The annual rainfall in a certain region is approximately normally distributed with mean 42.2 inches andstandard deviation 5.3 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 39 inches?c) What percentage of years will have an annual rainfall of between 38 inches and 43 inches?%

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Answer

(a) 63.3%

(b) 72.7%

(c) 28.7%

Explanation

Given:

[tex]\begin{gathered} Mean,\mu=42.2\text{ }inches \\ \\ Standard\text{ }deviation,\sigma=5.3\text{ }inches \end{gathered}[/tex]

(a) What percentage of years will have an annual rainfall of less than 44 inches?

First, we standardize 44 inches by changing x to z:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Here,

[tex]\begin{gathered} x=44,\mu=42.2,and\text{ }\sigma=5.3 \\ \\ z=\frac{44-42.2}{5.3}=\frac{1.8}{5.3} \\ \\ z=0.34 \end{gathered}[/tex]

So,

[tex]P(x<44)=P(z<0.34)[/tex]

Interpreting the result in a normal curve, we have:

[tex]\begin{gathered} 0.5+P(z=0.34) \\ \\ 0.5+0.1331=0.6331 \end{gathered}[/tex]

Therefore, the percentage to the nearest tenth will be:

[tex]0.6331\times100\%=63.3\%[/tex]

(b) What percentage of years will have an annual rainfall of more than 39 inches?

[tex]\begin{gathered} x=39,\mu=42.2,and\text{ }\sigma=5.3 \\ \\ z=\frac{39-42.2}{5.3}=\frac{-3.2}{5.3} \\ \\ z=-0.603 \end{gathered}[/tex]

So,

[tex]P(x>5.3)=P(z>-0.603)[/tex]

Interpreting the result in a normal curve, we have:

[tex]\begin{gathered} =0.5-P(z=-0.603) \\ \\ =0.5-(-0.2268) \\ \\ =0.5+0.2268 \\ \\ =0.7268 \end{gathered}[/tex]

Therefore, the percentage to the nearest tenth will be:

[tex]\begin{gathered} 0.7268\times100\% \\ \\ =72.7\% \end{gathered}[/tex]

(c) What percentage of years will have an annual rainfall of between 38 inches and 43 inches?

We need to find P(39 ≤ x ≤ 43).

Standardizing x to z by applying:

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ \end{gathered}[/tex]

Here,

[tex]\begin{gathered} P(\frac{39-42.2}{5.3}\leq z\leq\frac{43-42.2}{5.3}) \\ \\ P(\frac{-3.2}{5.3}\leq z\leq\frac{0.8}{5.3}) \\ \\ P(-0.604\leq z\leq0.151) \end{gathered}[/tex]

Also, interpreting the result in a normal curve, we have:

[tex]\begin{gathered} P(z=0.151)-P(z=-0.604) \\ \\ P(z=0.151)+P(z=0.604) \\ \\ =0.060+0.2271 \\ \\ =0.2871 \end{gathered}[/tex]

Hence, the percentage to the nearest tenth will be:

[tex]\begin{gathered} 0.2871\times100\% \\ \\ =28.7\% \end{gathered}[/tex]

In summary,

(a) 63.3%

(b) 72.7%

(c) 28.7%