Chapter 7, Problem 1PS

Given a normal distribution with $\mu =100\text{and}\sigma =10,$ if you select a sample of $n=25,$ what is the probability that $\bar{X}$ is

a. less than 95?

b. between 95 and 97.5?

c. above 102.2?

d. There is a $65\%$ chance that $\bar{X}$ is above what values?

To determine

Determine the probability that $\overline{X}$ is less than 95.

Answer to Problem 1PS

The probability is 0.0062.

Explanation of Solution

Given X follows a normal distribution with mean $(\mu )$ 100 and standard deviation $(\sigma )$ 10. Also, 25 (n) samples are selected.

The sample mean $\left(\overline{X}\right)$ will follow a normal distribution with mean $\left({\mu }_{\overline{X}}\right)$ 100 and standard deviation can be calculated as:

$\begin{array}{c}\text{Standard deviation (}{\sigma }_{\overline{X}}\text{)}=\frac{\sigma }{\sqrt{n}}\\ =\frac{10}{\sqrt{25}}\\ =2\end{array}$

The probability that the sample mean takes a value less than 95 needs to be calculated.

Then, z-value corresponding to 95 is,

$\begin{array}{c}z=\frac{\overline{X}-{\mu }_{\overline{X}}}{\sigma }\\ =\frac{95-100}{2}\\ =-2.5\end{array}$

This implies $P\left(Z<-2.5\right)$

From the standard normal cumulative table given in E.2, the area corresponding $z=-2.5$ is 0.0062

Hence, the probability of the sample mean being less than 95 is 0.0062.

To determine

Determine the probability that $\overline{X}$ lies between 95 and 97.5.

Answer to Problem 1PS

The probability is 0.0994.

Explanation of Solution

The probability that the sample mean lies between 95 and 97.5 needs to be calculated.

Z-value corresponding to 97.5 is,

$\begin{array}{c}z=\frac{\overline{X}-{\mu }_{\overline{X}}}{\sigma }\\ =\frac{97.5-100}{2}\\ =-1.25\end{array}$

This implies that the required probability is $P\left(-2.5<Z<-1.25\right)$

From the standard normal cumulative table given in E.2, the area corresponding to $z=-1.25$ is 0.1056.

Therefore,

$\begin{array}{c}P\left(95<\overline{X}<97.5\right)=P\left(-2.5<Z<-1.25\right)\\ =0.1056-0.0062\\ =0.0994\end{array}$

Hence, the probability that $\overline{X}$ lies between 95 and 97.5 is 0.0994.

To determine

Determine the probability that $\overline{X}$ is above 102.2.

Answer to Problem 1PS

The probability is 0.1357.

Explanation of Solution

The probability that the sample mean takes a value above 102.2 needs to be calculated.

Z-value corresponding to 102.2 is:

$\begin{array}{c}z=\frac{\overline{X}-{\mu }_{\overline{X}}}{\sigma }\\ =\frac{102.2-100}{2}\\ =1.10\end{array}$

This implies that the required probability is $P\left(Z>1.10\right)$

From the standard normal cumulative table given in E.2, the area corresponding to $z=1.10$ is 0.8643

This implies

$\begin{array}{c}P\left(\overline{X}>102.2\right)=P\left(Z>-1.5\right)\\ =1-0.8643\\ =0.1357\end{array}$

Hence, the probability of the sample mean being greater than 102.2 is 0.1357.

To determine

Determine the value of $\overline{X},$ if the probability of getting $\overline{X}$ is above $65\%$ .

Answer to Problem 1PS

The required value of $\overline{X}$ is 22.

Explanation of Solution

Consider $a$ be the value of $\overline{X}$ for which the probability that $\overline{X}$ will be greater than $a$ be $65\%$

Therefore,

$\begin{array}{c}P\left(\overline{X}>a\right)=0.65\\ 1-P\left(\overline{X}\le a\right)=0.65\\ P\left(\overline{X}\le a\right)=0.35\end{array}$

From the standard normal cumulative table given in E.2, the value of Z corresponding to the area equal to 0.35 is $-0.39$

Therefore, the required $\overline{X}$ can be calculated as:

$\begin{array}{l}\overline{X}={\mu }_{\overline{X}}+\sigma Z\\ \overline{X}=100+2(-0.39)\\ \overline{X}=99.22\end{array}$

Hence, the value of $\overline{X}$ above which the area is $65\%$ is 99.22.