Selina Chapter 24 Measures of Central Tendency ICSE Solutions Class 10 Maths

ICSE Solutions for Selina Concise Chapter 24 Measure of Central Tendency Class 10 Maths

Exercise 24(A)

1. Find the mean of the following set of numbers:

(i) 6, 9, 11, 12 and 7

(ii) 11, 14, 23, 26, 10, 12, 18 and 6

Solution

(i) By definition, we know

Mean = ∑x/n

Here, n = 5

Thus,

Mean = (6 + 9 + 11 + 12 + 7)/ 5 = 45/5 = 9

(ii) By definition, we know

Mean = ∑x/n

Here, n = 8

Thus,

Mean = (11 + 14 + 23 + 26 + 10 + 12 + 18 + 6)/8

= 120/8

= 15

2. Marks obtained (in mathematics) by 9 student are given below:

60, 67, 52, 76, 50, 51, 74, 45 and 56

(a) find the arithmetic mean

(b) if marks of each student be increased by 4; what will be the new value of arithmetic mean.

Solution

(a) Mean = ∑x/n

Here, n = 9

Thus,

Mean = (60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56)/ 9 = 531/9 = 59

(b) If the marks of each student be increased by 4 then new arithmetic mean will be = 59 + 4 = 63

3. Find the mean of the natural numbers from 3 to 12.

Solution

The numbers between 3 to 12 are 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

Here n = 10

Mean = ∑x/n

= (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/ 10 = 75/10 = 7.5

4. (a) Find the mean of 7, 11, 6, 5, and 6

(b) If each number given in (a) is diminished by 2, find the new value of mean.

Solution

(a) Mean = ∑x/n , here n = 5

= (7 + 11 + 6 + 5 + 6)/ 5 = 35/5 = 7

(b) If 2 is subtracted from each number, then the mean will he changed as 7 – 2 = 5

5. If the mean of 6, 4, 7, ‘a’ and 10 is 8. Find the value of ‘a’

Solution

Given,

No. of terms (n) = 5

Mean = 8

Sum of all terms = 8×5 = 40 …(i)

But, sum of numbers = 6 + 4 + 7 + a + 10 = 27 + a …(ii)

On equating (i) and (ii), we get

27 + a = 40

Thus, a = 13

6. The mean of the number 6, ‘y’, 7, ‘x’ and 14 is 8. Express ‘y’ in terms of ‘x’.

Solution

Given,

No. of terms (n) = 5 and mean = 8

So, the sum of all terms = 5×8 = 40 …(i)

but sum of numbers = 6 + y + 7 + x + 14 = 27 + y + x …(ii)

On equating (i) and (ii), we get

27 + y + x = 40

⇒ x + y = 13

Hence, y = 13 – x

7. The ages of 40 students are given in the following table:

Age( in yrs)	12	13	14	15	16	17	18
Frequency	2	4	6	9	8	7	4

Find the arithmetic mean.

Solution

Age in yrs (xi)	Frequency (fi)	fixi
12	2	24
13	4	52
14	6	84
15	9	135
16	8	128
17	7	119
18	4	72
Total	40	614

Mean = ∑f_i x_i/∑f_i = 614/40 = 15.35

Exercise 24(B) 

1. The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

Age – Years	16 – 18	18 – 20	20 – 22	22- 24	24-26
No. of Students	2	7	21	17	3

Solution

Age in years (C.I.)	xi	Number of students (fi)	xifi
16 – 18	17	2	34
18 – 20	19	7	133
20 – 22	21	21	441
22 – 24	23	17	391
24 – 26	25	3	75
Total		50	1074

Mean = ∑f_i x_i/∑f_i = 1074/50 = 21.48

2. The following table gives the weekly wages of workers in a factory.

Weekly Wages (Rs)	No. of Workers
50-55	5
55-60	20
60-65	10
65-70	10
70-75	9
75-80	6
80-85	12
85-90	8

Calculate the mean by using:

