Mathematics, 18.01.2020 03:31, maytce7237

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

0 children = 27 women

1 child = 22 women

2 children = 30 women

3 children = 12 women

4 children = 7 women

5 chuldren = 2 women

a. find the sample mean number of children

b. find the sample standard deviation of the number of children.

c. find the sample median of the number of children

d. what is the first quartile of the number of children

e. what proportion of women had more then the mean number of children?

f. for what proportion of the women was the number of children more than one standard deviation greater than the mean?

g. for what proportion of the women was the number of children within one standrd deviation of the mean?

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answered: hope4517

The given data where the number of observation (frequency) for a

specified number of children is a frequency distribution.

The correct responses are;

a. 1.56b. 1.305c. 2d. 0e. 0.51f. 0.21g. 0.52

Reason:

a. The given data is presented in the following frequency table as follows;

$\begin{tabular}{c|l|c|}x&f& x \times f\\0&27&0\\1&22&22\\2&30&60\\3&12&36\\4&7&28\\5&2&10\end{array}\right]$

∑(x × f) = 0 + 22 + 60 + 36 + 28 + 10 = 156

∑f = 100

$\displaystyle The \ mean \ number \ of \ children, \ \overline x = \mathbf{\frac{\sum (f\times x)}{\sum f}} = \frac{156}{100} = 1.56$

The sample mean number of children, $\overline x$ = 1.56

b. The standard deviation of a frequency distribution is given as follows;

$\displaystyle s =\sqrt{\dfrac{\sum \left f\times (x_i-\overline x \right )^{2} }{ \sum f - 1}} = \sqrt{\frac{168.64}{100-1} } \approx \mathbf{ 1.305}$

The sample standard deviation of the number of children, s ≈ 1.305

c. The median of the number of children is given by the number of children of the 50th percentile, which is, x = 2

The median number of children is; x = 2

d. The first quartile of the number of children is the number of children in the 25th percentile, which is, x = 0

The first quartile of the number of children is; x = 0

e. The number of women that have more than the mean number of children is; 30 + 12 + 7 + 2 = 51

Therefore;

$\displaystyle Proportion \ of \ women \ having \ more \ children \ than \ the \ mean = \frac{51}{100} = 0.51$

The proportion of women having more children than the mean = 0.51

f. The number of children given by one standard deviation greater than the mean = 1.56 + 1.305 = 2.865

Number of women with more children than 2.865 = 12 + 7 + 2 = 21

$\displaystyle Proportion \ of \ women \ with \ more \ children \ than \ 2.865 = \frac{21}{100} = \mathbf{0.21}$

The proportion of women that have more children than one standard deviation greater than the mean is 0.21

g. Number of children within one standard deviation from the mean is found as follows;

1.56 - 1.305 ≤ x ≤ 1.56 + 1.305

Which gives;

0.255 ≤ x ≤ 2.865

Number of women having children in the above range = 22 + 30 = 52

$\displaystyle Proportion \ of \ women = \frac{52}{100} = \mathbf{0.52}$

The proportion of women that have the number of children within one standard deviation of the mean is 0.52

Learn more about frequency distribution here:

link

answered: sotoamerica0814

a. Sample mean number of children is 1.56

b. The sample standard deviation of the number of children is 1.3052

c. The sample median of the number of children is 2

d. The first quartile of the number of children is 0

e. The proportion of women having more then the mean number of children is 0.51

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

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

Step-by-step explanation:

a) The sample mean number of children:

$\begin{aligned}\bar{X} &=\frac{\sum_{i=1}^{n} X_{i}}{n} \\&=\frac{(0)(27)+(1)(22)+(2)(30)+(3)(12)+(4)(7)+(5)(2)}{100}\end{aligned}$

= 1.56

b) The sample standard deviation of the number of children:

$\begin{aligned}s &=\sqrt{\frac{1}{n-1}\left(\sum_{i=1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)} \\&=\sqrt{\frac{1}{100-1}\left(27(0)^{2}+22(1)^{2}+30(2)^{2}+12(3)^{2}+7(4)^{2}+2(5)^{2}-100(1.56)^{2}\right)}\end{aligned}$

= 1.3052

c) The sample median of the number of children:

Given n = 100 which is a even number.

A sample medium is the average of the numbers in positions $\frac{n}{2}$ and $\frac{n}{2}+1$ = 100/2 + 1 = 51

Therefore the sample median is $\frac{2+2}{2}$ =2

d) The first quartile of the number of children:

The first quartile is defined as the average of the 25th and 26th data when the sample is in ascending or increasing order.

(0.25)(n+1) = (0.25)(100+1) = 25.25

Thus the first quartile is $\frac{0+0}{2}$ = 0

e) The proportion of women which had more then the mean number of children:

2, 3, 4 and 5 are the numbers greater than the mean value 1.56.

The number of women having 2, 3, 4 and 5 children are 30 + 12 + 7 + 2 = 51

The proportion of these women are

51/100 = 0.51

f) Standard deviation greater than the mean:

$\bar{X}+s$ = 1.56 + 1.305

= 2.865

Thus this is more than one standard deviation greater than the mean of 3, 4 and 5

The number of women having 3, 4 and 5 children = 12 + 7 + 2 = 21

Therefore the proportion of women having the number of children more than one standard deviation greater than the mean,

= 21/100

= 0.21

g) 1 and 2 are within one standard deviation

( $\bar{X}-s$ , $\bar{X}+s$ ) = (1.56 - 1.305, 1.56 + 1.305) = (0.255, 2.865)

The number of women having 1 or 2 children = 22 + 30 = 52

Therefore the proportion of women having the number of children within one standard deviation of the mean = 52/100 = 0.52

answered: Guest

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

answered: Guest

Wish i could u but i just don’t know the answer