Let us work out 26.1 | WB Class 10

Ganit Prakash - Class-X - Statistics: Mean

Let us work out 26.1 Measures of Central Tendency

📘 Exercise 26.1 Solutions (Q1 - Q14)

Chapter 26

Question 1

I have written ages of my 40 friends in the table given below.

Age (years)	15	16	17	18	19	20
Number of friends	4	7	10	10	5	4

Let us find average age of my friends by direct method.

Explanation (Direct Method):

Age in years ($x_i$)	Number of friends ($f_i$)	$f_i x_i$
15	4	60
16	7	112
17	10	170
18	10	180
19	5	95
20	4	80
Total	$\sum f_i = 40$	$\sum f_i x_i = 697$

Using the Direct Method formula for Mean ($\bar{x}$):

\[\begin{array}{l} \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \\ \bar{x} = \frac{697}{40} \\ \bar{x} = 17.425 \end{array}\]

Answer: The average age of the friends is 17.425 years.

Question 2

I have written member of 50 families of our village in the table given below.

Number of members	2	3	4	5	6	7
Number of families	6	8	14	15	4	3

Let us write the average member of 50 families by the method of assumed mean.

Explanation (Assumed Mean Method):

Let the assumed mean ($a$) be 4.

No. of members ($x_i$)	No. of families ($f_i$)	Deviation ($d_i = x_i - a$)	$f_i d_i$
2	6	$2 - 4 = -2$	-12
3	8	$3 - 4 = -1$	-8
4	14	$4 - 4 = 0$	0
5	15	$5 - 4 = 1$	15
6	4	$6 - 4 = 2$	8
7	3	$7 - 4 = 3$	9
Total	$\sum f_i = 50$		$\sum f_i d_i = 12$

Using the Assumed Mean Method formula:

\[\begin{array}{l} \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i} \\ \bar{x} = 4 + \frac{12}{50} \\ \bar{x} = 4 + 0.24 \\ \bar{x} = 4.24 \end{array}\]

Answer: The average number of members per family is 4.24.

Question 3

If the arithmetic mean of the data given below is 20.6, let us find the value of 'a'.

Variable ($x_i$)	10	15	a	25	35
Frequency ($f_i$)	3	10	25	7	5

Explanation:

Variable ($x_i$)	Frequency ($f_i$)	$f_i x_i$
10	3	30
15	10	150
a	25	25a
25	7	175
35	5	175
Total	$\sum f_i = 50$	$\sum f_i x_i = 530 + 25a$

We are given that the Mean ($\bar{x}$) = 20.6.

\[\begin{array}{l} \frac{\sum f_i x_i}{\sum f_i} = \text{Mean} \\ \frac{530 + 25a}{50} = 20.6 \\ 530 + 25a = 20.6 \times 50 \\ 530 + 25a = 1030 \\ 25a = 1030 - 530 \\ 25a = 500 \\ a = \frac{500}{25} = 20 \end{array}\]

Answer: The value of $a$ is 20.

Question 4

If the arithmetic mean of the distribution given below is 15, let us find the value of p.

Variable	5	10	15	20	25
Frequency (f)	6	p	6	10	5

Explanation:

Variable ($x_i$)	Frequency ($f_i$)	$f_i x_i$
5	6	30
10	p	10p
15	6	90
20	10	200
25	5	125
Total	$\sum f_i = 27 + p$	$\sum f_i x_i = 445 + 10p$

We are given that the Mean ($\bar{x}$) = 15.

\[\begin{array}{l} \frac{\sum f_i x_i}{\sum f_i} = \text{Mean} \\ \frac{445 + 10p}{27 + p} = 15 \\ 445 + 10p = 15(27 + p) \\ 445 + 10p = 405 + 15p \\ 445 - 405 = 15p - 10p \\ 40 = 5p \\ p = \frac{40}{5} = 8 \end{array}\]

Answer: The value of $p$ is 8.