(i) Direct Method

(ii) Short – Cut Method

Solution

(i) Direct Method

Weekly Wages (Rs)	Mid-Value (xi)	No. of Workers (fi)	fixi
50-55	52.5	5	262.5
55-60	57.5	20	1150.0
60-65	62.5	10	625.0
65-70	67.5	10	675.0
70-75	72.5	9	652.5
75-80	77.5	6	465.0
80-85	82.5	12	990.0
85-90	87.5	8	700.0
Total		80	5520.00

Mean = ∑f_i x_i/ ∑f_i = 5520/80 = 69

(ii) Short – cut method

Weekly wages (Rs)	No. of workers (fi)	Mid-value (xi)	A = 72.5 (di = x – A)	fidi
50-55	5	52.5	-20	-100
55-60	20	57.5	-15	-300
60-65	10	62.5	-10	-100
65-70	10	67.5	-5	-50
70-75	9	A = 72.5	0	0
75-80	6	77.5	5	30
80-85	12	82.5	10	120
85-90	8	87.5	15	120
Total	80			-280

Here, A = 72.5

3. The following are the marks obtained by 70 boys in a class test:

Marks	No. of boys
30 – 40	10
40 – 50	12
50 – 60	14
60 – 70	12
70 – 80	9
80 – 90	7
90 – 100	6

Calculate the mean by:

(i) Short – cut method

(ii) Step – deviation method

Solution

(i) Short – cut method

Marks	No. of boys (fi)	Mid-value xi	A = 65 (di = x – A)	fidi
30 – 40	10	35	-30	-300
40 – 50	12	45	-20	-240
50 – 60	14	55	-10	-140
60 – 70	12	A = 65	0	0
70 – 80	9	75	10	90
80 – 90	7	85	20	140
90 – 100	6	95	30	180
Total	70			-270

Here, A = 65

(ii) Step – deviation method

Marks	No. of boys (fi)	Mid-value xi	A = 65 ui = (xi – A)/h	fiui
30 – 40	10	35	-3	-30
40 – 50	12	45	-2	-24
50 – 60	14	55	-1	-14
60 – 70	12	A = 65	0	0
70 – 80	9	75	1	9
80 – 90	7	85	2	14
90 – 100	6	95	3	18
Total	70			-27

Here, A = 65 and h = 10

4. Find mean by step – deviation method:

C. I.	63-70	70-77	77-84	84-91	91-98	98-105	105-112
Freq	9	13	27	38	32	16	15

Solution

C. I.	Frequency (fi)	Mid-value xi	A = 87.50 ui = (xi – A)/h	fiui
63 – 70	9	66.50	-3	-27
70 – 77	13	73.50	-2	-26
77 – 84	27	80.50	-1	-27
84 – 91	38	A = 87.50	0	0
91 – 98	32	94.50	1	32
98 – 105	16	101.50	2	32
105 – 112	15	108.50	3	45
Total	150			29

Here, A = 87.50 and h = 7

5. The mean of the following frequency distribution is 21(1/7) . Find the value of ‘f’.

C. I.	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
frequency	8	22	31	f	2

Solution

Given,

C. I.	frequency	Mid-value (xi)	fixi
0-10	8	5	40
10-20	22	15	330
20-30	31	25	775
30-40	f	35	35f
40-50	2	45	90
Total	63 + f		1235 + 35f

Exercise 24(C) 

1. A student got the following marks in 9 questions of a question paper.

3, 5, 7, 3, 8, 0, 1, 4 and 6.

Find the median of these marks.

Solution

Arranging the given data in descending order:

8, 7, 6, 5, 4, 3, 3, 1, 0

Clearly, the middle term is 4 which is the 5^th term.

Hence, median = 4

2. The weights (in kg) of 10 students of a class are given below:

21, 28.5, 20.5, 24, 25.5, 22, 27.5, 28, 21 and 24.

Find the median of their weights.

Solution

Arranging the given data in descending order:

28.5, 28, 27.5, 25.5, 24, 24, 22, 21, 21, 20.5

It’s seen that,

The middle terms are 24 and 24, 5^th and 6^th terms

Thus,

Median = (24 + 24)/2 = 48/2 = 24

3. The marks obtained by 19 students of a class are given below:

27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35 and 28.

