Chapter 11.5, Problem 24HP

To determine

To find: Whether the statement is true or false and explain.

Answer to Problem 24HP

The given statement is true.

Explanation of Solution

Given:

Consider the given statement.

Calculation:

Consider the random variable $X$ which is normal distributed with mean $\mu$ and standard deviation $\sigma$ . Therefore, the probability density function of $X$ is given by, $f\left(x\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}$ for $-\infty <x<\infty$ .

Consider the probability of $P\left(\mu -\sigma \le X\le \mu +\sigma \right)=\frac{1}{\sqrt{2\pi }\sigma }{\displaystyle {\int }_{\mu -\sigma }^{\mu +\sigma }{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}}dx}$ .

Consider $z=\frac{x-\mu }{\sigma }$ that is $\sigma dz=dx$ hence,

  $\begin{array}{c}P\left(\mu -\sigma \le X\le \mu +\sigma \right)=\frac{1}{\sqrt{2\pi }}{\displaystyle {\int }_{-1}^1{e}^{-\frac{z^2}{2}}dz}\\ =\frac{2}{\sqrt{2\pi }}{\displaystyle {\int }_0^1{e}^{-\frac{z^2}{2}}dz}\end{array}$

Use the calculator to solve the above integral.

  $\begin{array}{c}P\left(\mu -\sigma \le X\le \mu +\sigma \right)=\frac{2}{\sqrt{2\pi }}\left(0.8556\right)\\ =0.68\end{array}$

Hence the probability of getting the data in one standard deviation from mean is $0.68$ .

Therefore, $68\%$ of data lies in the interval $\left(\mu -\sigma ,\mu +\sigma \right)$ .

Thus, in normal distribution, according to the Empirical rule, $50\%$ of the data lies to the left of the mean $\mu$ and $50\%$ of the data lies to the right of the mean $\mu$ , and, $34\%$ of the data lies between $\mu$ and $\mu -\sigma$ and $34\%$ of the data lies between $\mu$ and $\mu +\sigma$ . It means that most of the data lies within one standard deviation of the mean.

Therefore, the given statement is true.