OP Malhotra Class 11 Maths Solutions Chapter 28 Statistics Ex 28(b)

Effective ISC Class 11 Maths Solutions S Chand Chapter 28 Statistics Ex 28(b) can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 28 Statistics Ex 28(b)

Question 1.
Find the median of the following sets of data:
(i) 2, 3, 5, 7, 9
(ii) 4, 8, 12, 16, 20, 23, 28, 32
(iii) 60, 33, 63, 61, 44, 48, 51
(iv) 13, 22, 25, 8, 11, 19, 17, 31, 16, 10
Solution:
(i) arranging the given data in ascending order; we have 2, 3, 5, 7, 9
Here, no. of observations = n = 5 (odd)
∴ M_d = size of \(\left(\frac{n+1}{2}\right)\)th observation = size of \(\left(\frac{5+1}{2}\right)\)th term = size of 3rd term = 5

(ii) The given data is already in ascending order.
Here no. of observations = n = 8 (even)
∴ Median = \(\frac{\left(\frac{n}{2}\right) \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }}{2}\) = \(\frac{4 \text { th term }+5 \text { th term }}{2}\) = \(\frac{16+20}{2}\) = 18

(iii) Arranging the given data in ascending order; we have, 33, 44, 48, 51, 60, 61, 63
Here no. of observations = n = 7 (odd)
∴ Median = size of \(\left(\frac{n+1}{2}\right)\)th term = size of \(\left(\frac{7+1}{2}\right)\)th term = size of 4th term = 51

(iv) Arranging the given data in ascending order, we get, 8, 10, 11, 13, 16, 17, 19, 22, 25, 31
Here no. of observations = n = 10 (even)
∴ Median = size of =

Question 2.
Find the median of the following data:
41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?
Solution:
Arranging the given data in ascending order; we get
41, 43, 57, 58, 61, 71, 92, 99, 127,
Here no. of observations = n = 9 (odd)
∴ Median = size of \(\left(\frac{n+1}{2}\right)\)th term = size of \(\left(\frac{9+1}{2}\right)\)th term = size of 5th term = 61
When observation 58 is replaced by 85,
Then data arranged in ascending order is an under : 41, 43, 57, 58, 61, 71, 92, 99, 127,
Then median = size of \(\left(\frac{9+1}{2}\right)\)th term = size of 5th term = 71

Question 3.
Calculate the median of the following incomes.

Wages in ₹	20	21	22	23	24	25	26	27	28
No. of workers	8	10	11	16	20	25	15	9	6

Solution:
The table of values is given as under:

Wages in (₹)	No. of workers	c.f.	Wages in ₹	No. of workers	c.f
20	8	8	25	25	90
21	10	18	26	15	105
22	11	29	27	9	114
23	16	45	28	6	120
24	20	65

Here M_d = \(\left(\frac{n+1}{2}\right)\)th item = size of \(\left(\frac{120+1}{2}\right)\)th item = size of 60.5th term = 24
Thus median income of workers be ₹ 24.

Question 4.
Compute the median of the following distributions:

f	2	3	4	5	6	7
x	3	8	10	12	16	14

Solution:
The table of values is given as under:

x	3	8	10	12	16	14
f	2	3	4	5	6	7	Σf = N = 27
c.f	2	5	9	14	20	27

Here Σf = = 27 ∴ M_d = size of \(\left(\frac{N+1}{2}\right)\)th item = size of 14th item = 12

Question 5.
Marks obtained by 38 students are given below. Calculate the median marks:

Marks	20	90	50	70	60	95
No. of students	4	5	8	10	6	5

Solution:
The table of values is given as under:

Marks	Frequency	c.f
20	4	4
90	5	9
50	8	17
70	10	27
60	6	33
95	5	38
	Σf = N = 38

Here M_d = size of \(\left(\frac{N+1}{2}\right)\)th item = size of \(\left(\frac{38+1}{2}\right)\)th item = size of 19.5th item = 70 [since value 19.5 lies under c.f column in 27]

