Question

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?%

Answer 1

(a) 63.3%

(b) 72.7%

(c) 28.7%

Step-by-step explanation

Given:

`\begin{gathered} Mean,\mu=42.2\text{ }inches \\ \\ Standard\text{ }deviation,\sigma=5.3\text{ }inches \end{gathered}`

(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:

`z=(x-\mu)/(\sigma)`

Here,

`\begin{gathered} x=44,\mu=42.2,and\text{ }\sigma=5.3 \\ \\ z=(44-42.2)/(5.3)=(1.8)/(5.3) \\ \\ z=0.34 \end{gathered}`

So,

`P(x<44)=P(z<0.34)`

Interpreting the result in a normal curve, we have:

`\begin{gathered} 0.5+P(z=0.34) \\ \\ 0.5+0.1331=0.6331 \end{gathered}`

Therefore, the percentage to the nearest tenth will be:

`0.6331*100\%=63.3\%`

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

`\begin{gathered} x=39,\mu=42.2,and\text{ }\sigma=5.3 \\ \\ z=(39-42.2)/(5.3)=(-3.2)/(5.3) \\ \\ z=-0.603 \end{gathered}`

So,

`P(x>5.3)=P(z>-0.603)`

Interpreting the result in a normal curve, we have:

`\begin{gathered} =0.5-P(z=-0.603) \\ \\ =0.5-(-0.2268) \\ \\ =0.5+0.2268 \\ \\ =0.7268 \end{gathered}`

Therefore, the percentage to the nearest tenth will be:

`\begin{gathered} 0.7268*100\% \\ \\ =72.7\% \end{gathered}`

(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:

`\begin{gathered} z=(x-\mu)/(\sigma) \\ \end{gathered}`

Here,

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

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

`\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}`

Hence, the percentage to the nearest tenth will be:

`\begin{gathered} 0.2871*100\% \\ \\ =28.7\% \end{gathered}`

In summary,

(a) 63.3%

(b) 72.7%

(c) 28.7%