## anonymous one year ago According to records, the amount of precipitation in a certain city on a November day has a mean of .1 inches, with a standard deviation of .6 inches. What is the probability that the mean daily precipitation will be .098 inches or more for a random sample of 40 November days (taken over many years)?

1. DebbieG

This is asking about a probability related to a SAMPLE MEAN, not just a data value. Thuse, you need to use the Central Limit Theorem formula to get the z-score: $\large z=\frac{ \bar{x} -\mu}{ \frac{ \sigma }{ \sqrt{n}} }$ You have $\bar{x}=0.098$$\mu=0.1$$\sigma=0.6$and n = 40. Compute the z-score, then use a normal table (or a calculator) to find the area to the right of the z-score.

2. anonymous

i get an answer of 0.021

3. anonymous

wait thats just the z score right?

4. DebbieG

Yes. And you should have gotten z=-0.021, since x-bar<mu. So now you need the standard normal probability for z>-0.021.

5. anonymous

so on the calculator i would just do P(Z<=-.021) correct?

6. anonymous

or would it be 1 - that?

7. DebbieG

What is the probability that the mean daily precipitation will be .098 inches OR MORE... so you want the P(Z>-0.021) ---- correct, you can get the area to the left, and subtract from 1. Or get the area to the right.

8. DebbieG

Or, if you understand how to take advantage of the symmetry of the distribution, you can even use P(Z<0.021) which is = P(Z>-0.021). :)

9. anonymous

awesome I think i am following you