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)?

Question

debbieg · Answer

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.

anonymous · Answer

i get an answer of 0.021

anonymous · Answer

wait thats just the z score right?

debbieg · Answer

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.

anonymous · Answer

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

anonymous · Answer

or would it be 1 - that?

debbieg · Answer

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.

debbieg · Answer

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).  :)

anonymous · Answer

awesome I think i am following you