Chapter 6.8, Problem 12E

(a) .

To determine

To Calculate: The probability that the value of a random variable, having a normal distribution, is within a given interval.

Answer to Problem 12E

An adult male chosen at random for the survey is between $65$ and $73$ inches tall with a probability of $0.847$

Explanation of Solution

Given Information:

• The mean of the normally distributed random variable is $\text{μ}=69$ and the standard deviation is $\text{σ}=2.8$
• The range for which the probability of occurrence is to be calculated is $65\le X\le 73$

Concept Used:

• The probability distribution function of a normally distributed random variable with mean $\text{μ}$ and standard deviation $\text{σ}$ is given by
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$
• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
• $e^{-x^2}$ does not have an anti-derivative. So, use a calculator or computer to calculate the definite integrals of such functions.

Calculation: Put $\text{a}=65$ and $\text{b}=73$ in the above integral for probability to write

  $\text{P}\left(65\le X\le 73\right)=\underset{65}{\overset{73}{{\displaystyle \int }}}\frac{1}{2.8\sqrt{2\text{π}}}{e}^{-\frac{{\left(x-69\right)}^2}{2\times {2.8}^2}}$

Input the values of lower limit $\text{a}$ , upper limit $\text{b}$ , the mean $\text{μ}$ and the standard deviation $\text{σ}$ in a calculator to get

  $\text{P}\left(65\le X\le 73\right)=0.847$

Therefore, an adult male chosen at random for the survey is between $65$ and $73$ inches tall with a probability of $0.847$

(b).

To determine

To Calculate: The probability that the value of a random variable, having a normal distribution, is greater than a certain number.

Answer to Problem 12E

The probability that an adult male chosen at random for the survey is more than $6$ feet tall $0.142$

Explanation of Solution

Given Information:

• The mean of the normally distributed random variable is $\mu =69$ and the standard deviation is $\text{σ}=2.8$
• The probability to be calculated is for a height of more than $6$ feet, that is $72$ inches.

Concept Used:

• The probability distribution function of a normally distributed random variable with mean $\text{μ}$ and standard deviation $\text{σ}$ is given by
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-{\left(x-\mu \right)}^2}{2{\sigma }^2}}$
• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$
• $e^{-x^2}$ does not have an anti-derivative. So, use a calculator or computer to calculate the definite integrals of such functions.

Calculation: Put $\text{a}=72$ and $\text{b}=\infty$ in the above integral for probability to write

  $\text{P}\left(72\le X<\infty \right)=\underset{72}{\overset{\infty }{{\displaystyle \int }}}\frac{1}{2.8\sqrt{2\text{π}}}{e}^{-\frac{{\left(x-69\right)}^2}{2\times {2.8}^2}}$

Enter the values of lower limit $\text{a}$ , upper limit $\text{b}$ , the mean $\text{μ}$ and the standard deviation $\text{σ}$ in a calculator to get

  $\text{P}\left(72\le X\le \infty \right)=0.142$

Note: While using a calculator, avoid the improper integral by replacing $\text{b}=\infty$ with $\text{b}=100$ as people more than $100$ inches tall are very rare.

Instead of $100$ , even higher numbers can be entered but the observed difference will be miniscule.

Therefore, probability that an adult male chosen at random for the survey is more than $6$ feet tall $0.142$