Chapter 6.2, Problem 19E

a.

To determine

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

Answer to Problem 19E

The population mean $\mu$ and standard deviation $\sigma$ are 75.4 and 5.1225.

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,

$\begin{array}{c}\mu =\frac{\text{Sum of temperature}}{\text{Population size}}\\ =\frac{69+75+79+83+71}{5}\\ =\frac{377}{5}\\ =75.4\end{array}$

Thus, the population mean $\mu$ is 75.4.

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 as75.4 and N as5.

That is,

$\begin{array}{c}\sigma =\sqrt{\frac{{\left(69-75.4\right)}^2+{\left(75-75.4\right)}^2+{\left(79-75.4\right)}^2+{\left(83-75.4\right)}^2+{\left(71-75.4\right)}^2}{5}}\\ =\sqrt{\frac{{\left(-6.4\right)}^2+{\left(-0.4\right)}^2+{\left(3.6\right)}^2+{\left(7.6\right)}^2+{\left(4.4\right)}^2}{5}}\\ =\sqrt{\frac{40.96+0.16+12.96+57.76+19.36}{5}}\\ =\sqrt{\frac{131.2}{5}}\end{array}$

$\begin{array}{c}=\sqrt{26.24}\\ =5.1225\end{array}$

Thus, the standard deviation $\sigma$ is 5.1225.

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 {69, 75, 79, 83, 71}.

Thedifferent possible samples of size 2 are {(69, 69), (69,75), (69, 79), (69, 83), (69, 71), (75, 69), (75, 75), (75, 79), (75, 83), (75, 71), (79, 69), (79, 75), (79, 79), (79, 83), (79, 71), (83, 69), (83, 75), (83, 79), (83, 83), (83, 71), (71, 69), (71, 75), (71, 79), (71, 83), (71, 71)}.

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 19E

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

Sample	Sample mean $\left(\overline{x}\right)$
69, 69	69
69, 75	72
69, 79	74
69, 83	76
69, 71	70
75, 69	72
75, 75	75
75, 79	77
75, 83	79
75, 71	73
79, 69	74
79, 75	77
79, 79	79
79, 83	81
79, 71	75
83, 69	76
83, 75	79
83, 79	81
83, 83	83
83, 71	77
71, 69	70
71, 75	73
71, 79	75
71, 83	77
71, 71	71
Total	1,885

The mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$ of the samples means are 75.4 and 3.62215.

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 (69, 69):

$\begin{array}{c}\overline{x}=\frac{69+69}{2}\\ =\frac{138}{2}\\ =69\end{array}$

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 (69, 69):

Substitute $\overline{x}$ as 69 and n as 2 in sample standard deviation

That is,

$\begin{array}{c}\text{Sample variance}=\frac{{\left(69-69\right)}^2+{\left(69-69\right)}^2}{2-1}\\ =\frac{{\left(0\right)}^2+{\left(0\right)}^2}{1}\\ =\frac{0+0}{1}\\ =0\end{array}$

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
69, 69	69	0
69, 75	72	18
69, 79	74	50
69, 83	76	98
69, 71	70	2
75, 69	72	18
75, 75	75	0
75, 79	77	8
75, 83	79	32
75, 71	73	8
79, 69	74	50
79, 75	77	8
79, 79	79	0
79, 83	81	8
79, 71	75	32
83, 69	76	98
83, 75	79	32
83, 79	81	8
83, 83	83	0
83, 71	77	72
71, 69	70	2
71, 75	73	8
71, 79	75	32
71, 83	77	72
71, 71	71	0
Total	1,885	656

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

$\begin{array}{c}{\mu }_{\overline{x}}=\frac{\text{Sum of sample means}}{\text{Total number of samples}}\\ =\frac{1,885}{25}\\ =75.4\end{array}$

Thus, the mean ${\mu }_{\overline{x}}$ of the sample means is 75.4.

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

$\begin{array}{c}{\sigma }_{\overline{x}}=\sqrt{\frac{{\sigma }^2}{2n}}\\ =\sqrt{\frac{656}{50}}\\ =3.62215\end{array}$

Thus, the standard deviation ${\sigma }_{\overline{x}}$ of the sample means is 3.62215.

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 75.4 and 5.1225.

Substitute 75.4 for $\mu$ in ${\mu }_{\overline{x}}=\mu$

${\mu }_{\overline{x}}=75.4$

Substitute 5.1225 for $\sigma$ and 2 for n in ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{5.1225}{\sqrt{2}}\\ =\frac{5.1225}{1.4142135}\\ =3.62215\end{array}$

From part (c), the mean ${\mu }_{\overline{x}}$ and standard deviation ${\sigma }_{\overline{x}}$ of the samples means are 75.4 and 3.62215.

Thus, the mean and standard deviation both are equal.

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