Question 6.
Find whether the following statements are true or false:
(i) The median of a frequency distribution is the most commonly occurring value.
(ii) The median of a discrete ungrouped frequency distribution containing a number of items is the value of the middle item, the data being arranged in ascending or descending order.
Solution:
(i) False, Mode of a frequency distribution is the most commonly occurring value.
(ii) True, M_d = size of \(\left(\frac{n+1}{2}\right)\)th item

Question 7.
In a school examination it is decided that exactly half the pupils will pass. Name the measure of central tendency that is used.
Solution:
Since by definition of median, equal number of items above and below the median. Thus median’s exactly the middle point of array of given numbers. Since it is given that exactly half the pupils will pass. Thus the measure of central tendency that is used be median.

Question 8.
(1, 2, 3, 6, 8) is a set of five positive integers whose mean is 4 and median is 3. Write down two other sets of five positive integers, each having the same mean and median as this set.
Solution:
For given five observations, mean is 4 and median is 3.
Let the observations 1, 2, 3, 5, 9
∴ Mean = \(\frac{1+2+3+5+9}{5}\) = 4 and M_d = size of \(\left(\frac{5+1}{2}\right)\)th item = size of 3rd item = 3
Let the observations are 1, 2, 3, 4, 10
Here Mean = \(\frac{1+2+3+4+10}{5}\) = 4 and M_d = 3

Question 9.
Find the median from the following distribution:

Marks	10-25	25-40	40-55	55-70	70-85	85-100
Frequency	6	20	44	26	3	1

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 100 }{ 2 }\) 50 (even) which lies in median class 40-55
∴ l = 40; f = 44; C = 26 and i = 15

Here the required median marks are 48.18 marks.

Question 10.

Class Interval	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Frequency	7	10	13	26	35	22	11	6

Solution:
Here class intervals are discontinuous, we make it continuous by using
adjustment factor = \(\frac { 16-15 }{ 2 }\) = 0.5 which is to be subtracting from lower limit and adding to upper limit of each class interval.
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 130 }{ 2 }\) = 65 which lies in median class 30.5 – 35.5
Thus, l = 30. 5; f = 35; c = 56; i = 5

Question 11.

Value	Frequency	Value	Frequency
Less than 10	4	Less than 50	96
Less than 20	16	Less than 60	112
Less than 30	40	Less than 70	120
Less than 40	76	Less than 80	125

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 125 }{ 2 }\) = 62.5 which lies in median class 30 – 40.
Thus, l = 30; f = 36; c = 40; i = 10

Question 12.

Size	Frequency	Size	Frequency
More than 50	0	More than 20	123
More than 40	40	More than 10	165
More than 30	98

Solution:
The table of values is given as under:

Here \(\frac { N }{ 2 }\) = \(\frac { 165 }{ 2 }\) = 82.5 which lies in median class 30 – 40.
Thus, l = 30; f = 58; c = 67; i = 10

Question 13.
Find the median and first and third quartile for the following data:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Solution:
Given data is already in ascending order: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
Here number of observations = n = 11 (odd)
∴ M_d = size of \(\left(\frac{n+1}{2}\right)\)th item = size of \(\left(\frac{11+1}{2}\right)\)th item = size of 6th item = 12
Q₁ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{11+1}{4}\right)\)th item = size of 3rd item = 6
and Q₃ = size of \(3\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{11+1}{4}\right)\)th item = size of 9th item = 18

Question 14.
Compute Q₁, Q₃, D₃, D₆ and D₈ for the following data:
14, 7, 13, 12, 13, 17, 8, 10, 6, 15, 18, 21, 20
Solution:
Arranging the given data in ascending order: 6, 7, 8, 10, 12, 13, 14, 15, 17, 18, 20, 21
Here no. of observations = n = 13 (odd)
Q₁ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(\left(\frac{13+1}{4}\right)\)th item = size of 3,5th item = 3rd item + 0.5 (4th item – 3rd item) = 8 + 0.5 (10 – 8) = 8 + 1 = 9

