Unit 3 | unit 3 dispersion and skewness

Unit 3

Dispersion and Skewness

3.1 Concept of Dispersion

Central tendency do not reveal the variability present in the data. Dispersion measures the scatteredness of the data series around its average. It tells the variation of the data from one another and gives a clear idea about the distribution of the data.

Definition

In statistics, dispersion is extent to which a distribution is stretched or squeezed.

Characteristics of measures of dispersion

It should be rigidly defined
It should be easy to understand and calculate
Must be based on all observation of the data
Must be less affected by sampling fluctuation

3.2 Methods of Measuring Dispersion

Types of measures of dispersion

Algebraic measures – It includes the mathematical way to calculate the measures of dispersion.

Graphical measures – The way to calculate the measures of dispersion by graphs and figures.

Absolute measures of dispersion – It gives an idea about the amount of dispersion in a set of observations. It measures the dispersion in the same units as the units of original data. Absolute measures cannot be used for comparison of two or more data set variations.

Relative measures of dispersion – the relative measures of distribution are used for comparing the distribution of two or more data sets.

Coefficients of dispersion are used to compare two series with different measurement of unit.

Types of absolute measures of dispersion

Range – Range defines the difference between the maximum value and the minimum value given in a data set. More the range , group is more variable. The smaller the range the more homogenous is the group.

R = H – L

Example 1 – 5, 10, 15, 20, 7, 9, 17, 13, 12, 16, 8, 6

Range = H-L

=20 – 5 = 15

Coefficient of range:

Coefficient of range = (15/(20+5))*100 = 60

Example 2 – what is the range for the following set of numbers?

15,21,57,43,11,39,56,83,77,11,64,91,18,37

Solution:

Range = H-L

= 91 – 11 = 80

Therefore, the range is 80

Example 3 – the frequency table shows the number of goals the lakers scored in their last twenty matches. What was the range?

No. Of goals	Frequency
0	2
1	3
2	3
3	6
4	3
5	1
6	1
7	1

Solution:

The range is the difference between the lowest and highest values.

The highest value was 7 (They scored 7 goals on 1 occasion)

The lowest value was 0 (They scored 0 goals on 2 occasions)

Therefore, the range = 7 - 0 = 7

Example 4 – the following table shows the sales of DVD players made by a retail store each month last year

Month	No. Of sales
January	25
Feb	43
March	39
April	28
May	29
June	35
July	32
August	46
September	28
October	43
November	51
December	63

Solution:

The range is the difference between the lowest and highest values.

The lowest number of sales = 25 in January

The highest number of sales = 63 in December

So, the range = 63 - 25 = 38

Example 5 – what is the range for the following set of numbers?

57, -5, 11, 39, 56, 82, -2, 11, 64, 18, 37, 15, 68

Solution:

The range is the difference between the lowest and highest values.

The highest value is 82.

The lowest value is -5.

Therefore, the range = 82 - (-5) = 82+5 = 87

Merits

Simple and easy to understand
It gives a quick answer

Demerits

It is not based on all observation
Affected by sampling fluctuations
It cannot be calculated in open ended distributions

2. Interquartile range - The interquartile range measures the range of the middle 50% of the values only. It is calculated as the difference between the upper and lower quartile.

Interquartile range = upper quartile – lower quartile

= Q3 – Q1

Examples 1– find the interquartile range for 1, 2, 18, 6, 7, 9, 27, 15, 5, 19, 12.

Solution:

Arrange the numbers in ascending order

1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27

Find the median

Median = 9

(1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27)

Q1 as median in the lower half and Q3 as median in the upper half

Q1 = median in (1, 2, 5, 6, 7)

Q1 = 5

Q3 = median in (12, 15, 18, 19, 27)

Q3 = 18

Interquartile range = 18 – 5 = 13

Example 2 – find the interquartile for the following data set: 3, 5, 7, 8, 9, 11, 15, 16, 20, 21.

