Chapter EP, Problem 12.5.2EP

a.

To determine

To find the percent of fittings that have the diameter between $7$ centimeters and $8$ centimeters

Answer to Problem 12.5.2EP

  $\text{47\%}$

Explanation of Solution

Given:

The diameters have a mean of $\mu =7.5$ and standard deviation $\sigma =8.0$ centimeters.

Calculation:

The probability that number of fittings that have the diameter between $7$ centimeters and $8$ centimeters is represented by by $P(7<X<8)$ . Compute the corresponding z-value for $X=7$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{7-7.5}{8}\\ =\frac{-0.5}{8}\\ =-0.0625\end{array}$

Compute the corresponding z-value for $X=8$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{8-7.5}{8}\\ =\frac{0.5}{8}\\ =0.0625\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(7<X<8)=P(-0.0625<z<0.0625)\\ =P(z<0.0625)-P(z<-0.0625)\\ =\text{0}\text{.475082}\end{array}$

b.

To determine

To find the percent of fittings that have the diameter between $7.5$ centimeters and $8$ centimeters

Answer to Problem 12.5.2EP

  $\text{3\%}$

Explanation of Solution

Given:

The diameters have a mean of $\mu =7.5$ and standard deviation $\sigma =8.0$ centimeters.

Calculation:

The probability that number of fittings that have the diameter between $7.5$ centimeters and $8$ centimeters is represented by by $P(7.5<X<8)$ . Compute the corresponding z-value for $X=7.5$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{7.5-7.5}{8}\\ =0\end{array}$

Compute the corresponding z-value for $X=8$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{8-7.5}{8}\\ =\frac{0.5}{8}\\ =0.0625\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(7.5<X<8)=P(0<z<0.0625)\\ =P(z>0)-P(z<0.0625)\\ =\text{0}\text{.5}-\text{0}\text{.475082}\\ \text{=0}\text{.03}\end{array}$

c.

To determine

To find the percent of fittings that have the diameter greater than $6.5$ .

Answer to Problem 12.5.2EP

  $\text{45\%}$

Explanation of Solution

Given:

The diameters have a mean of $\mu =7.5$ and standard deviation $\sigma =8.0$ centimeters.

Calculation:

The probability that number of fittings that have the diameter greater than $6.5$ centimeters is represented by by $P(X>6.5)$ . Compute the corresponding z-value for $X=6.5$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{6.5-7.5}{8}\\ =\frac{-1}{8}\\ =-0.125\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(X>6.5)=P(Z>-0.125)\\ =\text{0}\text{.450262}\end{array}$

d.

To determine

To find the number of fittings that have the diameter between $6.0$ centimeters and $8.5$ centimeters.

Answer to Problem 12.5.2EP

  $\text{30}$

Explanation of Solution

Given:

The diameters have a mean of $\mu =7.5$ and standard deviation $\sigma =8.0$ centimeters. The number of fittings are $100$ .

Calculation:

The probability that number of fittings that have the diameter between $6.0$ centimeters and $8.5$ centimeters is represented by by $P(6<X<8.5)$ . Compute the corresponding z-value for $X=6$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{6-7.5}{8}\\ =\frac{-1.5}{8}\\ =-0.1875\end{array}$

Compute the corresponding z-value for $X=8.5$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{8.5-7.5}{8}\\ =\frac{1}{8}\\ =0.125\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(6<X<8.5)=P(-0.1875<z<0.125)\\ =P(z<0.125)-P(z<-\mathrm{0.1875.}\\ =\text{0}\text{.450262}-\text{0}\text{.425634}\\ \text{=0}\text{.03}\end{array}$

Since there are $100$ fittings to the number of fittings will be $100\times 0.03=30$