Question

2. Below is a frequency distribution of earnings of 50 contractors in a country. Regarding this distribution. Frequency Class Interval 0-10 10-20 20-30 30-40 40-50 50-60 Compute the mean, variance, the standard deviation and the Coefficient of variation (CV) of the earnings of these contractors. Compute the skewness of the distribution of earnings of these 50 contractors. Is the distribution positively or negatively skewed? (iii)

Answer 1

a)

The table to calculate mean and variance can be summarize as

Interval	mid value (X)	Frequency(f)	Freq*X	(X-mean)	(X-mean)2	f*(X-mean)2
0 - 10	5	2	10	-27.6	761.76	1523.52
10 - 20	15	7	105	-17.6	309.76	2168.32
20 - 30	25	12	300	-7.6	57.76	693.12
30 - 40	35	15	525	2.4	5.76	86.4
40 - 50	45	8	360	12.4	153.76	1230.08
50 - 60	55	6	330	22.4	501.76	3010.56
Total	-	50	1630	-	-	8712

$\bar X = \frac{\sum_{i=1}^{n}X_i}{n}$

1630 50

X = 32,6

g? - Liai f(X - mean 2 n

8712 50

o2 = 174.24

A=0

o= V(174.24)

0 = 13.2

Coefficient of variation = CV

CV * 100

$CV = \frac{13.2}{32.6}*100$

CV = 40.49

Interval	mid value (X)	Frequency(f)	Freq*X	(X-mean)	(X-mean)2	f*(X-mean)2	f*(X-mean)3
0 - 10	5	2	10	-27.6	761.76	1523.52	-42049.152
10 - 20	15	7	105	-17.6	309.76	2168.32	-38162.432
20 - 30	25	12	300	-7.6	57.76	693.12	-5267.712
30 - 40	35	15	525	2.4	5.76	86.4	207.36
40 - 50	45	8	360	12.4	153.76	1230.08	15252.992
50 - 60	55	6	330	22.4	501.76	3010.56	67436.544
Total	-	50	1630	-	-	8712	-2582.4

$\mu_3 = \frac{\sum_{i=1}^{n}f(X - mean)^3}{n}$

H3 = 2582.4 50

$\mu_3 =-51.648$

Coefficient of skewness = $\frac{\mu_3}{\mu_2^{3/2}}$

Coefficient of skewness = $\frac{\ -51.648}{13.2^3}$

Coefficient of skewness = $\frac{\ -51.648}{2299.968}$

Coefficient of skewness =

Since, the sign is negative hence skewness is negative