Chapter 11.5, Problem 26HP

To determine

To find: The relation between the z value, the position of the interval of X the normal distribution and the area under the normal curve and the probability of the interval occurring then use the example to explain the reasoning.

Answer to Problem 26HP

The relation between the z value, the position of the interval of X the normal distribution and the area under the normal curve and the probability of the interval occurring is shown in Table and the example for it is obtained.

Explanation of Solution

Consider the random variable X that has the random variable X and have the normal distribution o $\mu$ with the standard deviation $\sigma$ . As per the empirical rule of the normal distribution of the text book the area between $\mu -\sigma \le X\le \mu +\sigma$ is 0.68, is 0.95 and $\mu -2\sigma \le X\le \mu +2\sigma$ the area between $\mu -2\sigma \le X\le \mu +3\sigma$ is 0.997.

Consider the standard normal distribution of $Z$ is,

  $Z=\frac{X-\mu }{\sigma }$

Consider $\mu -\sigma \le X\le \mu +\sigma$ is,

  $\begin{array}{l}\mu -\sigma -\mu \le X\le \mu +\sigma -\mu \\ -1\le \frac{X-\mu }{\sigma }\le 1\\ -1\le Z\le 1\end{array}$

Similarly consider $\mu -2\sigma \le X\le \mu +2\sigma$ and $\mu -2\sigma \le X\le \mu +3\sigma$ respectively between the $-2\le Z\le 2$ and $-3\le Z\le 3$ .

The relationship between the position of the interval of X in the normal distribution and the z value, then the probability of the interval that occurs and the area under the normal curve is shown below.

Table 1

Normal	Standard Normal	Area	Probability
$\mu -\sigma \le X\le \mu +\sigma$	$-1\le Z\le 1$	68%	0.68
$\mu -2\sigma \le X\le \mu +2\sigma$	$-2\le Z\le 2$	95%	0.95
$\mu -2\sigma \le X\le \mu +3\sigma$	$-3\le Z\le 3$	99.7%	0.997

Consider the example.

Consider that the data is on scores per team in each game of first round of the NBA play-off.

Consider the mean of the scores per team is 11.06. Then, according to the empirical normal distribution of the textbook the middle 68% of the distribution lie on the interval of,

  $\begin{array}{l}\left(96.56-11.06\right)<X<\left(96.56+11.06\right)\\ 85.50<X<107.62\end{array}$

Then, the middle 95% lie in the interval.

  $\begin{array}{l}\left(96.56-2\left(11.06\right)\right)<X<\left(96.56+2\left(11.06\right)\right)\\ 74.44<X<118.68\end{array}$

Then, the middle 99.7% lie in the interval.

  $\begin{array}{l}\left(96.56-3\left(11.06\right)\right)<X<\left(96.56+3\left(11.06\right)\right)\\ 63.38<X<129.74\end{array}$