Find:

(i) median

(ii) lower quartile

(iii) upper quartile

(iv) interquartile range

Solution

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) The middle term is 10^th term i.e. 29

Hence, median = 29

(ii) Lower quartile

(iii) Upper quartile =

(iv) Interquartile range = q₃ – q₁ =35 – 26 = 9

4. From the following data, find:

(i) Median

(ii) Upper quartile

(iii) Inter-quartile range

25, 10, 40, 88, 45, 60, 77, 36, 18, 95, 56, 65, 7, 0, 38 and 83

Solution

Arranging the given data in ascending order, we have:

0, 7, 10, 18, 25, 36, 38, 40, 45, 56, 60, 65, 77, 83, 88, 95

(i) Median is the mean of 8^th and 9^th term

Thus, median = (40 + 45)/ 2 = 85/2 = 42.5

(ii) Upper quartile =

(iii) Interquartile range is given by,

q₁ = 16^th/4 term = 18; q₃ = 65

Interquartile range = q₃ – q₁

Thus, q₃ – q₁ = 65 – 18 = 47

5. The ages of 37 students in a class are given in the following table:

Age (in years)	11	12	13	14	15	16
Frequency	2	4	6	10	8	7

Find the median.

Solution

Age (in years)	Frequency	Cumulative Frequency
11	2	2
12	4	6
13	6	12
14	10	22
15	8	30
16	7	37

Number of terms (n) = 37

Median =

And, the 19^th term is 14

∴ the median = 14

Exercise 24(D)

1. Find the mode of the following data:

(i) 7, 9, 8, 7, 7, 6, 8, 10, 7 and 6

(ii) 9, 11, 8, 11, 16, 9, 11, 5, 3, 11, 17 and 8

Solution

(i) It’s seen that 7 occurs 4 times in the given data.

Hence, mode = 7

(ii) Mode = 11

As 11 occurs 4 times in the given data.

2. The following table shows the frequency distribution of heights of 50 boys:

Height (cm)	120	121	122	123	124
Frequency	5	8	18	10	9

Find the mode of heights.

Solution

Clearly, Mode is 122 cm because it has occurred the maximum number of times i.e. frequency is 18.

3. Find the mode of following data, using a histogram:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	12	20	9	4

Solution

Clearly, Mode is in 20-30, because in this class there are 20 frequencies.

4. The following table shows the expenditure of 60 boys on books. Find the mode of their expenditure:

Expenditure (Rs)	No. of students
20-25	4
25-30	7
30-35	23
35-40	18
40-45	6
45-50	2

Solution

Clearly, Mode is in 30-35 because it has the maximum frequency.

Exercise 24(E)

1. The following distribution represents the height of 160 students of a school.

Height (in cm)	No. of Students
140 – 145	12
145 – 150	20
150 – 155	30
155 – 160	38
160 – 165	24
165 – 170	16
170 – 175	12
175 – 180	8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine:

i. The median height.

ii. The interquartile range.

iii. The number of students whose height is above 172 cm.

Solution

Height (in cm)	No. of Students	Cumulative frequency
140 – 145	12	12
145 – 150	20	32
150 – 155	30	62
155 – 160	38	100
160 – 165	24	124
165 – 170	16	140
170 – 175	12	152
175 – 180	8	160
	N = 160

Now, let’s draw an ogive taking height of student along x-axis and cumulative frequency along y-axis.

(i) So, Median = 160/2 = 80^th term

Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.

(ii) As, the number of terms = 160

Lower quartile (Q₁) = (160/4) = 40^th term = 152

Upper quartile (Q₃) = (3 x 160/4) = 120^th term = 164

Inner Quartile range = Q₃ – Q₁

= 164 – 152

= 12

(iii) Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.

Now, The number of students whose height is above 172 cm

= 160 – 144 = 16

2. Draw ogive for the data given below and from the graph determine: (i) the median marks.

(ii) the number of students who obtained more than 75% marks.