Solution:

Arrange the numbers in ascending order

3, 5, 7, 8, 9, 11, 15, 16, 20, 21

Make a mark in the center of the data:

(3, 5, 7, 8, 9,) | (11, 15, 16, 20, 21)

Find the median

Q1 = 7

Q3 = 16

Interquartile range = 16 – 7 = 9

Example 3 - Find the interquartile for the following data set: 1, 3, 4, 5, 5, 6, 7, 11

Make a mark in the center of the data:

(1, 3, 4, 5,) (5, 6, 7, 11)

Find the median

Q1 = (3+4)/2 = 3.5

Q3 = (6+7)/2 = 6.5

Interquartile range = 6.5 – 3.5 = 3

Example 4 -

Find the interquartile range for odd sample size

63,64,64,70,72,76,77,81,81

Solution:

Make a mark in the center of the data:

(63,64,64,70,)72,(76,77,81,81)

Find the median

Q1 = (64+64)/2 = 64

Q3 = (77+81)/2 = 79

Interquartile range = 79 – 64 = 15

3.3 Quartile Deviation

Quartile deviation is the product of half of the difference between the upper and the lower quartiles.

QD = (Q3 - Q1) / 2

Coefficient of Quartile Deviation = (Q3 – Q1) / (Q3 + Q1)

Quartile deviation for ungrouped data

Examples 1

Day	Frequency
1	20
2	35
3	25
4	12
5	10
6	23
7	18
8	14
9	30
10	40

Solution:

Arrange the frequency data in ascending order

Day	Frequency
1	10
2	12
3	14
4	18
5	20
6	23
7	25
8	30
9	35
10	40

First quartile (Q1)

Qi= [i * (n + 1) /4] th observation

Q1= [1 * (10 + 1) /4] th observation

Q1 = 2.75 th observation

Thus, 2.75 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 12 and 14

First quartile (Q1) is calculated as

Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)

Q1 = 12 + 0.75 * (14 – 12) = 13.50

Third quartile (Q3)

Qi= [i * (n + 1) /4] th observation

Q3= [3 * (10 + 1) /4] th observation

Q3 = 8.25 th observation

So, 8.25 th observation lies between the 8th and 9th value in the ordered group, between frequency 30 and 35

Third quartile (Q3) is calculated as

Q3 = 8th observation +0.25 * (9th observation – 8th observation)

Q3 = 30 + 0.25 * (35 – 30) = 31.25

Now using the quartiles values Q1 and Q3, we will calculate the quartile deviation.

QD = (Q3 - Q1) / 2

QD = (31.25 – 13.50) / 2 = 8.875

Coefficient of Quartile Deviation = (Q3 – Q1) / (Q3 + Q1)

= (31.25 – 13.50) /(31.25 + 13.50) = 0.397

Example 2– Calculate quartile deviation from the following test scores:

Sl. N o	Test scores
1	17
2	17
3	26
4	27
5	30
6	30
7	31
8	37

Solution:

First quartile (Q1)

Qi= [i * (n + 1) /4] th observation

Q1= [1 * (8 + 1) /4] th observation

Q1 = 2.25 th observation

Thus, 2.25 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 17 and 26

First quartile (Q1) is calculated as

Q1 = 2nd observation +0.75 * (3rd observation - 2nd observation)

Q1 = 17 + 0.75 * (26 – 17) = 23.75

Third quartile (Q3)

Qi= [i * (n + 1) /4] th observation

Q3= [3 * (8 + 1) /4] th observation

Q3 = 6.75 th observation

So, 6.75 th observation lies between the 6th and 7th value in the ordered group, between frequency 30 and 31

Third quartile (Q3) is calculated as

Q3 = 6th observation +0.25 * (7th observation – 6th observation)

Q3 = 30 + 0.25 * (31 – 30) = 30.25

Now using the quartiles values Q1 and Q3, we will calculate the quartile deviation.

QD = (Q3 - Q1) / 2

QD = (30.25 – 23.75) / 2 = 3.25

Quartile deviation for grouped data

Where,

l = lower boundary of quartile group

h = width of quartile group

f = frequency of quartile group

N = total number of observations

C= cumulative frequency preceding quartile group

Example 3

Age in years	40 -44	45 – 49	50 – 54	55 - 59	60 – 64	65 - 69
Employees	5	8	11	10	9	7

Solutions:

In the case of Frequency Distribution, Quartiles can be calculated by using the formula:

Class interval	F	Class boundaries	CF
40 -44	5	39.5 – 44.5	5
45 – 49	8	44.5 – 49.5	13
50 – 54	11	49.5 – 54.5	24
55 – 59	10	54.5 – 59.5	34
60 – 64	9	59.5 – 64.5	43
65 – 69	7	64.5 – 69.5	50
Total	50

