Chapter 11.1, Problem 36E

To determine

To find: the probability of $P\left(\left|x-\mu \right|\ge 15\right)$

Answer to Problem 36E

The probability that the person late for meeting is $0.1336$

Explanation of Solution

Given:

Mean is $75$

Standard deviation is $10$ .

Calculation:

Given that average scores on a standard test are normally distributed with a mean of $75$ and standard deviation of $10$ . That is $\mu =75$ and $\sigma =10$ .

Find $P\left(\left|x-\mu \right|\ge 15\right)$

So, remember that the $z-score$ corresponding to $x-value$ is given by.

  $z=\frac{x-\mu }{\sigma }$

Now, consider

  $\left|x-\mu \right|\ge 15$

Now, since $\sigma =10$ is a positive integer so divide both sides of above inequality by it without affecting the inequality

  $\begin{array}{l}\Rightarrow \frac{\left|x-\mu \right|}{\sigma }\ge \frac{15}{\sigma }\\ \Rightarrow \left|\frac{x-\mu }{\sigma }\right|\ge \frac{15}{10}\\ \Rightarrow \left|z\right|\ge 1.5\end{array}$

Hence, our problem is now reduced to finding

  $P\left(\left|z\right|\ge 1.5\right)=P\left(z\ge 1.5\right)+P\left(z\le -1.5\right)$

Now, use the standard normal table to find $P\left(z\ge 1.5\right)$ and $P\left(z\le -1.5\right)$ .

To find $P\left(z\ge 1.5\right)$ using standard normal table, find the value where row $1$ and column $5$ intersects that is $.9332$ . This value is the probability $P\left(z\le 1.5\right)$ .

And to find $P\left(z\le -1.5\right)$ using standard normal table, find the value where row $-1$ and column $5$ intersects that is $.0668$ . This value is the probability $P\left(z\le -1.5\right)$ .

Standard Normal Table
$z$	$.0$	$.1$	$.2$	$.3$	$.4$	$.5$	$.6$	$.7$	$.8$	$.9$
$-3$	$.0013$	$.0010$	$.0007$	$.0005$	$.0003$	$.0002$	$.0002$	$.0001$	$.0001$	$.0000+$
$-2$	$.0228$	$.0179$	$.0139$	$.0107$	$.0082$	$.0062$	$.0047$	$.0035$	$.0026$	$.0019$
$-1$	$.1587$	$.1357$	$.1151$	$.0968$	$.0808$	$.0668$	$.0548$	$.0446$	$.0359$	$.0287$
$-0$	$.5000$	$.4602$	$.4207$	$.3821$	$.3446$	$.3085$	$.2743$	$.2420$	$.2119$	$.1841$
$0$	$.5000$	$.5398$	$.5793$	$.6179$	$.6554$	$.6915$	$.7257$	$.7580$	$.7881$	$.8159$
$1$	$.8413$	$.8643$	$.8849$	$.9032$	$.9192$	$.9332$	$.9452$	$.9554$	$.9641$	$.9713$
$2$	$.9772$	$.9821$	$.9861$	$.9893$	$.9918$	$.9938$	$.9953$	$.9965$	$.9974$	$.9981$
$3$	$.9987$	$.9990$	$.9993$	$.9995$	$.9997$	$.9998$	$.9998$	$.9999$	$.9999$	$1.000-$

Thus, $P\left(z\le 1.5\right)=.9332$

Now,

The probability that the person late for meeting is given by

  $\begin{array}{l}P\left(z\le 1.5\right)=.9332\\ P\left(z\ge 1.5\right)=1-P\left(z\le 1.5\right)\\ =1-0.9332\\ =.0668\end{array}$

Hence,

  $\begin{array}{l}P\left(\left|x-\mu \right|\ge 15\right)=P\left(\left|z\right|\ge 1.5\right)\\ =P\left(z\ge 1.5\right)+P\left(z\le -1.5\right)\\ =0.0668+0.0668\\ =0.1336\end{array}$

Conclusion:

Hence,the probability that the person late for meeting is is $0.1336$