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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)?

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  1. DebbieG
    • one year ago
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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.

  2. anonymous
    • one year ago
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    i get an answer of 0.021

  3. anonymous
    • one year ago
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    wait thats just the z score right?

  4. DebbieG
    • one year ago
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    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
    • one year ago
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    so on the calculator i would just do P(Z<=-.021) correct?

  6. anonymous
    • one year ago
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    or would it be 1 - that?

  7. DebbieG
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    awesome I think i am following you

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