First quartile (Q1)

Qi= [i * (n ) /4] th observation

Q1 = [1*(50)/4]th observation

Q1 = 12.50th observation

So, 12.50th value is in the interval 44.5 – 49.5

Group of Q1 = 44.5 – 49.5

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q1 = (44.5 + ( 5/ 8)* (1* (50/4) – 5)

Q1 = 49.19

Third quartile (Q3)

Qi= [i * (n) /4] th observation

Q3= [3 * (50) /4] th observation

Q3 = 37.5th observation

So, 37.5th value is in the interval 59.5 – 64.5

Group of Q3 = 59.5 – 64.5

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q3 = (59.5 + ( 5/ 9)* (3* (50/4) – 34)

Q3 = 61.44

QD = (Q3 - Q1) / 2

QD = (61.44 – 49.19) / 2 = 6.13

Coefficient of Quartile Deviation = (Q3 – Q1) / (Q3 + Q1)

= (61.44 – 49.19) /(61.44 + 49.19) = 0.11

Example 4 – Computation of quartile deviation for grouped test scores

Class	Frequency
9.3-9.7	22
9.8-10.2	55
10.3-10.7	12
10.8-11.2	17
11.3-11.7	14
11.8-12.2	66
12.3-12.7	33
12.8-13.2	11

Solution:

Class	Frequency	Class boundaries	CF
9.3-9.7	2	9.25-9.75	2
9.8-10.2	5	9.75-10.25	2 + 5 = 7
10.3-10.7	12	10.25-10.75	7 + 12 = 19
10.8-11.2	17	10.75-11.25	19 + 17 = 36
11.3-11.7	14	11.25-11.75	36 + 14 = 50
11.8-12.2	6	11.75-12.25	50 + 6 = 56
12.3-12.7	3	12.25-12.75	56 + 3 = 59
12.8-13.2	1	12.75-13.25	59 + 1 = 60

First quartile (Q1)

Qi= [i * (n ) /4] th observation

Q1 = [1*(60)/4]th observation

Q1 = 15th observation

So, 15th value is in the interval 10.25-10.75

Group of Q1 = 10.25-10.75

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q1 = (10.25 + ( 0.5/ 12)* (1* (60/4) – 7)

Q1 = 10.58

Third quartile (Q3)

Qi= [i * (n) /4] th observation

Q3= [3 * (60) /4] th observation

Q3 = 45th observation

So, 45th value is in the interval 11.25-11.75

Group of Q3 = 11.25-11.75

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q3 = (11.25 + ( 0.5/ 14)* (3* (60/4) – 36)

Q3 = 11.57

QD = (Q3 - Q1) / 2

QD = (11.57 – 10.58) / 2 = 0.495

Example 5 – Calculate quartile deviation from the following data

CI	F
10 – 15	6
15 – 20	10
20 – 25	15
25 – 30	22
30 – 40	12
40 – 50	9
50 – 60	4
60 - 70	2

Solution:

CI	F	Cf
10 – 15	6	6
15 – 20	10	16
20 – 25	15	31
25 – 30	22	53
30 – 35	12	65
35 – 40	9	74
45 – 50	4	78
55 – 60	2	80

First quartile (Q1)

Qi= [i * (n ) /4] th observation

Q1 = [1*(80)/4]th observation

Q1 = 20th observation

So, 20th value is in the interval 20 - 25

Group of Q1 = 20 - 25

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q1 = (20 + ( 5/ 15)* (1* (80/4) – 16)

Q1 = 21.33

Third quartile (Q3)

Qi= [i * (n) /4] th observation

Q3= [3 * (80) /4] th observation

Q3 = 60th observation

So, 60th value is in the interval 30 - 35

Group of Q3 = 30 - 35

Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3

Q3 = (30 + ( 5/ 12)* (3* (80/4) – 53)

Q3 = 32.91

QD = (Q3 - Q1) / 2

QD = (32.91 – 21.33) / 2 = 5.79

Merits

It provide better result than range mode
It is not affected by extreme values

Demerits

It is completely dependent on central item
All items are not taken onto consideration

3.4 Mean Deviation

The average of the absolute values of deviation from the mean, median or mode is called mean deviation. This method removes shortcoming of range and QD.