Q₃ = size of \(\left(\frac{n+1}{4}\right)\)th item = size of \(3\left(\frac{13+1}{4}\right)\)th item = size of 10.5th item = 10th item + 0.5 (11th item – 10th item) = 17 + 0.5 (18 – 17) = 17.5

D₃ = size of \(\left(\frac{n+1}{10}\right)\)th item = size of \(3\left(\frac{13+1}{10}\right)\)th item = size of 4.2th item = 4th item + 0.2 (5th item – 4th item) = 10 + 0.2 (12 – 10) = 10 + 0.4 = 10.4

D₆ = size of \(\left(\frac{n+1}{10}\right)\)th item = size of \(6\left(\frac{13+1}{10}\right)\)th item = size of 8.4th item = 8th item + 0.4 (9th item – 8th item) = 14 + 0.4 (15 – 14) = 14.4

D₈ = size of \( 8\left(\frac{n+1}{10}\right)\)th item = size of 11.2th item = 11 th item + 0.2 (12th – 11th) = 18 + 0.2 (20 – 18) = 18.4

Question 15.
Following are the scores of 12 students in a class test of 30 marks:
18, 20, 9, 15, 21, 26, 14, 13, 27, 22, 16, 28 Find D₇ and P₃₃.
Solution:
Arranging the data in ascending order : 9, 13, 14, 15, 16, 18, 20, 21, 22, 26, 27, 28
Here no. of observations = n = 12
∴ D₇ = size of 7\(\left(\frac{n+1}{10}\right)\)th item = size of 7 \(\left(\frac{12+1}{10}\right)\)th item = size of 9.1th item = 9th item + 0.1 (10th – 9th item) = 22 + 0.1 (26 – 22) = 22.4
P₃₃ = size of 33 \(\left(\frac{n+1}{100}\right)\)th item = size of 33 \(\left(\frac{12+1}{100}\right)\)th item = size of 4.29 th item = 4th item + 0.29 (5th item – 4th item) = 15 + 0.29 (16 – 15) = 15.29

Question 16.
Compute Q₁, Q₃, Q₆ and P₄₅ for the following data:

x	18	19	20	21	22	23	24	25	26	27
f	15	18	25	27	40	25	19	16	8	7

Solution:
We construct the following table:

xi	18	19	20	21	22	23	24	25	26	27
fi	15	18	25	27	40	25	19	16	8	7	Σfi = 200
Cumulative freq.	15	33	58	85	125	150	169	185	193	200

Q₁ = \(\left(\frac{n+1}{4}\right)\)th value = \(\left(\frac{200+1}{4}\right)\)th value = 50.25th value
= 50th value + 0.25 (51th value – 50th value) = 20 + 0.25 (20 – 20) = 20 [both 51th and 50th values are same]

Q₃ = 3 \(\left(\frac{n+1}{4}\right)\)th value = 3 \(\left(\frac{200+1}{4}\right)\)th value = 150.75th value = 150th value + 0.75 (151th value – 150th value) = 23 + 0.75 (24 – 23) = 23.75

D₆ = 6 \( \left(\frac{n+1}{10}\right)\)th value = 6 \(\left(\frac{200+1}{10}\right)\)th value = 120.6th value = 120th value + 0.6 (121th value – 120th value) = 22 + 0.6(22 – 22) = 22

P₄₅ = 45 \(\left(\frac{n+1}{100}\right)\)th value = 45 \(\left(\frac{200+1}{100}\right)\)th value = 90.45th value = 90th value + 0.45 (91th value – 90th value) =22 + 0.45 (22 – 22) = 22 (since 91th and 90th values are same)

Question 17.
Following are the different sizes and number of shoes in a shoe shop. Calculate median, first quartile, third quartile, 6th decile and 80th percentile.

