Question

A sample of 100 adult women was taken, and each was asked how many children she had. The results were as follows: Children Number of Women 0 27 21

a. Find the sample standard deviation of the number of children.

b. What proportion of the women had more than the mean number of children? Round the answer to two decimal places.

c. For what proportion of the women was the number of children more than one standard deviation greater than the mean? Round the answer to two decimal places.

d. For what proportion of the women was the number of children within one standard deviation of the mean? Round the answer to two decimal places.

Answer 1

Solution:

a. Find the sample standard deviation of the number of children.

Answer:

The sample standard deviation is given below:

b. What proportion of the women had more than the mean number of children? Round the answer to two decimal places.

Answer: $(33+9+8+2)$ women had more than the mean number of children.

Therefore, the proportion of the women had more than the mean number of children is:

$\boldsymbol{\frac{52}{100}=0.52}$

c. For what proportion of the women was the number of children more than one standard deviation greater than the mean? Round the answer to two decimal places.

Answer: $Mean +Standard-deviation = 1.56+1.31=2.87$

$(9+8+2)$women had more than 2.87 children.

$\therefore proportion = \frac{(9+8+2)}{100}=\frac{19}{100}=0.19$

d. For what proportion of the women was the number of children within one standard deviation of the mean? Round the answer to two decimal places.

Answer: $Mean -Standard-deviation =1.56-1.31=0.25$

$(21+33)$ women had within 0.25 and 2.87 children.

$\therefore proportion =\frac{21+33}{100}=\frac{54}{100}=0.54$