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Mila [183]
3 years ago
11

The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 90 inches comma and a standard

deviation of 12 inches. What is the probability that the mean annual snowfall during 36 randomly picked years will exceed 92.8 inches question mark Round your answer to four decimal places.
Mathematics
1 answer:
Debora [2.8K]3 years ago
3 0

Step-by-step answer:

standard deviation (sigma) = 12 inches

mean (mu) = 90 inches.

For 36 years, the standard error of the mean is

e = sigma / sqrt(n) = 12" / sqrt(36) = 2".

Deviation from the mean

= 92.8 -90 = 2.8

Since the standard error of the mean follows the normal distribution, we have

Z=(92.8-90)/2 = 1.4

Looking up the normal distribution tables, or check R, we find

P(Z>1.4) = 1-P(Z<1.4) = 1-0.9192433 = 0.08075666

Answer: P(snowfall>92.8) = 0.0808 (to four decimal places)

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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

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Step-by-step explanation:

<u>Step 1: Define Equation</u>

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<u>Step 2: Solve for </u><em><u>x</u></em>

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<u>Step 3: Check</u>

<em>Plug in x into the original equation to verify it's a solution.</em>

  1. Substitute in <em>x</em>:                  7(2(4) - 5) = 21
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Answer:

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Not sure if you're looking for the fraction or something else but to find the fraction, divide 15 by 36.

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This converts to a percentage of 41.6 percent. (multiply by 100 to get percent when dealing with decimals)

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