Normal distribution central limit theorum

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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. 

Mar 14th, 2015

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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