Unit -III
Unit -III
Measures of Dispersion
1. What do you understand by Range? What are its merits and demerits ?
Answer - 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
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. What is the range for the following
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
3. 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
4. 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
5. 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
6. 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
7. What do you understand by Quartile Deviation ? What are its merits and demerits ?
Answer - 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
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
8. Calculate the following-
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
9. 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
10. Calculate –
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
11. 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
12. 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
13. What is Mean Absolute Deviation? What are its merits and demerits?
Answer - 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.
OR
= 
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
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
14. Computation of mean deviation in ungrouped data
X = 55, 45, 39, 41, 40, 48, 42, 53, 41, 56
Solution:
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 |
|
|
Mean = 460/10 = 46
MD = 56/10 = 5.6
15. Peter 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 |
|
|
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
16. Computation of Mean deviation in grouped data
Class interval | 15 – 19 | 20 – 24 | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 - 49 |
Frequency | 1 | 4 | 6 | 9 | 5 | 3 | 2 |
Solution -
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 |
|  |
Mean =960/30 = 32
MD = 170 / 30 = 5.667
Coefficient of mean deviation
Coefficient of mean deviation = (5.67/32)*100 = 17.71
17. 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
7
MD from median is 390.32/32 = 12.91
18. 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 |
 |
|
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
19. Explain Standard deviation and state its merits and demerits.
Answer - 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
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
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
20. Standard deviation in ungrouped data
- Direct method
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 |
 |  |  |
= 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 |
 |  |  |
SD from short cut method = 2.9
21. 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 |
|
|
|
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
22. Standard deviation in grouped data
Direct method
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
23. 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 |
|
|
|
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
24. 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 |
|
|
|
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
25. Briefly explain the measures of relative dispersion.
Answer - Measures of relative 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.
The following are the different measure of relative dispersion.
- Coefficient of range:
2. Coefficient of Quartile Deviation = (Q3 – Q1) / (Q3 + Q1)
3. Coefficient of mean deviation
4. 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
Unit -III
Measures of Dispersion
1. What do you understand by Range? What are its merits and demerits ?
Answer - 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
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. What is the range for the following
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
3. 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
4. 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
5. 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
6. 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
7. What do you understand by Quartile Deviation ? What are its merits and demerits ?
Answer - 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
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
8. Calculate the following-
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
9. 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
10. Calculate –
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
11. 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
12. 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
13. What is Mean Absolute Deviation? What are its merits and demerits?
Answer - 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.
OR
= 
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
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
14. Computation of mean deviation in ungrouped data
X = 55, 45, 39, 41, 40, 48, 42, 53, 41, 56
Solution:
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 |
|
|
Mean = 460/10 = 46
MD = 56/10 = 5.6
15. Peter 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 |
|
|
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
16. Computation of Mean deviation in grouped data
Class interval | 15 – 19 | 20 – 24 | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 - 49 |
Frequency | 1 | 4 | 6 | 9 | 5 | 3 | 2 |
Solution -
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 |
|  |
Mean =960/30 = 32
MD = 170 / 30 = 5.667
Coefficient of mean deviation
Coefficient of mean deviation = (5.67/32)*100 = 17.71
17. 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
7
MD from median is 390.32/32 = 12.91
18. 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 |
 |
|
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
19. Explain Standard deviation and state its merits and demerits.
Answer - 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
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
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
20. Standard deviation in ungrouped data
- Direct method
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 |
 |  |  |
= 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 |
 |  |  |
SD from short cut method = 2.9
21. 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 |
|
|
|
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
22. Standard deviation in grouped data
Direct method
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
23. 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 |
|
|
|
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
24. 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 |
|
|
|
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
25. Briefly explain the measures of relative dispersion.
Answer - Measures of relative 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.
The following are the different measure of relative dispersion.
- Coefficient of range:
2. Coefficient of Quartile Deviation = (Q3 – Q1) / (Q3 + Q1)
3. Coefficient of mean deviation
4. 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
