Chapter 7.5, Problem 31E

i.

To determine

To calculate: To solve the proportion of the adult females in New York City are Less than $63.4$ inches tall.

Answer to Problem 31E

The proportion of the adult females in New York City are $0.5$ .

Explanation of Solution

Given information: The given values are as follows:

  $\mu =63.4,\sigma =3.2$ .

Calculation:

Since , the function is normally distributed, the density function is given by:

  $f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where $\mu =5$ denotes the mean of time it takes to answer a call and $\sigma =2$ its standard deviation.

For calculating the distribution for Less than $63.4$ inches tall.

Here, It can be solve as:

  $\begin{array}{c}f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}\\ {\displaystyle {\int }_0^{63.4}f\left(x\right)}dx={\displaystyle {\int }_0^{63.4}\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}}dx\\ {\displaystyle {\int }_0^{63.4}f\left(x\right)}dx={\displaystyle {\int }_0^{63.4}\frac{1}{3.2\sqrt{2\pi }}{e}^{-\frac{{\left(x-63.4\right)}^2}{2{\left(3.2\right)}^{}}}}dx\\ {\displaystyle {\int }_0^{63.4}f\left(x\right)}dx=0.5\end{array}$ .

ii.

To determine

To calculate: To solve the proportion of the adult females in New York City are Between $63and65$ inches tall.

Answer to Problem 31E

The proportion of the adult females in New York City are $0.24120$ .

Explanation of Solution

Given information: The given values are as follows:

  $\mu =63.4,\sigma =3.2$ .

Calculation:

Since , the function is normally distributed, the density function is given by:

  $f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where $\mu =5$ denotes the mean of time it takes to answer a call and $\sigma =2$ its standard deviation.

For calculating the distribution for Less than $63.4$ inches tall.

Here, it can solve as:

Between $63\text{and}65$ inches tall.

Here, we can solve as:

  $\begin{array}{c}f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}\\ {\displaystyle {\int }_{63}^{65}f\left(x\right)}dx={\displaystyle {\int }_{63}^{65}\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}}dx\\ {\displaystyle {\int }_{63}^{65}f\left(x\right)}dx={\displaystyle {\int }_{63}^{65}\frac{1}{3.2\sqrt{2\pi }}{e}^{-\frac{{\left(x-63.4\right)}^2}{2{\left(3.2\right)}^{}}}}dx\\ {\displaystyle {\int }_{63}^{65}f\left(x\right)}dx=0.24120\end{array}$ .

iii.

To determine

To calculate: To solve the proportion of the adult females in New York City are Taller than $6$ feet.

Answer to Problem 31E

The proportion of the adult females in New York City are $0.00359$ .

Explanation of Solution

Given information: The given values are as follows:

  $\mu =63.4,\sigma =3.2$ .

Calculation:

Since, the function is normally distributed, the density function is given by:

  $f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where $\mu =5$ denotes the mean of time it takes to answer a call and $\sigma =2$ its standard deviation.

For calculating the distribution for Less than $63.4$ inches tall.

Here, It can solve as:

For, Taller than $6$ feet.

  $\begin{array}{c}\text{6}\text{feet}=6\times 12\\ =\text{72}\text{inches}\end{array}$ .

Here, It can solve as:

  $\begin{array}{c}f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}\\ {\displaystyle {\int }_{72}^{\infty }f\left(x\right)}dx={\displaystyle {\int }_{72}^{\infty }\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}}dx\\ {\displaystyle {\int }_{72}^{\infty }f\left(x\right)}dx={\displaystyle {\int }_{72}^{\infty }\frac{1}{3.2\sqrt{2\pi }}{e}^{-\frac{{\left(x-63.4\right)}^2}{2{\left(3.2\right)}^{}}}}dx\\ {\displaystyle {\int }_{72}^{\infty }f\left(x\right)}dx=0.00359\end{array}$ .

iv.

To determine

To calculate: To solve the proportion of the adult females in New York City are Exactly $5$ feet tall.

Answer to Problem 31E

The proportion of the adult females in New York City are $0$ .

Explanation of Solution

Given information: The given values are as follows:

  $\mu =63.4,\sigma =3.2$ .

Calculation:

Since, the function is normally distributed, the density function is given by:

  $f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$

Where $\mu =5$ denotes the mean of time it takes to answer a call and $\sigma =2$ its standard deviation.

For calculating the distribution for Less than $63.4$ inches tall.

Here, It can solve as:

As Exactly $5$ feet tall.

  $\begin{array}{c}\text{5}\text{feet=5×12}\\ \text{=60}\text{inches}\end{array}$

Here, we can solve as:

  $\begin{array}{c}f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}\\ {\displaystyle {\int }_{60}^{60}f\left(x\right)}dx={\displaystyle {\int }_{60}^{60}\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}}dx\\ {\displaystyle {\int }_{60}^{60}f\left(x\right)}dx={\displaystyle {\int }_{60}^{60}\frac{1}{3.2\sqrt{2\pi }}{e}^{-\frac{{\left(x-63.4\right)}^2}{2{\left(3.2\right)}^{}}}}dx\\ {\displaystyle {\int }_{60}^{60}f\left(x\right)}dx=0\end{array}$