Chapter 2, Problem R2.6RE

a.

To determine

To find: The percent of test that are earing scores below 500 or above 900.

Answer to Problem R2.6RE

The percentage is 7.47%.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population}\text{mean }(\mu )=694\\ \text{Standard}\text{deviation}\left(\sigma \right)=112\end{array}$

Calculation:

Here, X be the random variable that exhibits the earning score is following the normal distribution.

The percent of test that are earing scores below 500 or above 900 can be calculated as:

  $\begin{array}{c}P\left(X<500\text{ or }X>900\right)=P\left(\frac{x-\mu }{\sigma }<\frac{500-\mu }{\sigma }\right)+P\left(\frac{x-\mu }{\sigma }>\frac{900-\mu }{\sigma }\right)\\ =P\left(Z<\frac{500-694}{112}\right)+P\left(Z>\frac{900-694}{112}\right)\\ =P\left(Z<-1.73\right)+P\left(Z>1.84\right)\\ =0.0747\end{array}$

Thus, the required percentage is 7.47%.

b.

To determine

To find: The 99^th percentile on the test

Answer to Problem R2.6RE

The percentile is 954.96.

Explanation of Solution

Using the standard normal table, the z-score for the probability 0.99 is 2.33.

The 99^th percentile on the test can be calculated as:

  $\begin{array}{c}x=\mu +Z\times \sigma \\ =694+\left(2.33\right)\times 112\\ =954.96\end{array}$

Therefore, the required percentile is 954.96.