Where, ∑ is total of;

X is the score, X is the mean, and N is the number of scores

D = Deviation of individual scores from mean

Example 1 –

Computation of mean deviation in ungrouped data

X = 55, 45, 39, 41, 40, 48, 42, 53, 41, 56

Solution:

X	(X – X )	Absolute deviation (signed ignored)
55	55 - 46 = 9	9
45	45 – 46 = -1	1
39	-7	7
41	-5	5
40	-6	6
48	2	2
42	-4	4
53	7	7
41	-5	5
56	10	10
∑X = 460		∑ X – X = 56

Mean = 460/10 = 46

MD = 56/10 = 5.6

Example 2- Peterdid a surveyed the number of pets owned by his classmates, with the following result. What is the mean deviation of the number of pets?

No. Of pets	Frequency
0	4
1	12
2	8
3	2
4	1
5	2
6	1

Solution:

X	F	Fx	(X – X )	f(X – X )
0	4	0	1.8	7.2
1	12	12	0.8	9.6
2	8	16	0.2	1.6
3	2	6	1.2	2.4
4	1	4	2.2	2.2
5	2	10	3.2	6.4
6	1	6	3.2	4.2
	30	54	4.2	33.6

Mean = 54/30 = 1.8

MD = 33.6/30 = 1.12

Computation of Mean deviation in grouped data

Example 3 -

Class interval	15 – 19	20 – 24	25 – 29	30 – 34	35 – 39	40 – 44	45 - 49
Frequency	1	4	6	9	5	3	2

Class Interval	F	X	FX	D	FD
15 – 19	1	17	17	15	15
20 – 24	4	22	88	10	40
25 – 29	6	27	162	5	30
30 - 34	9	32	288	0	0
35 - 39	5	37	185	5	25
40 - 44	3	42	126	10	30
45 - 49	2	47	94	15	30
	N = 30		∑fx = 960		= 170

Mean =960/30 = 32

MD = 170 / 30 = 5.667

Coefficient of mean deviation

Coefficient of mean deviation = (5.67/32)*100 = 17.71

Example 4 – Calculate mean deviation from the median

Class	5 -15	15 - 25	25 - 35	35 - 45	45 – 55
Frequency	5	9	7	3	8

Solution:

x	f	Cf	Mid-point x	x –median	F(x-m)
5 -15	5	5	10	17.42	87.1
15 -25	9	14	20	7.42	66.78
25 -35	7	21	30	2.58	18.06
35 -45	3	24	40	12.58	37.74
45- 55	8	32	50	22.58	180.64
	32				390.32

Since n/2 = 32/2 = 16, therefore the class is 25 – 35 is the median.

Median =

Median = 25+16-14 *10 = 27.42

MD from median is 390.32/32 = 12.91

Example 5 – calculate the mean deviation from continuous frequency distribution

Age group	15 - 25	25 - 35	35 - 45	45 - 55
No. Of people	25	54	34	20

Solution:

Age group (X)	Number of people (f)	Midpoint x	Fx	X -	f( x – x )
15 – 25	25	20	500	13.684	324.1
25 – 35	54	30	1620	3.684	198.936
35 – 45	34	40	1360	6.316	214.744
45 - 55	20	50	1000	16.316	352.32
	133				1090.1

Mean = 4480/133 = 33.684

MD = 1090.1/133 = 8.196

Merits

It is easy to calculate
It helps in making comparison
It is not affected by extreme items

Demerits

It ignores algebraic sign. And are not used for mathematical treatment
It is not reliable

3.5 Standard Deviation

Standard deviation is calculated as square root of average of squared deviations taken from actual mean. It is also called root mean square deviation. This measure suffers from less drawbacks and provides accurate results. It removes the drawbacks of ignoring algebraic sign. We square the deviation to make them positive.

Two ways of computing SD

Direct method

2. Shortcut method

http://cdn.yourarticlelibrary.com/wp-content/uploads/2016/10/clip_image0126_thumb.jpg

d = Deviation of the score from an assumed mean, say AM; i.e. d = (X – AM). AM is assumed mean

d2 = the square of the deviation.

∑d = the sum of the deviations.

∑d2 = the sum of the squared deviations.

