NIOS Class 12 Economics Chapter 9 Measures of Dispersion

NIOS Class 12 Economics Chapter 9 Measures of Dispersion, Solutions to each chapter is provided in the list so that you can easily browse through different chapters NIOS Class 12 Economics Chapter 9 Measures of Dispersion and select need one. NIOS Class 12 Economics Chapter 9 Measures of Dispersion Question Answers Download PDF. NIOS Study Material of Class 12 Economics Notes Paper 318.

NIOS Class 12 Economics Chapter 9 Measures of Dispersion

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Economics Chapter 9 Measures of Dispersion, NIOS Senior Secondary Course Economics Solutions for All Chapters, You can practice these here.

Measures of Dispersion

Chapter: 9

Module – IV Statistical Tools

TEXT BOOK QUESTIONS WITH ANSWERS

INTEXT QUESTIONS 9.1.

Q.1. The difference between the largest and the smallest data values is the:

(a) Variance.

(b) Inter – quartile range.

(c) Range.

(d) Coefficient of variation.

Ans. (c) Range.

Q.2. A researcher has collected the following sample data. The mean of the sample is 5. 3, 5, 12, 3, 2. The range is:

(a) 1

(b) 2

(c) 10

(d) 12

Ans. (c) 10 Range = L-S = 12 – 2 = 10

INTEXT QUESTIONS 9.2.

Q.1. If the first quartile is 104 and Quartile deviation is 18. Find the third quartile.

Ans. Third quartile is 140.

INTEXT QUESTIONS 9.3.

Q.1. Which of the following is a measure of dispersion?

(a) Percentiles.

(b) Quartiles.

(c) Interquartile range.

(d) All of the above are measures of dispersion.

Ans. (c) Interquartile range.

Q.2. The interquartile range is:

(a) The 50th percentile.

(b) Another name for the standard deviation.

(c) The difference between the largest and smallest values.

(d) The difference between the third quartile and the first quartile.

Ans. (b) Another name for the standard deviation.

Q.3. Which of the following limitation of the range is overcome by the interquartile range?

(a) The sum of the range variances is zero.

(b) The range is difficult to compute.

(c) The range is influenced too much by extreme values.

(d) The range is negative.

Ans. (c) The range is influenced too much by extreme values.

Q.4. A researcher has collected the following sample data. The mean of the sample is 5.

3 5 12 3 2

The inter – quartile range is:

(a) 1

(b) 2

(c) 10

(d) 12

Ans. (d) 12.

INTEXT QUESTIONS 9.4.

Q.1. Calculated mean deviation and coefficient of mean deviation from median.

No. of tomatoes per plant	0	1	2	3	4	5	6	7	8	9	10
No. of plants	2	5	7	11	18	24	12	8	6	4	3

Ans. 

X	f	Cf	|D|	f|D|
0	2	2	5	10
1	5	7	4	20
2	7	14	3	21
3	11	25	2	22
4	18	43	1	18
5	24	67	0	0
6	12	79	1	12
7	8	87	2	16
8	6	93	3	18
9	4	97	4	16
10	3	100	5	15
			30	168

Median = Size of (N + 1/2)th items

= 100 + 1/2 = 101/2 = 55.5th items

M.D = Σf |D|/N = 168/100 = 1.68

Coefficient of M.D = MD/ Media = 1.68/5 = 0.336

Q.2. Calculate mean deviation from mean:

Class	3-4	4-5	5-6	6-7	7-8	8-9	9-10
Frequency	3	7	22	60	85	32	8

Ans. 

Class (X)	Frequency (f)	Mid point (M)	d’=M-6.5/31	fd’	|D|=M-7.5	f|D|
3-4	3	3.5	0.09	0.27	4	12
4-5	7	4.5	0.06	0.42	3	21
5-6	22	5.5	0.03	0.63	2	44
6-7	60	6.5	0	0	1	60
7-8	85	7.5	0.03	2.55	0	0
8-9	32	8.5	0.06	1.92	1	32
9-10	8	9.5	0.09	0.72	2	16
	217					185

X̅ = ΣX/N = 217/7 = 31 

A = 6.5

X̅ = A + Σfd’/Σf x C

= 6.5 + 6.96/217 x 31 = 6.5 + 215.75/217 = 6.5 + 0.99 = 7.49 ≈ 7.5

MD = Σf|D|/N = 185/217 = 0.85

INTEXT QUESTIONS 9.5.

Q.1. Sona, Karina, Omar, Mustafa and Anil obtained marks of 6, 7, 3, 7, 2 on a standardized test respectively. Find the standard deviation of their scores.

Ans. 

Students	Scores	X-X̅	(X-X̅)²
1	6	1	1
2	7	2	4
3	3	-2	4
4	7	2	4
5	2	-3	9
	25		22

X̅ = 25/5 = 5

Standard deviation (S.D) =

INTEXT QUESTIONS 9.6.

Q.1. The hourly wages of a sample of 130 system analysts are given below:

mean = = 60

range = 20

mode = 73

variance = 324

median = 74

The coefficient of variation equals:

(a) 0.30%

(b) 30%

(c) 5.4%

(d) 54%

Ans. (b) 30%

Q.2. The measure of dispersion that is influenced most by extreme values is:

(a) The standard deviation.

(b) The range.

(c) The interquartile range.

Ans. (b) The range.

Q.3. The descriptive measure of dispersion that is based on the concept of a deviation about the mean is:

(a) The range.

(b) The interquartile range.

(c) The absolute value of the range.

(d) The standard deviation.

Ans. (d) The standard deviation.

Q.4. The numerical value of the standard deviation can never be:

(a) Zero.

(b) Negative.

(c) One.

Ans. (b) Negative.

Q.5. A researcher has collected the following sample data. The mean of the sample is 5.

3 5 12 3 2

The Standard deviation is:

(a) 8.944

(b) 4.062

(c) 13.2

(d) 16.5

Ans. (b) S.d is 4.062

Or

The coefficient of variation is:

(a) 72.66%

(c) 264%

(b) 81.24%

(d) 330%

Ans. (b) 81.24%

TERMINAL EXERCISE

RANGE

Q.1. The following are the prices of shares A, B Co. Ltd. from Monday to Saturday:

Days	Mon.	Tues.	Wed.	Thur.	Fri.	Sat.
Prices (in ₹)	200	210	208	160	220	250

Calculate the range and its coefficient.

Ans. Range = L – S

= 250 – 160 = 90

Coefficient of Range = L – S / L + S = 250 – 160/ 250 + 160 = 90/410 = 0.21

Q.2. Find the range of given data:

108, 107, 105, 106, 107, 104, 103, 101, 104.

Ans. Range = L – S

= 108 – 101 = 7

Q.3. Find the value of range of frequency distribution:

Age (in yrs)	14	15	16	17	18	19	20
No. of students	1	2	2	2	6	4	0

Ans. Range = L – S = 20 – 14 = 6

Hence, Range = 6

Q.4. Calculate the range for the distribution given below:

Height (in cms)	150	151	152	154	159	160	165	166
No. of boys	2	2	9	15	18	10	4	1

Ans. Range = L – S = 166 – 150 = 16

Q.5. Find the range of the following:

Profit (in 1000 ₹)	0-10	10-20	20-30	30-40	40-50
No. of firms	0	6	0	7	15

Ans. Range = L – S

= 50 – 0 = 50

Q.6. Find the range of the following distribution:

Class Interval	10-20	20-30	30-40	40-50	50-60
Frequency	8	10	15	18	19

Ans. Range = L – S = 60 – 10 = 50

Q.7. Calculate the QD for a group of data,

241, 521, 421, 250, 300, 365, 840, 958

Ans. The data: 241, 250, 300, 365, 421, 521, 840, 958

Here, N = 8

Q₁ = Size of (N + 1/4)th item = Size of (8 + 1/4)th item

= Size of 2.25th item

= 250+ (0.25) x (300 – 250)

= 250 + 12.5 = 262.5

Q₃ = Size of 3(N + 1) / 4th item = Size of 3(8 + 1)/4 = 27/4th item

= 6.25 item

= 521 + 0.25 x (840 – 521)

= 521 + 79.75 = 600.75

Q.D = Q₃ – Q₁/2 = 600.75 – 162.5/2 = 219.125 

Q.8. From the following figures find the quartile deviation and its coefficient:

Height (cms)	150	151	152	153	154	155	156	157	158
No. of Students	15	20	32	35	33	22	20	12	10

Ans. 

Height (cms)	No. of students (f)	Cumulative frequency
150	15	15
151	20	15 + 20 = 35
152	32	35 + 32 = 67
153	35	67 + 35 = 102
154	33	102 + 33 = 135
155	22	135 + 22 = 157
156	20	157 + 20 = 177
157	12	177 + 12 = 189
158	10	189 + 10 = 199

Here, N = 199

Q₁ = Size of (N+1/4) th item

= 199 + 1/4 = 200/4 = 50th

It is in 152 cms. Q₃ = Size of 3(N+1)/4th item = 3 (199 + 1)/4 = 600/4 = 150th item

It is in 155 cms. Q.D = Q₃ – Q₁/2 = 155 – 152/2 = 3/2 = 1.5 cm

Coefficient of Q.D =  Q₃ – Q₁/ Q₃ + Q₁ = 155 – 152/ 155 + 152 = 3/307 = 0.009

Q.9. Using quartile deviation, state which of the two variables – A and B is more variable:

Ans. A.

Mid – point	Frequency	cf
15	15	15
20	33	48
25	56	104
30	103	207
35	40	247
40	32	279
45	10	289

Q₁ = N + 1/4 = 289 + 1/4 = 290/4 = 72.5

It lies in 25 

Q₃ = 3 (N + 1)/4 = 3 (289 + 1)/4 = 3 × 210/4 = 870/4 = 217.5

It lies in 40

Q.D = Q₃ – Q₁/2 = 40 – 25/2 = 15/2 = 7.5

B.

Mid – point	Frequency	cf
100	340	340
150	492	832
200	890	1724
250	1420	3144
300	620	3764
350	360	4124
400	187	4311
450	140	4451

Q₁ = N + 1/4 = 4451 + 1/4 = 4452/4 = 2226

It lies in 250

Q₃ = 3 (N +1)/4 = 3(4451 + 1)/4 = 3 x 4452/4 = 3339

It lies in 300

Q.D = Q₃ – Q₁/1 = 300 – 250/2 = 50/2 = 25

Thus, B is more variable

Q.10. Find the quartile deviation from the following table:

Size	4-8	8-12	12-16	16-20	20-24	24-28	28-32	32-36	36-40
Frequency	6	10	18	30	15	12	10	6	2

Ans. 

Size	Frequency	cf
4-8	6	6
8-12	10	16
12-16	18	34
16-20	30	64
20-24	15	79
24-28	12	91
28-32	10	101
32-36	6	107
36-40	2	109

Q₁ = N + 1/4 = 109 + 1/4 = 110/4 = 27.5

Q₁ lies in 12 – 16

Q₃ = 3(N + 1)/4 = 3(109 + 1)/4 = 3 x 110/4 = 330/4 = 82.5

Q₃ lies in 24 – 28

Q₁ = l₂ + l₂ – l₁/f (N/4 – cf) = 12 + 4/18 (109/4 – 34) = 10.5

Q₃ = l₂ + l₂ – l₁/f (3N/4 – cf) = 24 + 4/12 (327/4 – 91) = 20.91

Q.D = Q₃ – Q₁/2 = 20.91 – 10.5/2 = 5.2

Q.11. Calculate the coefficient of quartile deviation from the following data:

Class Interval	Frequency
10-15	4
15-20	12
20-25	16
25-30	22
30-40	10
40-50	8
50-60	6
60-70	4
	8

Ans. 

Class Interval	Frequency	Cumulative Frequency
10-15	4	4
15-20	12	4+12=16
20-25	16	16+16=32
25-30	22	32+22=54
30-40	10	54+10=64
40-50	8	64+8=72
50-60	6	72+6=78
60-70	4	78+4=82

Q₁ = N/4 = 82/4 = 20.5

Q₁ lies in 20 – 25

Q₃ = 3N/4 = 3 x 82/4 = 246/4 = 61.5

Q₁ = l₁ + l₂ – l₁/f (N/4 – cf)

= 20 + 5/16 (20.5 – 32) = 20 + 5/16 (11.5) = 16.4

Q₃ = l₁ + l₂ – l₁/f (3N/4 – cf) = 30 + 40 – 30/10 (61.5 – 64)

= 30 + (-2.5) = 30 – 2.5 = 28.75

Coeff. of Q.D = Q₃ – Q₁/Q₃ + Q₁ = 28.75 – 16.4/28.75 + 16.4 = 0.273

Q.12. Determine the standard deviation of the following students test results percentage

92%, 66%, 99%, 75%, 69%, 51%, 89%, 75%, 54%, 45%, 69%

Ans. 

Students	Marks	X – X̅ = X – 71	(X – X̅)²
1	45	-26	676
2	51	20	400
3	54	-17	289
4	66	-5	25
5	69	-2	4
6	69	-2	4
7	75	4	16
8	75	4	16
9	89	18	324
10	92	21	441
11	99	28	784
			Σ (X – X̅)² = 2979

X̅ = ΣX/N = 784/11 = 71.27

Q.13. Calculate the coefficient of variation for the following data set: The price (in ₹) of a stock over five trading days was 52, 58, 55, 57, 59.

Ans. 

Days	Prices	X – X̅	(X – X̅)²
1	52	-4	16
2	55	-1	1
3	57	1	1
4	58	2	4
5	59	3	9
			31

Q.14. The frequency table of the monthly salaries of 20 people is shown below:

Salary (in ₹)	3500	4000	4200	4300
Frequency	5	8	5	2

(a) Calculate the mean of the salaries of the 20 people.

(b) Calculate the standard deviation of the salaries of the 20 people.

Ans. 

Salary in (₹)	Frequency (f)	X – X̅	(X – X̅)²
3500	5	-795	632025
4000	8	-792	627264
4200	5	-795	632025
4300	2	-798	636804
			2528118

X̅ = ΣX/N = 16000/20 = 800

The mean of the salaries of the 20 people is 800.

(b) 

Q.15. The following table shows the grouped data in classes, for the height of 50 people.

Height (in cm) – classes	Frequency
120 ≤ 130	2
130 ≤ 140	5
140 ≤ 150	25
150 ≤ 160	10
160 ≤ 170	8

(a) Calculate the mean of the salaries of the 50 people.

(b) Calculate the standard deviation of the salaries of the 50 people.

Ans. 

(a) Mean (X̅) = Σfx / Σf

= 7420/50

= 148.4

(b) Standard Deviation

Q.16. The following is the frequency distribution for the speeds of a sample of automobiles travelling on an interstate highway:

Speed Miles per Hour	Frequency
50-54	2
55-59	4
60-64	5
65-69	10
70-74	9
75-79	5
	35

Calculate the mean and the standard deviation of speed.

Ans. 

Mean (X̅) = Σfm/Σf = 2325/35 = 66.42

Standard deviation (S.D)

Q.17. In 2012, the average age of workers in a company was 22 with a standard deviation of 3.96. In 2013, the average age was 24 with a standard deviation of 4.08. In which year do the ages show a more dispersed distribution? Show your complete work and support your answer.

Therefore the year 2012 shows a more dispersed distribution.

Ans. In 2012, Mean = 22, σx = 3.96

In 2013, Mean = 24, σx = 4.08

In Comparison Distribution Reason

Average Magnitude II > I

X̅₁₁ = 25 > X̅₁ = 22

Variation I > II

CV₁ = 3.96/22 x 100 = 18

CV₁₁ = 4.08/24 x 100 = 18 = 17

Q.18. The following is a frequency distribution for the ages of a sample of employees at a local company

Age (in years)	Frequency
30-39	2
40-49	3
50-59	7
60-69	5
70-79	1

(a) Determine the average age for the sample.

(b) Compute the standard deviation.

(c) Compute the coefficient of variation.

Ans. 

(a) Average age of the sample = Σfm/Σf = 981/18 = 54.5

(b) 

Q.19. The population change between 1990 and 2000 for several small cities are shown below:

City	A	B	C	D	E	F	G	H
Population Change	3083	1466	-461	1113	-11	395	3290	437

For the above sample, determine the following measures

(a) The Mean.

(b) The standard deviation.

(c) The median.

Ans. 

City	Population	X – X̅	(X – X̅)²
A	3083	1919	3682561
B	1466	302	91204
C	-461	703	494209
D	1113	-51	2601
E	-11	1153	1329409
F	395	-769	591361
G	3290	2126	4519876
H	437	727	528529
	9312		11239750

(a) X̅ = Σfx / Σf = 9312/8 = 1164

(b) 

(c) In ascending order, the changes will be: -11, -461, 395, 437, 1113, 1466, 3083, 3290

Median = N/2 = 4th item

Hence it is in 437

Median = 437

Some Other Important Questions For Examinations

Very Short Answer Type Questions