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)?

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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)?

Mathematics
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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.
i get an answer of 0.021
wait thats just the z score right?

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Yes. And you should have gotten z=-0.021, since x-bar-0.021.
so on the calculator i would just do P(Z<=-.021) correct?
or would it be 1 - that?
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.
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). :)
awesome I think i am following you

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