Posted by lauren on .
Professor Smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of 9mm. Each of Smith's students took a random sample of size n and calculated the sample mean. Smith found that about 68% of the students had sample means between 48.5 and 51.5 mm. What was n? ( assume that n is large enogh that the Central Limiot Theorem is applicable)

statistics 
drwls,
The standard deviation of nsample means is sigma/sqrt(n). In your case sigma = 9 for the singlesample distribution and the 1.5 for the groups of n. Therefore
1.5 = 9/sqrt(n)
sqrt n = 9/1.5 = 6 and n = 36