Chapter 6.2, Problem 20E

a.

To determine

Find the population mean $\mu$ and standard deviation $\sigma$.

Answer to Problem 20E

The population mean $\mu$ and standard deviation $\sigma$ are 27.2 and 3.9699, respectively.

Explanation of Solution

Calculation:

The given data shows that the summer temperature for five days in July.

The formula to find the population mean is,

$\begin{array}{c}\mu =\frac{\text{Sum of temperature}}{\text{Population size}}\\ =\frac{29+30+26+31+20}{5}\\ =\frac{136}{5}\\ =27.2\end{array}$

Thus, the population mean $\mu$ is 27.2.

The formula to find the population standard deviation is,

$\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{N}}$

Here, x represents the population values,µ represents the population mean and N represents the population size.

Substitute µ as27.2 and N as5.

That is,

$\begin{array}{c}\sigma =\sqrt{\frac{{\left(29-27.2\right)}^2+{\left(30-27.2\right)}^2+{\left(26-27.2\right)}^2+{\left(31-27.2\right)}^2+{\left(20-27.2\right)}^2}{5}}\\ =\sqrt{\frac{{\left(1.8\right)}^2+{\left(2.8\right)}^2+{\left(-1.2\right)}^2+{\left(3.8\right)}^2+{\left(-7.2\right)}^2}{5}}\\ =\sqrt{\frac{3.24+7.84+1.44+14.44+51.84}{5}}\\ =\sqrt{15.76}\end{array}$

$=3.9699$

Thus, the standard deviation $\sigma$ is 3.9699.

b.

To determine

List all samples of size 2 drawn with replacement.

Explanation of Solution

Calculation:

The two values are randomly selected with replacement from the population {29, 30,26, 31, 20}.

Thedifferent possible samples of size 2 are {(29, 29), (29,30), (29, 26), (29, 31), (29, 20), (30, 29), (30, 30), (30, 26), (30, 31), (30, 20), (26, 29), (26, 30), (26, 26), (26, 31), (26, 20), (31, 29), (31, 30), (31, 26), (31, 31), (31, 20), (20, 29), (20, 30), (20, 26), (20, 31), (20, 20)}.

Thus, there are 25 different samples of size 2.

c.

To determine

Compute the sample mean $\overline{x}$ for each of the 25 samples of size 2. Also, compute the mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$ of the samples means.

Answer to Problem 20E

The sample mean $\overline{x}$ for each of the 25 samples of size 2 are,

Sample	Sample mean $\left(\overline{x}\right)$
29, 29	29
29, 30	29.5
29, 26	27.5
29, 31	30
29, 20	24.5
30, 29	29.5
30, 30	30
30, 26	28
30, 31	30.5
30, 20	25
26, 29	27.5
26, 30	28
26, 26	26
26, 31	28.5
26, 20	23
31, 29	30
31, 30	30.5
31, 26	28.5
31, 31	31
31, 20	25.5
20, 29	24.5
20, 30	25
20, 26	23
20, 31	25.5
20, 20	20
Total	680

The mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$ of the samples means are 27.2 and 2.8071.

Explanation of Solution

Calculation:

The formula for finding sample mean is $\overline{x}$ is,

$\overline{x}=\frac{\text{Sum of samples}}{\text{Sample size}}$

The sample mean for the sample (29, 29):

$\begin{array}{c}\overline{x}=\frac{29+29}{2}\\ =\frac{58}{2}\\ =29\end{array}$

The formula to find the sample variance is,

$\text{Sample standeard deviation}=\frac{{{\displaystyle \sum \left(x-\overline{x}\right)}}^2}{n-1}$

Where, x represents the sample values, $\overline{x}$ represents the sample mean and n represents the sample size.

The sample variance for the sample (29, 29):

Substitute $\overline{x}$ as 29 and n as 2 in sample standard deviation

That is,

$\begin{array}{c}\text{Sample variance}=\frac{{\left(29-29\right)}^2+{\left(29-29\right)}^2}{2-1}\\ =\frac{{\left(0\right)}^2+{\left(0\right)}^2}{1}\\ =\frac{0+0}{1}\\ =0\end{array}$

Similarly, the sample mean and sample standard deviation for the remaining samples of size 2 is obtained as given in below Table.

Sample	Sample mean $\left(\overline{x}\right)$	Sample variance
29, 29	29	0
29, 30	29.5	0.5
29, 26	27.5	4.5
29, 31	30	2
29, 20	24.5	40.5
30, 29	29.5	0.5
30, 30	30	0
30, 26	28	8
30, 31	30.5	0.5
30, 20	25	50
26, 29	27.5	4.5
26, 30	28	8
26, 26	26	0
26, 31	28.5	12.5
26, 20	23	18
31, 29	30	2
31, 30	30.5	0.5
31, 26	28.5	12.5
31, 31	31	0
31, 20	25.5	60.5
20, 29	24.5	40.5
20, 30	25	50
20, 26	23	18
20, 31	25.5	60.5
20, 20	20	0
Total	680	394

The mean ${\mu }_{\overline{x}}$ of the sample means is obtained as follows:

$\begin{array}{c}{\mu }_{\overline{x}}=\frac{\text{Sum of sample means}}{\text{Total number of samples}}\\ =\frac{680}{25}\\ =27.2\end{array}$

Thus, the mean ${\mu }_{\overline{x}}$ of the sample means is 27.2.

The standard deviation ${\sigma }_{\overline{x}}$ of the sample means is obtained as follows:

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{{\sigma }^2}{2n}}\\ =\sqrt{\frac{394}{50}}\\ =2.8071\end{array}$

Thus, the standard deviation ${\sigma }_{\overline{x}}$ of the sample means is 2.8071.

d.

To determine

Verify that ${\mu }_{\overline{x}}=\mu$ and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

Explanation of Solution

Calculation:

The two values are randomly selected with replacement. Therefore, n is 2.

From part (a), the population mean $\mu$ and standard deviation $\sigma$ are 27.2 and 3.9699.

Substitute 27.2for $\mu$ in ${\mu }_{\overline{x}}=\mu$

${\mu }_{\overline{x}}=27.2$

Substitute 3.9699 for $\sigma$ and 2 for n in ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{3.9699}{\sqrt{2}}\\ =\frac{3.9699}{1.4142135}\\ =2.8071\end{array}$

From part (c), the mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$ of the samples means are 27.2 and 2.8071.

Thus, the mean and standard deviation both are equal.

Hence ${\mu }_{\overline{x}}=\mu$ and ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.