Marks	10 – 19	20 -29	30 – 39	40 – 49	50 – 59	60 – 69	70 – 79	80 – 89	90 – 99
No. of students	14	16	22	26	18	11	6	4	3

Solution

Marks	No. of students	Cumulative frequency
9.5 – 19.5	14	14
19.5 – 29.5	16	30
29.5 – 39.5	22	52
39.5 – 49.5	26	78
49.5 – 59.5	18	96
59.5 – 69.5	11	107
69.5 – 79.5	6	113
79.5 – 89.5	4	117
89.5 – 99.5	3	120

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

(i) So, the median = 120/ 2 = 60^th term

Through mark 60, draw a parallel line to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Total marks = 100

75% of total marks = 75/100 ×100 = 75 marks

Hence, the number of students getting more than 75% marks = 120 – 111 = 9 students.

3. The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m – 1 and median q. Find p and q.

Solution

Mean of 1, 7, 5, 3, 4 and 4 = (1 + 7 + 5 + 3 + 4 + 4)/6

= 24/6

= 4

So, m = 4

Now, given that

The mean of 3, 2, 4, 2, 3, 3 and p = m -1 = 4 – 1 = 3

Thus, 17 + p = 3 x n …. ,where n = 7

⇒ 17 + p = 21

⇒ p = 4

Arranging the terms in ascending order, we have:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4^th term = 3

Hence, q = 3

4. In a malaria epidemic, the number of cases diagnosed were as follows:

Date (July)	1	2	3	4	5	6	7	8	9	10	11	12
Number	5	12	20	27	46	30	31	18	11	5	0	1

On what days do the mode and upper and lower quartiles occur?

Solution

Date	Number	C.f.
1	5	5
2	12	17
3	20	37
4	27	64
5	46	110
6	30	140
7	31	171
8	18	189
9	11	200
10	5	205
11	0	205
12	1	206

(i) Mode = 5^th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile = 206 ×(3/4) = 154.5^th = 7^th July

Lower quartile = 206 ×(1/4) = 51.5^th = 4^th July

5. The income of the parents of 100 students in a class in a certain university are tabulated below.

Income (in thousand Rs)	0 – 8	8 – 16	16 – 24	24 – 32	32 – 40
No. of students	8	35	35	14	8

 (i) Draw a cumulative frequency curve to estimate the median income.

(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.

(iii) Calculate the Arithmetic mean.

Solution

(i) Cumulative Frequency Curve

Income (in thousand Rs.)	No. of students (f)	Cumulative Frequency	Class mark (x)	fx
0 – 8	8	8	4	32
8 – 16	35	43	12	420
16 – 24	35	78	20	700
24 – 32	14	92	28	392
32 – 40	8	100	36	288
	∑f = 100			∑ fx = 1832

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

Here, N = 100

N/2 = 50

At y = 50, affix A.

Through A, draw a horizontal line meeting the curve at B.

Through B, a vertical line is drawn which meets OX at M.

OM = 17.6 units

Hence, median income = 17.6 thousands

(ii) 15% of 100 students = (15×100)/100 = 15

From c.f. 15, draw a horizontal line which intersects the curve at P.

From P, draw a perpendicular to x–axis meeting it at Q which is equal to 9.6

Thus, freeship will be awarded to students provided annual income of their parents is upto 9.6 thousands.

(ii) Mean = ∑ fx/∑ f = 1832/100 = 18.32

6. The marks of 20 students in a test were as follows:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.

Calculate:

(i) the mean

(ii) the median

(iii) the mode

Solution

Arranging the terms in ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20

∑x = 2 + 6 + 8 + 9 + 11 + 11 + 12 + 13 + 13 + 14 + 14 + 15 + 15 + 15 + 15 + 16 + 16 + 18 + 19 + 20 = 257

(i) Mean = ∑x/∑n = 257/ 20 = 12.85

(ii) Median = (10^th term + 11^th term)/2 = (13 + 14)/ 2 = 27/2 = 13.5

(iii) Mode = 15 since it has maximum frequencies i.e. 3

7. The marks obtained by 120 students in a mathematics test is given below:

Marks	No. of students
0-10	5
10-20	9
20-30	16
30-40	22
40-50	26
50-60	18
60-70	11
70-80	6
80-90	4
90-100	3

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:

(i) the median

(ii) the number of students who obtained more than 75% in test.