Question 5

Rahamatchacha will go to retail market for selling mangoes kept in 50 packing boxes. Let us write the number of boxes contained varying number of mangoes in the table given below.

Number of mangoes	50 - 52	52 - 54	54 - 56	56 - 58	58 - 60
Number of boxes	6	14	16	9	5

Let us write the mean number of mangoes kept in 50 packing boxes. [using any method]

Explanation (Using Direct Method):

First, we find the class mark ($x_i$) for each interval. $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

No. of mangoes (Class)	Class mark ($x_i$)	No. of boxes ($f_i$)	$f_i x_i$
50 - 52	51	6	306
52 - 54	53	14	742
54 - 56	55	16	880
56 - 58	57	9	513
58 - 60	59	5	295
Total		$\sum f_i = 50$	$\sum f_i x_i = 2736$

\[\begin{array}{l} \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \\ = \frac{2736}{50} \\ = 54.72 \end{array}\]

Answer: The mean number of mangoes is 54.72.

Question 6

Mohidul has written ages of 100 patients of village hospital. Let us write by calculating average age of 100 patients. [using any method]

Age (years)	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Number of patients	12	8	22	20	18	20

Explanation (Using Step-Deviation Method):

Let assumed mean ($a$) = 35 and class size ($h$) = 10.

Step deviation $u_i = \frac{x_i - a}{h} = \frac{x_i - 35}{10}$

Age ($C.I.$)	Class mark ($x_i$)	No. of patients ($f_i$)	$u_i$	$f_i u_i$
10 - 20	15	12	-2	-24
20 - 30	25	8	-1	-8
30 - 40	35	22	0	0
40 - 50	45	20	1	20
50 - 60	55	18	2	36
60 - 70	65	20	3	60
Total		$\sum f_i = 100$		$\sum f_i u_i = 84$

\[\begin{array}{l} \text{Mean } (\bar{x}) = a + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \\ = 35 + 10 \times \left( \frac{84}{100} \right) \\ = 35 + 10 \times 0.84 \\ = 35 + 8.4 = 43.4 \end{array}\]

Answer: The average age of the patients is 43.4 years.

Question 7

Let us find the mean of the following data by direct method.

(i)

Class interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	4	6	10	6	4

(ii)

Class interval	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
Frequency	10	16	20	30	13	11

Explanation for 7 (i):

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$f_i x_i$
0 - 10	5	4	20
10 - 20	15	6	90
20 - 30	25	10	250
30 - 40	35	6	210
40 - 50	45	4	180
Total		$\sum f_i = 30$	$\sum f_i x_i = 750$

\[\begin{array}{l} \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{750}{30} = \mathbf{25} \end{array}\]

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Explanation for 7 (ii):

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$f_i x_i$
10 - 20	15	10	150
20 - 30	25	16	400
30 - 40	35	20	700
40 - 50	45	30	1350
50 - 60	55	13	715
60 - 70	65	11	715
Total		$\sum f_i = 100$	$\sum f_i x_i = 4030$

\[\begin{array}{l} \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{4030}{100} = \mathbf{40.3} \end{array}\]

Question 8

Let us find the mean of the following data by assumed mean method.

(i)

Class interval	0 - 40	40 - 80	80 - 120	120 - 160	160 - 200
Frequency	12	20	25	20	13

(ii)

Class interval	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75
Frequency	4	10	8	12	6

Explanation for 8 (i):

Let Assumed Mean ($A$) = 100

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	Deviation ($d_i = x_i - A$)	$f_i d_i$
0 - 40	20	12	-80	-960
40 - 80	60	20	-40	-800
80 - 120	100	25	0	0
120 - 160	140	20	40	800
160 - 200	180	13	80	1040
Total		$\sum f_i = 90$		$\sum f_i d_i = 80$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i} \\ = 100 + \frac{80}{90} \\ = 100 + 0.888... \\ = \mathbf{100.89} \text{ (approx)} \end{array}\]

