# In a certain community, 512 unmarried women gave birth last year. The ages and numbers of unmarried mothers are presented in the frequency distribution table below.a) Approximate the mean age of...

In a certain community, 512 unmarried women gave birth last year. The ages and numbers of unmarried mothers are presented in the frequency distribution table below.

a) Approximate the mean age of the unmarried mothers?

b)Estimate the standard deviation of the age of the unmarried mothers?

c)What is the median age of the unmarried mothers?

d)State the class boundaries of the modal interval?

**INFORMATION**

10-14yrs (10)

15-19yrs (239)

20-24yrs(168)

25-29yrs(62)

30-34yrs(23)

35-39yrs(8)

40-44yrs(2)

### 2 Answers | Add Yours

(1) The standard way of estimating the mean is to use the midpoint of the classes as the representative value. Then multiply the midpoint of each class by its frequency, sum these products and divide by the cumulative frequency or `bar(x)=(sum(f_i X_m))/n`

The midpoints are 12,17,22,27,32,37,42 so we have:

`(12*10+17*239+22*168+27*62+32*23+37*8+42*2)/512~~20.8`

**Then an estimate for the mean is 20.8**

(2) To estimate the standard deviation we use the shortcut formula:

`s=sqrt((n(sum f*X_m^2)-(sum f*X_m)^2)/(n(n-1)))` where `X_m` is the midpoint of a class, f the frequency and n the cumulative frequency.

Here `s=sqrt((512(235053)-10669^2)/(512*511))~~4.99`

**So an estimate for the standard deviation is 4.99**

(3) We estimate the median by finding the class that the 256th entry is in; in this case it is the class 20-24 .

**An estimate for the median is 22.**

(4) The modal class is the class with the most entries.

**The modal class is 15-19**

thank you!!!