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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?

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\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=(x-\mu)/(\sigma) \\ z=(44-42.1)/(6.1) \\ z=(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=(x-\mu)/(\sigma) \\ z=(40-42.1)/(6.1) \\ z=(-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=(39-42.1)/(6.1) \\ z=-0.51 \\ \\ z=(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%.

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