Question

Calculate the Skewness by using the Karl-Pearson method for the following frequency distribution and interpret the result.

Ages(Years)

15-19

20-24

25-29

30-34

35-39

40-44

45-49

50-54

No.of Men

29

176

208

173

82

40

15

3

 

Answer 1

s = standard deviation = $\sqrt{\frac{1}{N}\sum x^2f-\bar x^2}=6.783875$

$N/2 =826/2=413,$ Hence median class = 24.5 - 29.5

l = lower boundary of median class = 24.5

F =cumulative frequency upto l =305

f = frequency of median class = 208

c = width of median class = 5

Hence median = $\bar x = l+\dfrac{N/2-F_t}{f}\cdot c = 24.5+\dfrac{413-305}{208}\times 5= 27.09615$

Now, for Mode

To find Mode Class

Here, maximum frequency is 276.

 The mode class is 19.5-24.5.

L=lower boundary point of mode class =19.5

$f_1$ = frequency of the mode class =276

$f_0$ = frequency of the preceding class =29

$f_2$ = frequency of the succeeding class =208

c = class length of mode class =5

$Z=L+(\dfrac{f_1-f_0}{2f_1-f_0-f_2})\cdot c\\\ \\Z=19.5+(\dfrac{276-29}{2\cdot 276-29-208})\cdot 5\\\ \\Z=19.5+(\dfrac{247}{315})\cdot 5\\\ \\Z= 23.4216$

Pearson's Coefficient of Skewness = $\dfrac{3(\bar x -Mo)}{s}=\dfrac{3(27.09615-23.4216)}{6.7838}=1.625$

Since value of Pearson's coefficient of skewness>0, the distribution is positively skewed.