Chapter 7.3, Problem 60E

(a)

To determine

To find: The mean and standard deviation of $\overline{x}$

Answer to Problem 60E

Mean $=2.4$

Standard deviation $=0.75895$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{Population mean}\left(\mu \right)=6\\ \text{Population standard deviation}\left(\sigma \right)=2.4\\ \text{Sample size}\left(n\right)=10\end{array}$

Formula used:

  $\begin{array}{l}{\mu }_{\overline{x}}=\mu \\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\end{array}$

As per central limit theorem

  $\overline{X}~N\left(\mu ,\frac{{\sigma }^2}{n}\right)$

Where

  $\begin{array}{l}\mu \text{ is the population mean}\\ \sigma \text{ is the population stnadard deviation}\\ \text{n is the sample size}\end{array}$

Calculation:

The mean and standard deviation are:

  $\begin{array}{l}{\mu }_{\overline{x}}=6\\ {\sigma }_{\overline{x}}=\frac{2.4}{\sqrt{10}}=0.75895\end{array}$

Thus, the mean and standard deviation are 6 and 0.75895.

(b)

To determine

To explain: The reason behind not be able to safely calculate $P\left(\overline{x}<5\right)$ with sample size 10.

Explanation of Solution

In order to use Central Limit theorem,

The population should follow normality and the sample size should be greater than or equal to 30.

Here, the distribution of population is unknown and the sample size is also less than 30.

Therefore, $\overline{x}$ cannot be approximately normal distributed.

(c)

To determine

To calculate: The probability of $\overline{x}<5$

Answer to Problem 60E

The probability is 0.0934.

Explanation of Solution

Calculation:

The probability can be calculated as follows

  $$

  $\begin{array}{c}P\left(\overline{x}<5\right)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{5-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ =P\left(Z<\frac{5-6}{\frac{2.4}{\sqrt{10}}}\right)\\ =P\left(Z<-1.32\right)\\ =0.0934\end{array}$

The probability is 0.0934.