Chapter 6.3, Problem 3E

Solution

To find: the indicated probability for a randomly selected x -value from the distribution.

  $P\left(x\le \overline{x}-\sigma \right)=0.16$

Given information:

The probability to be found is: $P\left(x\le \overline{x}-\sigma \right)$

It is given that the distribution is normal and has mean $\overline{x}$ and standard deviation $\sigma$ .

Concept Used:

Areas under a Normal Curve

A normal distribution with mean $\overline{x}$ and standard deviation $\sigma$ has the following properties:

• The total area under the related normal curve is 1.
• About 68% of the area lies within 1 standard deviation of the mean.
• About 95% of the area lies within 2 standard deviation of the mean.
• About 99.7% of the area lies within 3 standard deviation of the mean.

  

Explanation:

The objective is to find $P\left(x\le \overline{x}-\sigma \right)$

Now, the probability that a randomly selected x -value is less than or equal to $\overline{x}-\sigma$ is the shaded area under the normal curve as shown below:

  

So,

  $\begin{array}{c}P\left(x\le \overline{x}-\sigma \right)=0.0015+0.0235+0.135\\ =0.16\end{array}$