Chapter 6, Problem 21CQ

Heights of People The average height of a certain age group of people is 53 inches. The standard deviation is 4 inches. If the variable is normally distributed, find the probability that a selected individual’s height will be

a. Greater than 59 inches

b. Less than 45 inches

c. Between 50 and 55 inches

d. Between 58 and 62 inches

To determine

To find: The probability that a selected individual’s height will be greater than 59 inches.

Answer to Problem 21CQ

The probability that a selected individual’s height will be greater than 59 inches is 0.0668.

Explanation of Solution

Given info:

The random variable $X$ represents the height of certain age group of people and if follows the normal distribution with mean of 53 inches and standard deviation of 4 inches.

Calculation:

Use the z- score formula to convert the $X$ in to z-value.

The formula for the z- score is

$z=\frac{X-\mu }{\sigma }$

Substitute the values in the above formula is as follows:

$\begin{array}{c}z=\frac{59-53}{4}\\ =1.5\end{array}$

Thus, the value of $z$ is, 1.5.

Use the Excel formula, to find the probabilities for specified z-value.

$\begin{array}{c}P\left(z>1.5\right)=1-P\left(z\le 1.5\right)\\ =1-\left(NORMSDIST\left(1.5\right)\right)\\ =1-0.9332\\ =0.0668\end{array}$

Thus, the $P\left(X>59\right)=\mathrm{0.0668.}$

Conclusion:

There is a 0.0668 of chance that a selected individual’s height will be greater than 59 inches.

To determine

To find:  The probability that a selected individual’s height will be less than 45 inches.

Answer to Problem 21CQ

The probability that a selected individual’s height will be less than 45 inches is 0.0228.

Explanation of Solution

Calculation:

Substitute the values in the z-score formula is as follows:

$\begin{array}{c}z=\frac{45-53}{4}\\ =-2\end{array}$

Thus, the value of $z$ is, –2.

Use the Excel formula, to find the probabilities for specified z-value.

$\begin{array}{c}P\left(z<-2\right)=\left(NORMSDIST\left(1.5\right)\right)\\ =0.0228\end{array}$

Thus, the $P\left(X<45\right)=\mathrm{0.0228.}$

Conclusion:

There is a 0.0228 of chance that a selected individual’s height will be less than 45 inches.

To determine

To find: The probability that a selected individual’s height will be between 50 and 55 inches.

Answer to Problem 21CQ

The probability that a selected individual’s height will be Between 50 and 55 inches is 0.4648.

Explanation of Solution

Calculation:

Substitute the values in the z-score formula is as follows:

$\begin{array}{c}{z}_1=\frac{50-53}{4}\\ =-0.75\\ z_2=\frac{55-53}{4}\\ =0.5\end{array}$

Thus, the value of $z_1{\text{ and z}}_2$ are –0.75 and 0.5.

Use the Excel formula, to find the probabilities for specified z-value.

$\begin{array}{c}P\left(-0.75<z<0.5\right)=P\left(z<0.5\right)-P\left(z\le -0.75\right)\\ =0.6915-0.2266\left(\begin{array}{l}\text{Use the Excel formula,}\\ \left(=NORMSDIST\left(z\right)\right)\end{array}\right)\\ =0.4648\end{array}$

Thus, the $P\left(50<X<55\right)=0.4648$.

Conclusion:

There is a 0.4648 of chance that a selected individual’s height will be Between 50 and 55 inches.

To determine

To find: The probability that a selected individual’s height will between 58 and 62 inches.

Answer to Problem 21CQ

The probability that a selected individual’s height will be between 58 and 62 inches is 0.0934.

Explanation of Solution

Calculation:

Substitute the values in the z-score formula is as follows:

$\begin{array}{c}{z}_1=\frac{58-53}{4}\\ =1.25\\ z_2=\frac{62-53}{4}\\ =2.25\end{array}$

Thus, the value of $z_1{\text{ and z}}_2$ are 1.25 and 2.25.

Use the Excel formula, to find the probabilities for specified z-value.

$\begin{array}{c}P\left(1.25<z<2.25\right)=P\left(z<2.25\right)-P\left(z\le 1.25\right)\\ =0.9878-0.8944\left(\begin{array}{l}\text{Use the Excel formula,}\\ \left(=NORMSDIST\left(z\right)\right)\end{array}\right)\\ =0.0934\end{array}$

Thus, the $P\left(58<X<62\right)=\mathrm{0.0934.}$

Conclusion:

There is a 0.0934 of chance that a selected individual’s height will be between 58 and 62 inches.