Homework Help: statistics

Posted by lauren on Saturday, February 16, 2008 at 5:16pm.

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, Saturday, February 16, 2008 at 6:21pm
The standard deviation of n-sample means is sigma/sqrt(n). In your case sigma = 9 for the single-sample 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

Answer this Question

Related Questions

Statistics - Suppose a researcher wants to study the effectiveness of a new ...
Statistics - 1. In thinking about doing statistical analysis, the sample mean ...
statistics - If random samples, each with n = 36 scores, are selected from a ...
Statistics - In Professor Smith's statistics course, the correlation between...
Statistics - A computer repair service found that a random sample of 45 repair ...
Statistics - 1. The scores of students on the ACT college entrance examination ...
statistics - The scores of students on the ACT college entrance examination in ...
Statistics - A study was conducted in order to estimate u, the mean number of ...
statistics - T test for independent groups and dependent groups (Two-group ...
statistics - At a local high school students were randomly assigned to one of ...

For Further Reading

First Name:
School Subject:
Answer: