Chapter 1.2, Problem 10E

A sample of 100 adult women was taken, and each was asked how many children she had. The results were as follows:

1. a. Find the sample mean number of children.
2. b. Find the sample standard deviation of the number of children.
3. c. Find the sample median of the number of children.
4. d. What is the first quartile of the number of children?
5. e. What proportion of the women had more than the mean number of children?
6. f. For what proportion of the women was the number of children more than one standard deviation greater than the mean?
7. g. For what proportion of the women was the number of children within one standard deviation of the mean?

To determine

Find the sample mean number of children.

Answer to Problem 10E

The sample mean number of children is 1.56.

Explanation of Solution

Given info:

The data shows number of children and number of women.

Calculation:

The total of the results for the given data is,

Children	Number of Women
0	27
1	22
2	30
3	12
4	7
5	2
Total	100

Sample mean:

Let $X_1,X_2,\mathrm{..},X_n$ be the sample then the sample mean is,

$\overline{X}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{X}_i}$

The sample size n is 100.

The formula for finding the sample mean number of children is,

$\begin{array}{c}\overline{X}=\frac{1}{100}{\displaystyle \sum _{i=1}^{100}{X}_i}\\ =\frac{1}{100}\left(27\left(0\right)+22\left(1\right)+30\left(2\right)+12\left(3\right)+7\left(4\right)+2\left(5\right)\right)\\ =\frac{1}{100}\left(22+60+36+28+10\right)\\ =\frac{156}{100}\end{array}$

$=1.56$

Thus, the sample mean number of children is 1.56.

To determine

Find the sample standard deviation of the number of children.

Answer to Problem 10E

The sample standard deviation of the number of children is 1.3051.

Explanation of Solution

Calculation:

Sample variance:

Let $X_1,X_2,\mathrm{..},X_n$ be the sample then the sample variance is,

$s^2=\frac{1}{n-1}\left({\displaystyle \sum _{i=1}^n{X}_i^2-n{\overline{X}}^2}\right)$

The sample variance of the number of children is obtained as follows:

$\begin{array}{c}{s}^2=\frac{1}{100-1}\left({\displaystyle \sum _{i=1}^{100}{X}_i^2-100{\overline{X}}^2}\right)\\ =\frac{1}{99}\left(27{\left(0\right)}^2+22{\left(1\right)}^2+30{\left(2\right)}^2+12{\left(3\right)}^2+7{\left(4\right)}^2+2{\left(5\right)}^2-100{\left(1.56\right)}^2\right)\\ =\frac{1}{99}\left(22+120+108+112+50-243.36\right)\\ =\frac{168.64}{99}\end{array}$

$=1.7034$

Thus, the sample variance of the number of children is 1.7034.

The formula for finding the sample standard deviation of the number of children is,

$\begin{array}{c}s=\sqrt{s^2}\\ =\sqrt{1.7034}\\ =1.3051\end{array}$

Thus, the sample standard deviation of the number of children is 1.3051.

To determine

Find the sample median of the number of children.

Answer to Problem 10E

The sample median of the number of children is 2.

Explanation of Solution

Calculation:

Sample median:

If n numbers are ordered from smallest to largest:

• If n is odd, the sample is the number in position $\frac{n+1}{2}$.
• If n is even, the sample median is the average of the numbers in positions $\frac{n}{2}$ and $\frac{n}{2}+1$.

From the given data, it is observed that the sample median is the average of 50^th and 51^st value when arranged in order because the sample size n is 100.

Here, the value of 50^th and 51^st is 2.

Therefore, the average of 50^th and 51^st value is,

$\begin{array}{c}\text{Median}=\frac{2+2}{2}\\ =2\end{array}$

Thus, the sample median of the number of children is 2.

To determine

Find the first quartile of the number of children.

Answer to Problem 10E

The first quartile of the number of children is 0.

Explanation of Solution

Calculation:

The first quartile is the average of the 25^th and 26^th value when arranged in order.

From the given data, it is observed that the 25^th and 26^th values are 0.

Therefore, the average of 25^th and 26^th value is,

$\begin{array}{c}\text{First quartile}=\frac{0+0}{2}\\ =0\end{array}$

Thus, the first quartile of the number of children is 0.

To determine

Find the proportion of the women had more than the mean number of children.

Answer to Problem 10E

The proportion of the women had more than the mean number of children is 0.51.

Explanation of Solution

Calculation:

From part (a), the mean is 1.56 children.

The number of women had more than the mean number of children is 51 $\left(=30+12+7+2\right)$.

The sample size n is 100.

The formula for finding the proportion of the women had more than the mean number of children is,

$\begin{array}{c}\text{Proportion}=\frac{\text{Number of women had more than the mean number of children}}{n}\\ =\frac{51}{100}\\ =0.51\end{array}$

Thus, the proportion of the women had more than the mean number of children is 0.51.

To determine

Find the proportion of the women was the number of children more than one standard deviation greater than the mean.

Answer to Problem 10E

The proportion of the women was the number of children more than one standard deviation greater than the mean is 0.21.

Explanation of Solution

Calculation:

From part (a), the mean is 1.56 children and from part (b), the standard deviation is 1.7034.

The quantity that is one standard deviation greater than the mean is,

$\begin{array}{c}\overline{X}+s=1.56+1.3051\\ =2.8651\end{array}$

The number of women was the number of children more than one standard deviation greater than the mean is 21 $\left(=12+7+2\right)$.

The formula for finding the proportion of the women was the number of children more than one standard deviation greater than the mean is,

$\begin{array}{c}\text{Proportion}=\frac{\left(\begin{array}{l}\text{Number of women was the number of children more }\\ \text{than one standard deviation greater than the mean}\end{array}\right)}{n}\\ =\frac{21}{100}\\ =0.21\end{array}$

Thus, the proportion of the women was the number of children more than one standard deviation greater than the mean is 0.21.

To determine

Find the proportion of the women was the number of children within one standard deviation of the mean.

Answer to Problem 10E

The proportion of the women was the number of children within one standard deviation of the mean is 0.52.

Explanation of Solution

Calculation:

From part (a), the mean is 1.56 children and from part (b), the standard deviation is 1.7034.

The quantity that is one standard deviation with in the mean is,

$\begin{array}{c}\overline{X}-s=1.56-1.3051\\ =0.2549\end{array}$

$\begin{array}{c}\overline{X}+s=1.56+1.3051\\ =2.8651\end{array}$

The number of women was the number of children within one standard deviation of the mean is 52 $\left(=22+30\right)$.

The formula for finding the proportion of the women was the number of children within one standard deviation of the mean is,

$\begin{array}{c}\text{Proportion}=\frac{\left(\begin{array}{l}\text{Number of women was the number of children}\\ \text{ within one standard deviation of the mean}\end{array}\right)}{n}\\ =\frac{52}{100}\\ =0.52\end{array}$

Thus, the proportion of the women was the number of children within one standard deviation of the mean is 0.52.