# Normal Distribution

• Apr 6th 2007, 09:28 PM
harry
Normal Distribution
A random sample of the duration of 50 telphone calls handled by a local telephone company had a mean of 11.6 min and a standard devation of 3.8 Minutes. Find a 95% confidence interval for the true mean duration of calls handled by the company.
• Apr 6th 2007, 10:01 PM
CaptainBlack
Quote:

Originally Posted by harry
A random sample of the duration of 50 telphone calls handled by a local telephone company had a mean of 11.6 min and a standard devation of 3.8 Minutes. Find a 95% confidence interval for the true mean duration of calls handled by the company.

The random variable:

t = (SampleMean - PopulationMean)/(SampleSD/sqrt(n))

where n is the sample size, has a t distribution

With n=50, nu the number of degrees of freedom is 49, with this number of
degrees of freedom a 95% interval for t is ~=(-2.01, 2.01). So the 95%
confidence interval for the populaion mean is:

(SampleMean - 2.01*SampleSD/sqrt(n), SampleMean + 2.01*SampleSD/sqrt(n))

or: (10.52, 12.68).

If you used the normal approximation here the 2.01 would be replaced by 1.95
and the interval would be (10.55, 12.65)

RonL