided random variable `X_1 = ( X - mu ) / sigma ` will become standard normal, so we'll be able to use Z-scores.

Then, consider what is the probability that `X ` will be within one population mean from `mu . ` It is the same as the probability that `X_1 ` will be within `1 ` from zero, which is twice the probability that `X_1 ` will be between zero and `1 . ` This probability is the Z-score of `1 , ` which is about `0.34 ` (using the standard normal table).

This way, the confidence level that sample size of `1 ` gives is only `2*0.34 = 0.68 = 68% , ` which is not enough. To get `90% = 0.9 ` confidence, we need sample size such that the corresponding standard deviation will have Z-score of `0.9 / 2 = 0.45 , ` which is about `1.645 .`

Recall that the standard deviation of a sample with `n ` elements is `sqrt ( n ) ` times less than the population standard deviation, so we need `n ` such that `13.9 / sqrt ( n ) lt= 1.645 , ` or `n gt= ( 13.9 / 1.645 )^2 approx 71.4 .`

This means that the sample size must be at least **72**.

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**Further Reading**