Chapter PI, Problem 8RE

(a)

To determine

To draw and carefully label the model.

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, the model for it will be as:

(b)

To determine

To find out what percent of storms produce rainfall with pH over $\text{6}$ .

Answer to Problem 8RE

  $3.36\%$ .

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, let us first find the z -score as:

  $\begin{array}{l}z=\frac{6-4.9}{0.6}\\ =1.83\end{array}$

Thus, the percent of storms produce rainfall with pH over $\text{6}$ is as:

  $\begin{array}{l}P(z\ge 1.83)=normalcdf(1.83,E99,0,1)\\ =0.0336\\ =3.36\%\end{array}$

(c)

To determine

To find out what percent of storms produce rainfall with pH under $\text{4}$ .

Answer to Problem 8RE

  $6.68\%$ .

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, let us first find the z -score as:

  $\begin{array}{l}z=\frac{4-4.9}{0.6}\\ =-1.5\end{array}$

Thus, the percent of storms produce rainfall with pH under $\text{4}$ is calculated as:

  $\begin{array}{l}P(z<-1.5)=normalcdf(-E99,-1.5,0,1)\\ =0.0668\\ =6.68\%\end{array}$

(d)

To determine

To find out what is the pH level for the most acidic $\text{20\%}$ of all storms.

Answer to Problem 8RE

The $4.4$ pH level is for the most acidic $\text{20\%}$ of all storms.

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, we have,

  $invNorm(0.2,0,1)=-0.84$

Thus, the value will be calculated as:

  $\begin{array}{l}-0.84=\frac{y-4.9}{0.6}\\ \Rightarrow -0.84\times 0.6=y-4.9\\ \Rightarrow -0.504=y-4.9\\ \Rightarrow y=4.4\end{array}$

Thus, the $4.4$ pH level is for the most acidic $\text{20\%}$ of all storms.

(e)

To determine

To find out what is the pH level for the least acidic $\text{5\%}$ of all storms.

Answer to Problem 8RE

The $5.88$ pH level is for the least acidic $\text{5\%}$ of all storms.

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, we have,

  $invNorm(0.95,0,1)=1.64$

Thus, the value will be calculated as:

  $\begin{array}{l}1.64=\frac{y-4.9}{0.6}\\ \Rightarrow 1.64\times 0.6=y-4.9\\ \Rightarrow 0.984=y-4.9\\ \Rightarrow y=5.88\end{array}$

Thus, the $5.88$ pH level is for the least acidic $\text{5\%}$ of all storms.

(f)

To determine

To find out what is the IQR for the pH of rainfall.

Answer to Problem 8RE

  $0.804$ .

Explanation of Solution

In the question, it is given based on long-term investigation, researchers have suggested that the acidity of rainfall in the Shenandoah mountains can be described by the Normal model $\text{N(4}\text{.9,0}\text{.6)}$ . Thus, we have,

  $\begin{array}{l}invNorm(0.25,0,1)=-0.67\\ invNorm(0.75,0,1)=0.67\end{array}$

Therefore the IQR will be as:

  $\begin{array}{l}-0.67=\frac{y-4.9}{0.6}\\ \Rightarrow -0.67\times 0.6=y-4.9\\ \Rightarrow -0.402=y-4.9\\ \Rightarrow y=4.498\end{array}$

And also,

  $\begin{array}{l}0.67=\frac{y-4.9}{0.6}\\ \Rightarrow 0.67\times 0.6=y-4.9\\ \Rightarrow 0.402=y-4.9\\ \Rightarrow y=5.302\end{array}$

Thus, we have,

  $\begin{array}{l}IQR=5.302-4.498\\ =0.804\end{array}$