Question

1)     In a frequency distribution of 100 families given below the number of families corresponding to expenditure groups 20-40 and 60-80 are missing from the table. However, the median is known to be 50.

 Expenditure group

0-20

20-40

40-60

60-80

80-100

No. of families

14

f2

27

f4

15

a)         Find the missing frequencies.

b)        Find mean, mode, variance, standard deviation, range, mean deviation for mean and median

Answer 1

a) Let $f_2=$ the number of families in class $20-40,$ $f_4=$ the number of families in class

$60-80.$ Then

$$14+f_2+27+f_4+15=100$$

$$f_2+f_4=44$$

Estimated Median =$L+\dfrac{n/2-B}{G}\times w$  

where:

L is the lower class boundary of the group containing the median

n is the total number of values

B is the cumulative frequency of the groups before the median group

G is the frequency of the median group

w is the group width

Given $n=100, w=20, Median=M=50$

Then $L=20$ or $L=40$

Take $L=20$

$$20+\dfrac{100/2-14}{f_2}\times 20=50$$

$$f_2=24$$

But $20+24=44<100/2.$ We have a contradiction.

Take $L=40$

$$40+\dfrac{100/2-(14+f_2)}{27}\times 20=50$$

$$36-f_2=13.5$$

$$f_2=22.5$$

Take $f_2=23. f_4=21$

b)

$$10(14)+30(23)+50(27)+70(21)+90(15)=5000$$

$$mean=\dfrac{5000}{100}$$

$$mean=\bar{x}=50$$

Find Mode Class

Maximum frequency is 27.The mode class is 40-60.

$L=40$

$f_3=27$

$f_2=23$

$f_4=21$

$w=20$

$$mode=L+\dfrac{f_3-f_2}{2f_3-f_2-f_4}\times w$$

$$=40+\dfrac{27-23}{2(27)-23-21}\times 20=48$$

$$10^2(14)+30^2(23)+50^2(27)+70^2(21)+90^2(15)=$$

$$=314000$$

$$314000-\dfrac{(5000)^2}{100}=64000$$

$$Var(X)=s^2=\dfrac{64000}{100-1}=646.4646$$

$$s=\sqrt{s^2}=\sqrt{\dfrac{64000}{99}}=25.4257$$

Mean deviation of Mean

$$\delta\bar{x}=\dfrac{\sum_i|x_i-\bar{x}|f_i}{100}$$

$$\sum_i|x_i-\bar{x}|f_i$$ $$=40(14)+20(23)+0(27)+20(21)+40(15)=2040$$

$$\delta\bar{x}=\dfrac{\sum_i|x_i-\bar{x}|f_i}{100}=\dfrac{2040}{100}=20.4$$

Mean deviation of Median

$$\delta M=\dfrac{\sum_i|x_i-M|f_i}{100}$$

Since $\bar{x}=M$

$$\delta M=\delta\bar{x}=20.4$$

$$Range=90-10=80$$