N = No. Of the scores

Standard deviation in ungrouped data

Direct method

Example 1–

X = 12, 15, 10, 8, 11, 13, 18, 10, 14, 9

Mean = 120/10 = 12

Scores	d
12	12-12 = 0	0
15	15-12 = 3	9
10	10 -12 = -2	4
8	-4	16
11	-1	1
13	1	1
18	6	36
10	-2	4
14	2	4
9	-3	9
= 120	= 0	= 84

= 2.9

2. Shortcut method

Assumed mean (AM) = 11

Scores	D = (X- AM)
12	12-11 = 1	1
15	15-11 = 4	16
10	10 -11 = -1	1
8	-3	9
11	0	0
13	2	4
18	7	49
10	--1	1
14	3	9
9	-2	4
= 120	= 10	= 94

http://cdn.yourarticlelibrary.com/wp-content/uploads/2016/10/clip_image0166_thumb.jpg

SD from short cut method = 2.9

Example 2 –Ram did a survey of the number of pets owned by his classmates, with the following results

No. Of pets	Frequency
0	4
1	12
2	8
3	2
4	1
5	2
6	1

Solution:

x	f	Fx	X – x	(x – x ) 2	F (x – x ) 2
0	4	0	-1.8	3.24	12.96
1	12	12	-0.8	0.64	7.68
2	8	16	0.2	0.04	0.32
3	2	6	1.2	1.44	2.88
4	1	4	2.2	4.84	4.84
5	2	10	3.2	10.24	20.48
6	1	6	4.2	17.64	17.64
	30	54			66.80

Mean = 54/30 = 1.8

SD = √66.80/30 = 1.49

Standard deviation in grouped data

Direct method

Example 3 –

C.I.	0 - 2	3 - 5	6- 8	9-11	12-14	15 -17	18 - 20
F	1	3	5	7	6	5	3

Solution:

C.I	f	Mid-point x	Fx	d		Fd2
0-2	1	1	1	-10.1	102.01	102.01
3-5	3	4	12	-7.1	50.41	151.23
6-8	5	7	35	-4.1	16.81	84.05
9-11	7	10	70	-1.1	1.21	8.47
12-14	6	13	78	1.9	3.61	21.66
15-17	5	16	80	4.9	24.01	120.05
18-20	3	19	57	7.9	62.41	187.23
	30		333			674.70

Mean = 333/30 = 11.1

SD =

Shortcut method

C.I	f	Mid-point x	d(X-AM)	Fd	Fd2
0-2	1	1	-9	-9	81
3-5	3	4	-6	-18	108
6-8	5	7	-3	-15	45
9-11	7	10	0	0	0
12-14	6	13	3	18	54
15-17	5	16	6	30	180
18-20	3	19	9	27	243
	30			33	711

Assumed mean = 10

Step deviation method

C.I	f	Mid-point x	d	Fd	Fd2
0-2	1	1	-3	-3	9
3-5	3	4	-2	-6	12
6-8	5	7	-1	-5	5
9-11	7	10	0	0	0
12-14	6	13	1	6	6
15-17	5	16	2	10	20
18-20	3	19	3	9	27
	30			11	79

Here, d is calculated as (X –AM)/i, where i is length of class interval

d = (1 -10)/3 = -3 and so on

Coefficient of standard deviation

Coefficient of SD = (4.74/11.1)*100 = 42.70

Example 4 – calculate the standard deviation using the direct method

Class interval	Frequency
30 – 39	3
40 – 49	1
50 – 59	8
60 – 69	10
70 – 79	7
80 – 89	7
90 – 99	4

Solution:

Class interval	Frequency	Mid-point x	Fx	X – x	(x – x ) 2	F (x – x ) 2
30 – 39	3	34.5	103.5	-33.5	1122.25	3366.75
40 – 49	1	44.5	44.5	-23.5	552.25	552.25
50 – 59	8	54.5	436.0	-13.5	182.25	1458
60 – 69	10	64.5	645.0	-3.5	12.25	122.5
70 – 79	7	74.5	521.5	6.5	42.25	295.75
80 – 89	7	84.5	591.5	16.5	272.25	1905.75
90 – 99	4	94.5	378.0	26.5	702.25	2809
	40		2720			10510

Mean = 2720/40 = 68

SD = √10510/40 = 16.20

Example 5 - calculate the mean and standard deviation of hours spent watching television by the 220 students.

