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 annual precipitation amounts in a certain mountain range are normally distributed with a mean of 84 inches, and a standard deviation of 10 inches. What is the likelihood that the mean annual precipitation during
25 randomly picked years will be less than 86.8 inches?

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Quizplus · Accepted Answer

0.9192To find the likelihood that the mean annual precipitation during 25 randomly picked years will be less than 86.8 inches, we first calculate the z-score using the formula $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we get $z = \frac{86.8 - 84}{10/\sqrt{25}} = \frac{2.8}{2} = 1.4$. Looking up the z-score of 1.4 in the z-table gives us a value of approximately 0.9192, which means there is a 91.92% chance that the mean annual precipitation during 25 randomly picked years will be less than 86.8 inches.

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