Calculate the Skewness and Kurtosis is values

Class Interval Frequency

20-30 18

30-40 35

40-50 65

50-60 32

60-70 10

Arithmetic mean= = 43.8125

Median = 40 +[ ] = 40+4.153846154

= 44.153846154

Mode = 40+ [ ] = 40 +

= 40+4.761904762 = 44.761904762

Q3 = 50 + = 50 +

= 50+0.625

=50.625

Q1= 30 + = 30+

= 30+6.285714286 = 36.288714286

Standard deviation = Ö = Ö 110.4647875

= 10.510223

Class Interval

Frequency

(f)

Mid class

Value

(x)

xf

Less than cumulative frequency

x- ` x

F(x- ` x)

(x- ` x)2

F(x- ` x)2

(x- ` x)3

F(x- ` x)3

(x- ` x)4

F(x- ` x)4

20-30

18

25

450

18

-18.8125

-338.6250

353.9101

6370.3818

-6657.9348

-119842.8264

12525.3987

22545943.1770

30-40

35

35

1225

53

-8.8125

-308.4375

77.6601

2718.1035

-684.3801

-23953.3035

6031.0998

211088.4930

40-50

65

45

2925

118

1.1875

77.1875

1.4101

91.6565

1.6745

108.8425

1.9885

129.2525

50-60

32

55

1760

150

11.1875

358.0000

125.1601

4005.1232

1400.2292

44807.3344

15665.0647

501282.0704

60-70

10

65

650

160

21.1875

211.8750

448.9101

4489.1010

9511.2839

95112.8390

201520.3284

2015203.2840

Total

160

-

7010

-

-

0

-

17674.3660

-

3767.1140

-

25273646.26

S f(x- ` x) 0

m 1 = ------------------ = ---------------------- = 0

160 160

S f(x- ` x)2 17674.3660

m 2 = ------------------ = ---------------------- = 110.4647875

160 160

S f(x- ` x)3 -3767.1140

m 3 = ------------------ = ---------------------- = -23.5444425

160 160

S f(x- ` x)4 25273646.26

m 4 = ------------------ = ---------------------- = 157960.2891

160 160

1) Karl-pearsons coefficient of Skewness for mode

Arithmetic mean-mode

Coefficient of Skewness = --------------------------------------

Standard deviation

43.8125-44.761904762

= -----------------------------------------

10.510223

= -0.00593377975

2) Karl-Pearson’s coefficient of skewness for median

3( Arithmetic mean-median)

Coefficient of Skewness = --------------------------------------

Standard deviation

3 (43.8125- 44.153846154)

= -----------------------------------------

10.510223

-1.02403845

= ------------------------------- = -0.097432609

10.510223

3) Bowley’s coefficient of skewness

Q3+Q1-2(median)

Bowley’s Coefficient of Skewness = --------------------------------------

Q3 - Q1

50.625+36.285714286- 2(44.153846154)

= -----------------------------------------

50.625 – 36.285714286

86.91071428 – 88.3076923

= -------------------------------------------

14.33928572

1.39697802

= ------------------------------- = - 0.97423124

14.33928572

4) Skewness from moments

m 3 2

Skewness ( b 1 ) = ---------------------

m 2 3

(-25.54444625)2

= -----------------------------------------

(110.4647875)3

652.5195644

= ------------------------------- = - 0.0004840853652

1347943.176

5) Kurtosis

m 4

Kurtosis( b 2 ) = ---------------------

m 2 2

157960.2891

= -----------------------------------------

(110.4647875)2

157960.2891

= ------------------------------- = 12.94494463

12202.46928

Answers

1)Karl-pearson’s coefficient of Skewness for mode = - 0.00593377975

2) Karl-pearson’s coefficient of Skewness for median = - 0.097432609

3) Buorley’s cooefficient of Skewness = - 0.97423124

4)Skewness for moments = - 0.0004840853652

5)Kurtosis= 12.94494463

Last modified: Friday, 20 January 2012, 10:30 AM