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Explanation for 8 (ii):

Let Assumed Mean ($A$) = 50

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	Deviation ($d_i = x_i - A$)	$f_i d_i$
25 - 35	30	4	-20	-80
35 - 45	40	10	-10	-100
45 - 55	50	8	0	0
55 - 65	60	12	10	120
65 - 75	70	6	20	120
Total		$\sum f_i = 40$		$\sum f_i d_i = 60$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + \frac{\sum f_i d_i}{\sum f_i} \\ = 50 + \frac{60}{40} \\ = 50 + 1.5 \\ = \mathbf{51.5} \end{array}\]

Question 9

Let us find the mean of the following data by step deviation method.

(i)

Class interval	0 - 30	30 - 60	60 - 90	90 - 120	120 - 150
Frequency	12	15	20	25	8

(ii)

Class interval	0 - 14	14 - 28	28 - 42	42 - 56	56 - 70
Frequency	7	21	35	11	16

Explanation for 9 (i):

Let Assumed Mean ($A$) = 75, Class Size ($h$) = 30. Step deviation $u_i = \frac{x_i - 75}{30}$.

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$u_i$	$f_i u_i$
0 - 30	15	12	-2	-24
30 - 60	45	15	-1	-15
60 - 90	75	20	0	0
90 - 120	105	25	1	25
120 - 150	135	8	2	16
Total		$\sum f_i = 80$		$\sum f_i u_i = 2$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \\ = 75 + 30 \times \left( \frac{2}{80} \right) \\ = 75 + 30 \times 0.025 \\ = 75 + 0.75 = \mathbf{75.75} \end{array}\]

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Explanation for 9 (ii):

Let Assumed Mean ($A$) = 35, Class Size ($h$) = 14. Step deviation $u_i = \frac{x_i - 35}{14}$.

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$u_i$	$f_i u_i$
0 - 14	7	7	-2	-14
14 - 28	21	21	-1	-21
28 - 42	35	35	0	0
42 - 56	49	11	1	11
56 - 70	63	16	2	32
Total		$\sum f_i = 90$		$\sum f_i u_i = 8$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \\ = 35 + 14 \times \left( \frac{8}{90} \right) \\ = 35 + \frac{112}{90} \\ = 35 + 1.244... = \mathbf{36.24} \text{ (approx)} \end{array}\]

Question 10

If the mean of the following frequency distribution table is 24, let us find the value of p.

Class interval (number)	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Number of students	15	20	35	p	10

Explanation:

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$f_i x_i$
0 - 10	5	15	75
10 - 20	15	20	300
20 - 30	25	35	875
30 - 40	35	p	35p
40 - 50	45	10	450
Total		$\sum f_i = 80 + p$	$\sum f_i x_i = 1700 + 35p$

Given Mean = 24:

\[\begin{array}{l} \frac{\sum f_i x_i}{\sum f_i} = 24 \\ \frac{1700 + 35p}{80 + p} = 24 \\ 1700 + 35p = 24(80 + p) \\ 1700 + 35p = 1920 + 24p \\ 35p - 24p = 1920 - 1700 \\ 11p = 220 \\ p = \frac{220}{11} = 20 \end{array}\]

Answer: The value of $p$ is 20.

Question 11

Let us see the ages of the persons present in a meeting and determine their average age from the following table :

Age (Years)	30 - 34	35 - 39	40 - 44	45 - 49	50 - 54	55 - 59
Number of persons	10	12	15	6	4	3

Explanation (Using Assumed Mean Method):

Note: Even though classes are discontinuous, the class marks ($x_i$) will be the exact midpoints.

