Question

QuestionThe annual rainfall in a certain region is approximately normally distributed with mean 42.1 inches andstandard deviation 6.1 inches. Round answers to the nearest tenth of a percent.a) What percentage of years will have an annual rainfall of less than 44 inches?%b) What percentage of years will have an annual rainfall of more than 40 inches?c) What percentage of years will have an annual rainfall of between 39 inches and 43 inches?

Answer 1

$\begin{gathered} \text{Given} \\ \mu=42.1 \\ \sigma=6.1 \end{gathered}$

Part A: What percentage of years will have an annual rainfall of less than 44 inches?

First, solve for the z-score when x = 44 inches

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{44-42.1}{6.1} \\ z=\frac{1.9}{6.1} \\ z=0.31 \end{gathered}$

Next, find P(z < 0.31), by locating the probability to the left of the area of the z-table.

Multiply the probability by 100%

$0.62172\cdot100\%=62.172\%$

Rounding to the nearest tenth of a percent, the percentage is 62.2%.

Part B: What percentage of years will have an annual rainfall of more than 40 inches?

Solve for the z-score for x = 40 inches

$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{40-42.1}{6.1} \\ z=\frac{-2.1}{6.1} \\ z=-0.34 \end{gathered}$

Next, find the area to the left of z-score, and subtract it from 1.

$\begin{gathered} P(z>-0.34)=1-0.36693 \\ P(z>-0.34)=0.63307 \end{gathered}$

Multiply by 100%

$0.63307\cdot100\%=63.307\%$

Rounding to the nearest tenth of a percent, the percentage is 63.3%.

Part C: What percentage of years will have an annual rainfall of between 39 inches and 43 inches?

Find the z-score for both x = 39, and x = 43

$\begin{gathered} z=\frac{39-42.1}{6.1} \\ z=-0.51 \\ \\ z=\frac{43-42.1}{6.1} \\ z=0.9 \end{gathered}$

Find the area to the left of z-score.

Subtract 0.81594 by 0.30503, and multiply the result by 100%

[tex]\begin{gathered} P(-0.51Rounding to the nearest tenth of a percent, the percentage is 51.1%.