Chapter 6.5, Problem 43E

a.

To determine

Find the density function of  $\overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}$.

Answer to Problem 43E

The random variable $\overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}$ has normal distribution with mean $\mu$ and variance $\frac{{\sigma }^2}{n}$.

Explanation of Solution

Calculation:

From the given information, $Y_1,Y_2,\mathrm{...}{Y}_n$ are independent and identically distributed normal random variables with mean $\mu$ and variance ${\sigma }^2$.

From the Exercise 6.41, the random variable ${\displaystyle \sum _{i=1}^n{a}_i{Y}_i}$ has normal distribution with mean ${\displaystyle \sum _{i-1}^n{a}_i{\mu }_i}$ and variance ${\displaystyle \sum _{i-1}^n{a}_i^2{\sigma }_i^2}$.

The given random variable is $\overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}$. Here, $a_i=\frac{1}{n}$.

Therefore, the mean of $\overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}$  is ${\displaystyle \sum _{i-1}^n\frac{1}{n}\mu =\mu }$ and variance is $\frac{1}{n}{\sigma }^2\left({\displaystyle \sum _{i-1}^n{\left(\frac{1}{n}\right)}^2{\sigma }^2}=\frac{n}{n^2}{\sigma }^2\right)$.

 Thus, random variable $\overline{Y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}$ has normal distribution with mean $\mu$ and variance $\frac{{\sigma }^2}{n}$.

b.

To determine

Find the probability that the sample mean $\overline{Y}$ takes on a value that is within one unit of the population mean, $\mu$.

Answer to Problem 43E

The probability that the sample mean $\overline{Y}$ takes on a value that is within one unit of the population mean, $\mu$ is 0.7888.

Explanation of Solution

Calculation:

From the given information, ${\sigma }^2=16$ and $n=25$.

Hence, $\overline{Y}$ has normal distribution with mean $\mu$ and variance $\frac{16}{25}\left(=\frac{{\sigma }^2}{n}\right)$.

Consider,

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=P\left(-1\le \overline{Y}-\mu \le 1\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{\overline{Y}-\mu }{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{16}{25}}\right)}\le Z\le \frac{1}{\left(\sqrt{\frac{16}{25}}\right)}\right)\\ =P\left(\frac{-5}{4}\le Z\le \frac{5}{4}\right)\end{array}$

                           $\begin{array}{c}=P\left(-1.25\le Z\le 1.25\right)\\ =P\left(Z\ge -1.25\right)-P\left(Z\ge 1.25\right)\\ =1-P\left(Z\ge 1.25\right)-P\left(Z\ge 1.25\right)\\ =1-2P\left(Z\ge 1.25\right)\end{array}$

From the Table 4 in Appendix 3, the area right to 1.25 is 0.1056.

Therefore,

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=1-2P\left(Z\ge 1.25\right)\\ =1-\left(2\times 0.1056\right)\\ =0.7888\end{array}$

c.

To determine

Find the value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=36$.

Find the value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=64$.

Find the value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=81$.

Interpret the results of calculations.

Answer to Problem 43E

The value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=36$ is 0.8664.

The value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=64$ is 0.9544.

The value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ when $n=81$ is 0.9756.

The value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ is increasing with the sample size increases.

Explanation of Solution

Calculation:

If $n=36$:

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=P\left(-1\le \overline{Y}-\mu \le 1\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{\overline{Y}-\mu }{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{16}{36}}\right)}\le Z\le \frac{1}{\left(\sqrt{\frac{16}{36}}\right)}\right)\\ =P\left(\frac{-6}{4}\le Z\le \frac{6}{4}\right)\end{array}$

                           $\begin{array}{c}=P\left(-1.5\le Z\le 1.5\right)\\ =P\left(Z\ge -1.5\right)-P\left(Z\ge 1.5\right)\\ =1-P\left(Z\ge 1.5\right)-P\left(Z\ge 1.5\right)\\ =1-2P\left(Z\ge 1.5\right)\end{array}$

From the Table 4 in Appendix 3, the area right to 1.5 is 0.0668.

Therefore,

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=1-2P\left(Z\ge 1.5\right)\\ =1-\left(2\times 0.0668\right)\\ =0.8664\end{array}$

If $n=64$:

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=P\left(-1\le \overline{Y}-\mu \le 1\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{\overline{Y}-\mu }{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{16}{64}}\right)}\le Z\le \frac{1}{\left(\sqrt{\frac{16}{64}}\right)}\right)\\ =P\left(\frac{-8}{4}\le Z\le \frac{8}{4}\right)\end{array}$

                           $\begin{array}{c}=P\left(-2\le Z\le 2\right)\\ =P\left(Z\ge -2\right)-P\left(Z\ge 2\right)\\ =1-P\left(Z\ge 2\right)-P\left(Z\ge 2\right)\\ =1-2P\left(Z\ge 2\right)\end{array}$

From the Table 4 in Appendix 3, the area right to 2 is 0.0228.

Therefore,

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=1-2P\left(Z\ge 1.5\right)\\ =1-\left(2\times 0.0228\right)\\ =0.9544\end{array}$

If $n=81$:

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=P\left(-1\le \overline{Y}-\mu \le 1\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{\overline{Y}-\mu }{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\le \frac{1}{\left(\sqrt{\frac{{\sigma }^2}{n}}\right)}\right)\\ =P\left(\frac{-1}{\left(\sqrt{\frac{16}{81}}\right)}\le Z\le \frac{1}{\left(\sqrt{\frac{16}{81}}\right)}\right)\\ =P\left(\frac{-9}{4}\le Z\le \frac{9}{4}\right)\end{array}$

                           $\begin{array}{c}=P\left(-2.25\le Z\le 2.25\right)\\ =P\left(Z\ge -2.25\right)-P\left(Z\ge 2.25\right)\\ =1-P\left(Z\ge 2.25\right)-P\left(Z\ge 2.25\right)\\ =1-2P\left(Z\ge 2.25\right)\end{array}$

From the Table 4 in Appendix 3, the area right to 2.25 is 0.0122.

Therefore,

$\begin{array}{c}P\left(\left|\overline{Y}-\mu \right|\le 1\right)=1-2P\left(Z\ge 1.5\right)\\ =1-\left(2\times 0.0122\right)\\ =0.9756\end{array}$

Here, the value of $P\left(\left|\overline{Y}-\mu \right|\le 1\right)$ is increasing with the sample size increases.