Chapter 13.2, Problem 46E

To determine

To find: The probability for a randomly selected $x$ value from the distribution.

Answer to Problem 46E

The probability is approximately $0\cdot 4985$ .

Explanation of Solution

Given:

The mean of a normal distribution is $\overline{x}$ and standard deviation is $\sigma$ .

Calculation:

The normal distribution curve is shown below:

  

Probability for $\overline{x}-3\sigma \le x\le \overline{x}$ is the area under the curve lies in the interval between $\overline{x}-3\sigma$ and $\overline{x}$ .

Using the above diagram the area under the curve lies in the interval between $\overline{x}-3\sigma$ and $\overline{x}$ is given by:

  $\text{Area}\approx 2\cdot 35\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$0$}\right.+13\cdot 5\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$0$}\right.+34\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$0$}\right.$

So the probability for $\overline{x}-3\sigma \le x\le \overline{x}$ is given as:

  $\begin{array}{c}\text{P}\left(\overline{x}-3\sigma \le x\le \overline{x}\right)\approx 0\cdot 0235+0\cdot 135+0\cdot 34\\ =0\cdot 4985\end{array}$

So, the probability is approximately $0\cdot 4985$ .