Question

Applying the 68-95-99.7 rule. The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $2.10. You are taking a random sample of 28 lunches. The 68-95-99.7 rule says that the probability is about 0.95 that x with bar on top is within $_________ of the population mean μ. Fill in the blank (report to two decimal places)

Answer 1

Let X be the amount that you spend on any given lunch. Let the mean of X be dollars (the population mean of your spending on a lunch). The standard deviation of X is dollars. We do not know the distribution of X (normal or otherwise).

Let indicate the sample average of amount spend on lunch for a sample of size n=28 lunches

Using the central limit theorem, we know that has an approximate normal distribution with mean and standard deviation (or called standard error of mean) of

Now we use the 68-95-99.7 rule, which says that approximately 95% of all the values of a normally distributed population lie with in 2 standard deviations of mean.

That means for a given sample of 28 lunches, 95% of the times, the sample average would lie with in

dollars of mean of which is

ans: The 68-95-99.7 rule says that the probability is about 0.95 that is within $0.79 of the population mean.