# Thread: find sample mean and S.D

1. ## find sample mean and S.D

Suppose a random sample of n=36 observations is selected from a population that has normal distribution with mean 106 and standard deviation 12.

a) Give the mean and the standard deviation of the sample mean X bar.

b) find the probability that X bar exceeds 108.

c) Find the probability that the sample mean deviates from the population mean by less than 3.

I dun quite understand part a and part c, isn't the sample mean same a the population mean? How about the standard deviation, which formula shud I use?

For part b, I use P(X bar >= 108)= P(z>=108-106/(12/square root of 30)), is this correct?

2. ## Re: find sample mean and S.D

Part b) looks good but 36 is in the square root, not 30.

For part a) $\displaystyle \displaystyle \mu_{\bar{x}} = \mu , \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$

3. ## Re: find sample mean and S.D

Originally Posted by pickslides
Part b) looks good but 36 is in the square root, not 30.

For part a) $\displaystyle \displaystyle \mu_{\bar{x}} = \mu , \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
thanks a lot, so for part c, do I just follow part b to find: P(z<3)=p(<=2)?