(iii) the number of students who did not pass in the test if the pass percentage was 40.

(iv) the lower quartile

Solution

Marks	No. of students	c.f.
0-10	5	5
10-20	9	14
20-30	16	30
30-40	22	52
40-50	26	78
50-60	18	96
60-70	11	107
70-80	6	113
80-90	4	117
90-100	3	120

(i)  Median = (120 + 1)/2 = 60.5^th term

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A. From A draw a perpendicular to x-axis meeting it at B.

Then, the value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 – 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x = 40, y = 52)

(iv) Lower quartile = Q₁ = 120×(1/4) = 30^th term = 30

8. Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.

Weight	Frequency
40 – 45	5
45 – 50	17
50 – 55	22
55 – 60	45
60 – 65	51
65 – 70	31
70 – 75	20
75 – 80	9

Use your ogive to estimate the following:

(i) The percentage of students weighing 55 kg or more

(ii) The weight above which the heaviest 30% of the student fall

(iii) The number of students who are (a) underweight (b) overweight, if 55.70 kg is considered as standard weight.

Solution

Weight	Frequency	c. f.
40-45	5	5
45-50	17	22
50-55	22	44
55-60	45	89
60-65	51	140
65-70	31	171
70-75	20	191
75-80	9	200

(i) The number of students weighing more than 55 kg = 200 – 44 = 156

Thus, the percentage of students weighing 55 kg or more = (156/200)×100 = 78 %

(ii) 30% of students = (30×200)/100 = 60

Heaviest 60students in weight = 9 + 21 + 30 = 60

Weight = 65 kg (From table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx.) from graph

(b) overweight students when 55.70 kg is standard = 200 – 55.70 = 154 (approx.) from graph

9. The distribution, given below, shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained	5	6	7	8	9	10
No. of students	3	9	6	4	2	1

Solution

Marks obtained(x)	No. of students (f)	c.f.	fx
5	3	3	15
6	9	12	54
7	6	18	42
8	4	22	32
9	2	24	18
10	1	25	10
Total	25		171

Number of terms = 25

(i) Mean = 171/25 = 6.84

(ii) Median = (25 + 1)/2 ^th = 13^th term = 7

(iii) Mode = 6 since it has the maximum frequency i.e. 6

10. The mean of the following distribution is 52 and the frequency of class interval 30 – 40 is ‘f’. Find f.

Class Interval	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Frequency	5	3	f	7	2	6	13

Solution

C.I.	Frequency(f)	Mid value (x)	fx
10-20	5	15	75
20-30	3	25	75
30-40	f	35	35f
40-50	7	45	315
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
Total	36 + f		1940 + 35f

Mean = ∑ fx/∑ f = (1940 + 35f)/(36 + f) ...(i)

But, given mean = 52 …(ii)

From (i) and (ii), we have

(1940 + 35f)/ (36 + f) = 52

⇒ 1940 + 35f = 1872 + 52f

⇒ 17f = 68

Thus, f = 4

11. The monthly income of a group of 320 employees in a company is given below:

Monthly Income (thousands)	No. of employees
6 – 7	20
7 – 8	45
8 – 9	65
9 – 10	95
10 – 11	60
11 – 12	30
12 – 13	5

Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine:

(i) the median wage.

(ii) number of employees whose income is below Rs 8500.

(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.

(iv) the upper quartile.

Solution

Monthly Income (thousands)	No. of employees (f)	Cumulative frequency
6-7	20	20
7-8	45	65
8-9	65	130
9-10	95	225
10-11	60	285
11-12	30	315
12-13	5	320
Total	320

Number of employees = 320

(i) Median = 320/2 = 160^th term

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

(ii) The number of employees with income below Rs 8,500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11,500 = 305 (approx from the graph)

Thus, the number of employees (senior employees) = 320 – 305 = 15

(iv) Upper quartile = Q₃ = 320×(3/4) = 240^th term = 10.3 thousands = Rs 10,300