Solve the problem. Round the standard score to the nearest tenth before using the z-score tables. Express your answer as
a decimal rounded to four decimal places.
-The amount of snowfall falling in a certain mountain range is normally distributed with a mean of 72 inches, and a standard deviation of 12 inches. What is the likelihood that the mean annual snowfall during 36 randomly
Picked years will exceed 74.8 inches?

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Quizplus · Accepted Answer

To find the likelihood that the mean annual snowfall during 36 randomly picked years will exceed 74.8 inches, we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the values, we get $z = \frac{74.8 - 72}{\frac{12}{\sqrt{36}}} = \frac{2.8}{2} = 1.4$. Looking up the z-score of 1.4 in the z-table gives us the area to the left of the z-score, which is 0.9192. To find the probability of exceeding 74.8 inches, we subtract this value from 1, resulting in $1 - 0.9192 = 0.0808$.

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