Let Assumed Mean ($A$) = 42

Age (Years)	Class mark ($x_i$)	Number of persons ($f_i$)	Deviation ($d_i = x_i - 42$)	$f_i d_i$
30 - 34	32	10	-10	-100
35 - 39	37	12	-5	-60
40 - 44	42	15	0	0
45 - 49	47	6	5	30
50 - 54	52	4	10	40
55 - 59	57	3	15	45
Total		$\sum f_i = 50$		$\sum f_i d_i = -45$

\[\begin{array}{l} \text{Mean} = A + \frac{\sum f_i d_i}{\sum f_i} \\ = 42 + \left( \frac{-45}{50} \right) \\ = 42 - 0.9 \\ = 41.1 \end{array}\]

Answer: The average age of the persons is 41.1 years.

Question 12

Let us find the mean of the following data.

Class interval	5 - 14	15 - 24	25 - 34	35 - 44	45 - 54	55 - 64
Frequency	3	6	18	20	10	3

Explanation (Using Step-Deviation Method):

Let Assumed Mean ($A$) = 39.5, Class Size ($h$) = 10. Step deviation $u_i = \frac{x_i - 39.5}{10}$.

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$u_i$	$f_i u_i$
5 - 14	9.5	3	-3	-9
15 - 24	19.5	6	-2	-12
25 - 34	29.5	18	-1	-18
35 - 44	39.5	20	0	0
45 - 54	49.5	10	1	10
55 - 64	59.5	3	2	6
Total		$\sum f_i = 60$		$\sum f_i u_i = -23$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \\ = 39.5 + 10 \times \left( \frac{-23}{60} \right) \\ = 39.5 - \frac{23}{6} \\ = 39.5 - 3.833... = \mathbf{35.67} \text{ (approx)} \end{array}\]

Question 13

Let us find the mean of obtaining marks of girl students if their cumulative frequencies are as follows.

Class interval (marks)	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
Number of girl students	5	9	17	29	45

Explanation:

First, we convert the cumulative frequency distribution into a standard frequency distribution.

Let Assumed Mean ($A$) = 25, Class Size ($h$) = 10.

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$u_i = \frac{x_i - 25}{10}$	$f_i u_i$
0 - 10	5	5	-2	-10
10 - 20	15	9 - 5 = 4	-1	-4
20 - 30	25	17 - 9 = 8	0	0
30 - 40	35	29 - 17 = 12	1	12
40 - 50	45	45 - 29 = 16	2	32
Total		$\sum f_i = 45$		$\sum f_i u_i = 30$

\[\begin{array}{l} \text{Mean } (\bar{x}) = A + h \times \left( \frac{\sum f_i u_i}{\sum f_i} \right) \\ = 25 + 10 \times \left( \frac{30}{45} \right) \\ = 25 + 10 \times \left( \frac{2}{3} \right) \\ = 25 + 6.66... = \mathbf{31.67} \text{ (approx)} \end{array}\]

Question 14

Let us find the mean of the obtaining marks of 64 students from the table given below.

Class interval (marks)	1 - 4	4 - 9	9 - 16	16 - 17
Students.	6	12	26	20

Explanation (Using Direct Method):

Because the class sizes are unequal ($3, 5, 7, 1$), the direct method is the most straightforward to apply.

Class Interval	Class mark ($x_i$)	Frequency ($f_i$)	$f_i x_i$
1 - 4	$(1+4)/2 = 2.5$	6	$6 \times 2.5 = 15$
4 - 9	$(4+9)/2 = 6.5$	12	$12 \times 6.5 = 78$
9 - 16	$(9+16)/2 = 12.5$	26	$26 \times 12.5 = 325$
16 - 17	$(16+17)/2 = 16.5$	20	$20 \times 16.5 = 330$
Total		$\sum f_i = 64$	$\sum f_i x_i = 748$

\[\begin{array}{l} \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \\ = \frac{748}{64} \\ = \mathbf{11.6875} \end{array}\]

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WBBSE Board — Class 10 - Statistics: Mean, Median, Ogive, Mode - Let us work out 26.1 Solutions.

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