Size of shoes	No. of shoes	Size of shoes	No. of shoes
4.5	4	8	40
5	8	8.5	20
5.5	12	9	15
6	15	9.5	24
6.5	20	10	12
7	35	10.5	5
7.5	50	11	3

Solution:

∴ M_d = size of \(\left(\frac{\mathrm{N}+1}{2}\right)\)th item = size of \(\left(\frac{253+1}{2}\right)\)th item = size of 127 th item = 7.5
Q₁ = size of \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of \(\left(\frac{253+1}{4}\right)\)th item = size of 63.5 th item = 7.0
Q₃ = size of 3 \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of 3 \(\left(\frac{253+1}{4}\right)\)th item = size of 190.5 th item = 8.5
D₆ = size of 6 \(\left(\frac{N+1}{10}\right)\)th item = size of 6 \(\left(\frac{253+1}{10}\right)\)th item = size of 152.4 th item = 8
P₈₀ = size of 80 \(\left(\frac{\mathrm{N}+1}{100}\right)\)th item = size of 80\(\left(\frac{253+1}{100}\right)\)th item = 203.2th item = 9.0

Question 18.
Find out first quartile, third quartile and first decile.

Size of Item	f	Size of Item	f
0 – 10	2	40 – 50	34
10 – 20	18	50 – 60	20
20 – 30	30	60 – 70	6
30 – 40	45	70 – 80	3

Solution:
The table of values is given as under:

Question 19.
Calculate the median, 3rd decile and 20th percentile for the following data:

x	0-5	5-10	10-15	15-20	20-25
f	7	18	25	30	20

Solution:
The table of values is given as under:

Question 20.
Find the interquartile range, quartile deviation for the following data:

Age in years	20	30	40	50	60	70	80
No. of members	3	61	132	153	140	51	3

Solution:
The table of values is given as under:

For Q₁ : Q₁ = size of \(\left(\frac{N+1}{4}\right)\)th item = size of \(\left(\frac{543+1}{4}\right)\)th item = size of 136th item = 40

Q₃ = size of 3 \(\left(\frac{\mathrm{N}+1}{4}\right)\)th item = size of 3 \(\left(\frac{543+1}{4}\right)\)th item = size of 408 th item = 60
∴ Interquartile range = Q₃ – Q₁ = 60 – 40 = 20
∴ Q.D = \(\frac{Q_3-Q_1}{2}\) = \(\frac{60-40}{2}\) = 10

Question 21.
Find the interquartile range, semi-interquartile range, and coefficient of quartile deviation from the following frequency distribution.

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	60	45	120	25	90	80	120	60

Solution:
The table of values is given as under:

For Q₁ : \(\frac{N}{4}\) = \(\frac{600}{4}\) = 150, which lies in class interval 30 – 40
Here l = 30; f= 120; c = 105; i = 10
∴ Q₁ = l + \(\frac{\frac{N}{4}-C}{f}\) × i = 30 + \(\frac{150-105}{120}\) × 10 = 30 + 3.75 = 33.75

For Q₃ : \(\frac{3N}{4}\) = \(\frac{3 \times 600}{4}\) = 450 which lies in 70-80
Here l = 70; f= 120; c = 420; i = 10
Q₃ = l + \(\frac{\frac{3 N}{4}-C}{f}\) × i = 70 + \(\frac{450-420}{120}\) × 10 = 72.5
∴ Interquartile range = Q₃ – Q₁ = 72.5 – 33.75 = 38.75
and Semi-interquartile range = \(\frac{Q_3-Q_1}{2}\) = 19.375
and Coeff. of Q.D = \(\frac{Q_3-Q_1}{Q_3+Q_1}\) = \(\frac{72.5-33.75}{72.5+33.75}\) = \(\frac{38.75}{106.25}\) = 0.365