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:
P(65.5 < x < 66.5)
= P(1.506 < t(11) < 3.012)
= P(t(11) < 3.012) - P(t(11) < 1.506)
= t.dist(3.012,df=11) - t.dist(1.506,df=11)
= 0.994091003 - 0.919900545
= 0.0742 (to 4dp)
Mar 14th, 2015
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