Hours	No. Of students
10 – 14	2
15 – 19	12
20 – 24	23
25 – 29	60
30 – 34	77
35 – 39	38
40 - 44	8

Solution:

Hours	No. Of students	x	Fx	X – x	(x – x ) 2	F (x – x ) 2
10 – 14	2	12	24	-17.82	317.49	634.98
15 – 19	12	17	204	-12.82	164.31	1971.67
20 – 24	23	22	506	-7.82	61.12	1405.85
25 – 29	60	27	1620	-2.82	7.94	476.53
30 – 34	77	32	2464	2.18	4.76	366.55
35 – 39	38	37	1406	7.18	51.58	1959.98
40 - 44	8	42	336	12.18	148.40	1187.17
	220		6560			8002.73

Mean = 6560/220 = 29.82

SD = √8002.73/220 = 6.03

Merits

It takes into account all the items and are used for future statistical analysis
It is suitable for making comparison

Demerits

It is difficult to compute

3.6 Coefficient of Variation

Standard Variation is an absolute measure of dispersion. When comparison between two series has to be made, coefficient of variation is used. Coefficient of variation is a statistical measure of the dispersion of data with respect to mean.

Where, σ = Standard Deviation

µ = Mean

Example 1: Calculate Coefficient of Variation from the following data given below:

Standard Deviation = 4.5 Mean= 12.

Solution: We know,

Coefficient of Variation = x 100%

= x 100%

= 37.5%

Example 2: Calculate Standard Deviation and Co-efficient of Variation.

Class interval	Frequency
30 – 39	3
40 – 49	1
50 – 59	8
60 – 69	10
70 – 79	7
80 – 89	7
90 – 99	4

Solution:

Class interval	Frequency	Mid-point x	Fx	X – x	(x – x ) 2	F (x – x ) 2
30 – 39	3	34.5	103.5	-33.5	1122.25	3366.75
40 – 49	1	44.5	44.5	-23.5	552.25	552.25
50 – 59	8	54.5	436.0	-13.5	182.25	1458
60 – 69	10	64.5	645.0	-3.5	12.25	122.5
70 – 79	7	74.5	521.5	6.5	42.25	295.75
80 – 89	7	84.5	591.5	16.5	272.25	1905.75
90 – 99	4	94.5	378.0	26.5	702.25	2809
	40		2720			10510

Mean = 2720/40 = 68

SD = √10510/40 = 16.20

Coefficient of Variation = x 100%

= x 100%

= 23.82 %

3.7 Coefficient of Skewness:

Skewness is a measure of symmetry or lack of symmetry in a distribution. A distribution is symmetric if it looks same both its left and right side. The skewness for normal distribution is Zero. Negative values for skewness indicate that the data are skewed left and positive values for skewness indicate that the data are skewed right. For small data sets, this measure is unreliable. The below diagram shows how a normal distribution curve looks like in different situation of skewness:

http://surferhelp.goldensoftware.com/Resources/image/skew.png

The formula for measuring Coefficient of Skewness as given by Karl Pearson is as under:

Where, = Karl Pearson’s Coefficient of Skewness

= Standard Deviation

The formula for measuring Co-efficient of Skewness as given by Bowley is as under:

Example 1: From the following data, calculate Karl Pearson’s Co-efficient of Skewness:

Mean = 16, Mode = 38, Standard Deviation = 5

Solution: We know that

= -4.4

Example 2 – Below are the data of hours spent watching television by the 220 students. CalculateKarl Pearson’s Co-efficient of Skewness.

Hours	No. Of students
10 – 14	2
15 – 19	12
20 – 24	23
25 – 29	60
30 – 34	77
35 – 39	38
40 - 44	8

Solution:

Hours	No. Of students	x	Fx	X – x	(x – x ) 2	F (x – x )2
10 – 14	2	12	24	-17.82	317.49	634.98
15 – 19	12	17	204	-12.82	164.31	1971.67
20 – 24	23	22	506	-7.82	61.12	1405.85
25 – 29	60	27	1620	-2.82	7.94	476.53
30 – 34	77	32	2464	2.18	4.76	366.55
35 – 39	38	37	1406	7.18	51.58	1959.98
40 - 44	8	42	336	12.18	148.40	1187.17
	220		6560			8002.73