# Thread: Approximate probability of sample mean

1. ## Approximate probability of sample mean

A participant in a walk-a-thon has sponsor donations (in dollars) with a mean of 2000 and a standard deviation of 500. A random sample of size n=25 is taken from all of this participant's sponsor donations and the sample mean x is considered. Approximate P(X>2050). (Both x's should have a line above them to denote sample mean)

Thank you

2. $\displaystyle P(\bar X>2050) = P\biggl( {\bar X-2000\over 500/\sqrt{25}}> {2050-2000\over 500/\sqrt{25}}\biggr)$

then by the CLT $\displaystyle \approx P\biggl( Z> {2050-2000\over 500/\sqrt{25}}\biggr)=P( Z> .5)$.

3. Is this to be calulated by subtracting the normal distribution (0.9332) from 1? If so, I believe my answer should be 0.0668. Am I correct?

4. I don't think .9332 is correct. Look at the st normal tables again.

5. I'm sorry, i was looking at the wrong problem. I think the answer is just 0.3085.

6. that seems more reasonable

7. It is really that simple? The table basically gives you the answer, as long as you set up the equation correctly?