women's heights are normally distributed with a mean of 64.5 inches and a standard deviation of 2.3 inches. If a random sample of 12 women is selected, what will be the probability that the average height for the sample will be between 65.5 and 66.5 Inches? Give your answer as a decimal rounded to four places.

We're looking for the sample mean distribution here, so with a small sample size like 12 we should use the t-distribution with n-1 = 11 degrees of freedom.

We'll need to express 65.5 and 66.5 as t-values, which we'll get by finding their distance from the population mean in standard error units. And the standard error is given by standard deviation / sqrt(n) = 2.3/sqrt(12) = 0.66395281.

So our t-values are:

(65.5-64.5)/SEM and (66.5-64.5)/SEM

= 1/0.664 and 2/0.664

= 1.506131137 and 3.012262274.

The related p-value for these t-values will tell us their percentile - the probability of seeing a smaller value. So to find the probability between them, we'll subtract the smaller value from the larger: