RBSE Solutions for Class 11 Maths Chapter 13 Measures of Dispersion

RBSE Solutions for Class 11 Maths Chapter 13 Measures of Dispersion

Rajasthan Board RBSE Class 11 Maths Chapter 13 Measures of Dispersion Ex 13.1

Question 1.
Write the formula of quartile deviation.
Solution:
Coefficient of quartile deviation

Here, Q₃ = upper quartile, Q₁ = Lower quartile.

Question 2.
For any variable series Q₁ = 61 and Q₃ = 121. Find the quartile deviation.
Solution:
Quartile deviation (Q.D.) = Q3−Q1/2
= 121−61/2
= 60/2 = 30.

Question 3.
Find the quartile deviation and coefficient of quartile deviation of the following data.
3, 8, 11, 13, 17, 19, 20, 22, 23, 27, 31
Solution:
Given data is in assending order.
Total number of terms N = 11

Question 4.
Find the range and coefficient of range of the following table –

x	4.5	5.5	6.5	7.5	8.5	9.5	10.5	11.5
f	4	5	6	3	2	1	3	5

Solution:
Lower limit (S) = 4.5
Upper limit (L)= 11.5
Range = L – S
= 11.5 – 4.5 = 7

Thus range is 7 and coefficient of range is 0.44

Question 5.
Find the inter quartile range and its coefficient for the following series –

X	1	3	5	7
f	10	15	3	2

Solution:
We prepare a table using these datas:

X	f	c.f
1	10	10
3	15	25
5	3	28
7	2	30

Here, N = 30

Question 6.
Find the coefficient of range of the following data –

Size	10-15	15-20	20-25	25-30
Frequency	2	4	6	8

Solution:
Lower limit (S) = 10
Upper limit (L) = 30

Thus Range coefficient is 0-5.

Question 7.
Find the decile range and percentile range on the basis of following data :

X	f
0-10	3
10-20	9
20-30	8
30-40	5
40-50	7
50-60	5
60-70	7
70-80	6

Solution:

Class (x)	Frequency (f)	Cumulative frequency
0-10	3	3
10-20	9	12
20-30	8	20
30-40	5	25
40-50	7	32
50-60	5	37
60-70	7	44
70-80	6	50
	N = 50

Calculation of percentile range-

Cumulative frequency exactly greater than 45 is 50, whose corresponding class is 70 – 80

Cumulative frequency exactly greater than 5 is 12, whose corresponding class is 10-20

∴ Percentile range = 71.66 – 12.22 = 59.44
Thus Percentile range = 59.44

Question 8.
Find the decile range and percentile range of the following frequency distribution.

Marks (x)	0-10	10-20	20-30	30-40	40-50
Frequency(f)	5	8	20	14	3

Solution:

Class (x)	Frequency (f)	Cumulative frequency
0-10	5	5
10-20	8	13
20-30	20	33
30-40	14	47
40-50	3	50
	N = 30

Calculation of Decile Range –

Cumulative frequency exactly greater than 45 is 47, whose corresponding class is 30 – 40
∴ l = 30, f = 14, F = 33, h = 10

Cumulative frequency exactly greater than 5 is 13, whose corresponding class is 10 – 20
∴ l = 10, f = 8, F = 5, h = 10

Cumulative frequency exactly greater than 45 is 47, whose corresponding class is 30 – 40
∴ l = 30, f = 14, F = 33, h = 10

Cumulative frequency exactly greater than 5 is 13, whose corresponding class is 10 – 20
∴ l = 10, f = 8, F = 5, h = 10

Rajasthan Board RBSE Class 11 Maths Chapter 13 Measures of Dispersion Ex 13.2

Find the mean deviation from mean for the data given in Q1 and Q2.

Question 1.
4, 7, 8, 9, 10, 12, 13,17
Solution:

Question 2.
28, 60, 38,30, 32, 45, 53, 36,44, 34
Solution:

Find the mean deviation from median for the data given in Q. 3 and Q.4.

Question 3.
13,10, 12,13, 15, 18, 17, 11,14,16, 12
Solution:
Arranging is ascending order, value of varaible will be
10, 11, 12, 12, 13, 13, 14, 15, 16, 17, 18
Here n = 11
? Median (M) = Mean of values of 5th and 6th term
M = 13+13/2 = 26/2 = 13

xi	| xi – m|
13	0
10	3
12	1
13	0
15	2
18	5
17	4
11	2
14	1
16	3
12	1
N = 11	∑ | xi – m | = 22

Mean deviation from median

Question 4.
26, 32, 35, 39, 41, 62, 36, 50, 43.
Solution:
Arranging in ascending order, value of variable will be
26, 32, 35, 36, 39, 41, 43, 50, 62
Here number of total terms
N = 9

Table for calcuation of Mean deviation

Find the mean deviation from mode for the data given in Q. 5. and Q. 6.

Question 5.
2, 4, 6, 4, 8, 6, 4, 10, 4, 8
Solution:
Maximum time (4 times) in given data, 4 ocurs
Now, Mode (z) = 4

xi	|xi – z |
2	2
4	0
6	2
4	0
8	4
6	2
4	0
10	6
4	0
8	4
N = 10	∑ | xi – z | = 20

Thus, Mean deviation from mode,

Question 6.
2.2, 2.5, 2.1, 2.5, 2.9, 2.8, 2.5, 2.3
Solution:
In given data 2-5 occurs maximum times (3 times)
Thus Mode (z) = 2.5

xi	2.2	2.5	2.1	2.5	2.9	2.8	2.5	2.3	N = 8
| xi – z |	0.3	0	0.4	0	0.4	0.3	0	0.2	∑ |xi – z| = 1-6

Thus Mean deviation (δz) about mode

Find the mean deviation from mean for the data given in Q. 7 to Q. 8.

Question 7.

xi	5	10	15	20	25
fi	7	4	6	3	5

Solution:
Calculation table for mean deviation about mean

Question 8.

xi	20	40	60	80	100
fi	2	12	14	8	4

Solution:
Calculation table for mean deviation from mean.

Find the mean deviation from median for the data given in Q. 9. and Q. 10.

Question 9.

xi	5	7	9	10	12	15
fi	8	6	2	2	2	6

Solution:

Question 10.

xi	10	16	22	25	30
fi	3	5	6	7	8

Solution:

Find the mean deviation from mode for data given in Q. 11 to Q. 12.

Question 11.

xi	3	4	5	6	7	8
fi	2	4	6	3	2	1

Solution:

Question 12.

xi	10	20	30	40	50	60	70	80
fi	2	8	16	26	20	16	7	5

Solution:

Find the mean deviation from mean for data given in Q. 13. and Q. 14.

Question 13.

Income(daily)	Number
0-10	4
10-20	8
20-30	9
30-40	10
40-50	7
50-60	5
60-70	4
70-80	3

Solution:

Question 14.

Height (cm)	95-105	105-115	115-125	125-135	135-145	145-155
Number	9	13	26	30	12	10

Solution:

Find the mean deviation from median for data given in Q. 15. and Q. 16.

Question 15.

Marks	Number
10-20	3
20-30	4
30-40	7
40-50	8
50-60	2
60-70	1

Solution:

Question 16.

Age	Number
16-20	5
21-25	6
26-30	12
31-35	14
36-40	26
41-45	12
46-50	16
51-55	9

Solution:

Find the mean deviation from mode for data given in Q. 17 and Q. 18.

Question 17.

Class	20-30	30-40	40-50	50-60	60-70
Number	8	24	42	20	6

Solution:

Question 18.

Height (Inches)	52-55	55-58	58-61	61-64
No. of Students	10	20	35	10

Solution:

Rajasthan Board RBSE Class 11 Maths Chapter 13 Measures of Dispersion Ex 13.3

Find mean and variance for the data of Q. 1. and Q.2

Question 1.

xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3

Solution:

Question 2.

xi	82	83	87	88	92	94	99
fi	3	2	3	2	6	3	3

Solution:

Question 3.
Find mean and standard deviation by shortcut method.

xi	70	71	72	73	74	75	76	77	78
fi	2	1	12	29	25	12	10	4	5

Solution:
Let Assumed mean A = 73
Calculation Table

In question 4 and 5, find mean and variable for frequency distribution.

Question 4.

Class	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency	2	3	5	10	3	5	2

Solution:
Calculation table

Question 5.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Solution:
Calculation table

Question 6.
Find mean, variable and standard deviation by short-cut method.

Solution:
Table to find mean, variance, standard deviation :

Question 7.
Find the mean deviation of diameter of circles in given table.

Diameter (mm)	43-46	47-50	51-54	55-58	59-62
Number of circles	15	17	21	22	25

Solution:
Here groups not continuous. To make it continuous we will subtract 0-5 from lower limit of group and add 0-5 in upper limit.
Calculation table for mean deviation and mean diameter

Thus standard deviation of given circles = 5.55 mm

Question 8.
Calculate standard deviation, its coefficient and variable coefficient of given data.

Class	0-20	20-40	40-60	60-80	80-100
Frequency	2	5	15	7	1

Solution:
Calculate Table

Question 9.
Find the mean deviation from imaginary mean 35 for following distribution.
35, 25, 33, 50, 37, 35, 33, 37, 30
Solution:

Question 10.
Find mean deviation from mean, median, and mode and coefficient of mean deviation from the following series –

Monthly Wages (in Rs.)	Number of tenants
Less than 10	3
Less than 20	8
Less than 30	16
Less than 40	26
Less than 50	37
Less than 60	50
Less than 70	56
Less than 80	60

Solution:
Cumulative frequency – 1

Rajasthan Board RBSE Class 11 Maths Chapter 13 Measures of Dispersion Miscellaneous Exercise

Question 1.
Marks obtain by five students in math are 20, 25, 15, 35 and 30 its range will be :
(A) 15
(B) 20
(C) 25
(D) 30
Solution:
Range = Maximum value – minimum value = 35 – 15 = 20
Thus (B) is correct.

Question 2.
Formula of inter quartile range is –
(A) Q₃ + Q₁
(B) Q₃ – Q₁
(C) Q₃ – Q₂
(D) Q₃ – Q₄
Solution:
Thus (B) is correct.

Question 3.
If maximum cost of any times is ₹ 500 and minimum ₹ 75, then coefficient of range will be –
(A) 0.739
(B) 0.937
(C) 7.39
(D) 73.9
Solution:
Range coefficient

Thus (A) is correct.

Question 4.
Coefficient of range of variable series 10, 20, 30, 40, 50, 60 is –
(A) 3/2
(B) 5/6
(C) 7/5
(D) 5/7
Solution:
Coefficient of Range

Thus (D) is correct.

Question 5.
Mean deviation is lowest from –
(A) Mean
(B) Median
(C) Mode
(D) Origin
Solution:
(B) is correct.

Question 6.
Marks obtained by four students are 25, 35, 45 and 55 their mean deviation is –
(A) 10
(B) 1
(C) 0
(D) 40
Solution:
Mean of 25, 35, 45, 55

Question 7.
Mean deviation about median for distribution 2, 4, 5, 3, 8, 7, 8 is –
(A) 13/7
(B) 1/2
(C) 11/7
(D) 2
Solution:
Arrange in ascending order, obtained table is:

Question 8.
Mean of any variable series x¯¯¯ = 773 and mean deviation 64.4, then coefficient of mean deviation is –
(A) 0.065
(B) 12.003
(C) 0.083
(D) 0.073
Solution:
Mean deviation coefficient

Question 9.
Standard deviation of data 6,10,4,7,4,5 is –

Solution:

Question 10.
If standard deviation of marks obtained by students of a class is 1.4, then variable of distribution will be –
(A) 1.2
(B) 0.38
(C) 1.96
(D) 1.4
Solution:
Standard deviation
(σ) = 1.4
Variation (σ2)= (1.4)2 = 1.96
Thus (C) is correct.

Question 11.
If variance

then value of k is –
(A) 10
(B) 20
(C) 30
(D) 60
Solution:

Thus (C) is correct.

Question 12.
If coefficient of variance of a series is 30% and standard deviation is 15, then its means is –
(A) 0.5
(B) 5
(C) 2
(D) 50
Solution:

Thus (D) is correct.

Question 13.
In a series ∑x² = 100, n = 5 and ∑x = 20, then standard deviation is –
(A) 16
(B) 2
(C) 4
(D) 8
Solution:
Standard deviation

Thus (B) is correct.

Question 14.
Temperature of 7 days in a city are given in centigrade 18, 12, 6, -7, -12, 5, -4. Then range in centrigrade will be –
(A) 6
(B) 30
(C) 22
(D) 14
Solution:
Range = Max. Value – Min. Value
= 18 – (- 12) = 18 + 12 = 30
Thus (B) is correct.

Question 15.
If N = 10, ∑x = 120 and σ_x = 60, then variation coefficient is –
(A) 5
(B) 50
(C) 500
(D) 0.5
Solution:
Variation coefficient

Thus (C) is correct.

Question 16.
Algebraic sum of deviations from mean is :
(A) Negative
(B) Positive
(C) Different in each
(D) Zero
Solution:
(C) is correct.

Question 17.
If x¯¯¯ = 6, ∑x = 60 and ∑x² = 1000 then value σ is :
(A) 6
(B) 8
(C) 64
(D) 10
Solution:

Thus (B) is correct.

Question 18.
Coefficient of Range can be defined as –

Solution:
Thus (C) is correct.

Question 19.
If value of all the terms of a series are same, then find value of dispersion.
Solution:
Value of dispersion is zero because dispersion (deviation) cannot be find for same values.

Question 20.
Find the formula to find standard deviation in individual series.
Solution:

Question 21.
Standard deviatio of any distribution is 20.5 and arithmetic mean is 60, then find the coefficient of standard deviation.
Solution:

Question 22.
From the following distribution find inter quartile range, coefficients of inter quartile range, Quartile deviation and its coefficient.

More than digit	0	15	30	45	60	75	90	105
No. of students	150	140	100	80	70	30	14	0

Solution:
We prepare a table using these data:

Question 23.
Find the mean and standard deviation of the following frequency distribution by step deviation method.

Class	1-5	6-10	11-15	16-20	21-25	26-30	31-35
Frequency	5	7	18	25	20	4	1

Solution:

Question 24.
Find the mean deviation about mode and its coefficient from the following data –

Central size	6	7	8	9	10	11	12
Frequency	3	6	9	13	8	5	4

Solution:

Here Maximum frequency is 13 and its corresponding value is 9.
∴ Mode (z) = 9

Question 25.
Find the dispersion of the following data centred size –

Central size	32-38	38-44	44-50	50-56	56-62	62-68
No. of students	3	6	9	13	8	5

Solution:

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