NCERT Class 10 Maths – Exercise 13.1 Solutions

NCERT Class 10 Maths

Chapter 13 – Statistics | Exercise 13.1

(Rationalized Syllabus 2025-26)

💡 Formulas for Mean ($\bar{x}$)

Class Mark ($x_i$): $\frac{\text{Upper limit} + \text{Lower limit}}{2}$
Direct Method: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Step Deviation Method: $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$, where $u_i = \frac{x_i – a}{h}$

A survey was conducted by a group of students… regarding the number of plants in 20 houses. Find the mean number of plants per house. Which method did you use and why?

Using Direct Method

Since the values of $f_i$ and $x_i$ are small, we use the Direct Method.

No. of Plants	No. of Houses ($f_i$)	Class Mark ($x_i$)	$f_i x_i$
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
Total	20		162

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{162}{20} = 8.1$$

✔️ Mean number of plants = 8.1

Consider the following distribution of daily wages of 50 workers. Find the mean daily wages using an appropriate method.

Using Step Deviation Method

Assumed Mean $a = 550$, Class size $h = 20$.

Daily Wages	Workers ($f_i$)	$x_i$	$u_i = \frac{x_i – 550}{20}$	$f_i u_i$
500-520	12	510	-2	-24
520-540	14	530	-1	-14
540-560	8	550	0	0
560-580	6	570	1	6
580-600	10	590	2	20
Total	50			-12

$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 550 + \left(\frac{-12}{50}\right) \times 20$$

$$= 550 – 4.8 = 545.2$$

✔️ Mean daily wages = ₹ 545.20

The following distribution shows the daily pocket allowance. The mean pocket allowance is ₹ 18. Find the missing frequency $f$.

Using Direct Method

Given Mean $\bar{x} = 18$.

Allowance	Children ($f_i$)	$x_i$	$f_i x_i$
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	$f$	20	$20f$
21-23	5	22	110
23-25	4	24	96
Total	$44+f$		$752+20f$

$$18 = \frac{752 + 20f}{44 + f} \Rightarrow 18(44 + f) = 752 + 20f$$

$$792 + 18f = 752 + 20f \Rightarrow 2f = 40 \Rightarrow f = 20$$

✔️ Missing frequency $f = 20$

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded. Find the mean heartbeats per minute.

Using Step Deviation Method

Assumed Mean $a = 75.5$, $h = 3$.

Heartbeats	Women ($f_i$)	$x_i$	$u_i$	$f_i u_i$
65-68	2	66.5	-3	-6
68-71	4	69.5	-2	-8
71-74	3	72.5	-1	-3
74-77	8	75.5	0	0
77-80	7	78.5	1	7
80-83	4	81.5	2	8
83-86	2	84.5	3	6
Total	30			4

$$\bar{x} = 75.5 + \left(\frac{4}{30}\right) \times 3 = 75.5 + 0.4 = 75.9$$

✔️ Mean heartbeats = 75.9 per min

In a retail market, fruit vendors were selling mangoes kept in packing boxes. Find the mean number of mangoes.

Using Step Deviation Method

Note: Class intervals are not continuous (50-52, 53-55). However, the class marks ($x_i$) are 51, 54, 57… which have a consistent gap of 3. We can proceed with $a=57, h=3$.

Mangoes	Boxes ($f_i$)	$x_i$	$u_i$	$f_i u_i$
50-52	15	51	-2	-30
53-55	110	54	-1	-110
56-58	135	57	0	0
59-61	115	60	1	115
62-64	25	63	2	50
Total	400			25

$$\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3 = 57 + 0.1875$$

✔️ Mean number of mangoes = 57.19

Find the mean daily expenditure on food of 25 households in a locality.

Using Step Deviation Method

$a = 225, h = 50$.

Expenditure	$f_i$	$x_i$	$u_i$	$f_i u_i$
100-150	4	125	-2	-8
150-200	5	175	-1	-5
200-250	12	225	0	0
250-300	2	275	1	2
300-350	2	325	2	4
Total	25			-7

$$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50 = 225 – 14 = 211$$

✔️ Mean expenditure = ₹ 211

To find out the concentration of SO2 in the air (in ppm), data was collected for 30 localities. Find the mean.

Using Direct Method

Conc. SO2	Freq ($f_i$)	$x_i$	$f_i x_i$
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
Total	30		2.96

$$\bar{x} = \frac{2.96}{30} \approx 0.099$$

✔️ Mean concentration = 0.099 ppm

A class teacher has the following absentee record of 40 students. Find the mean number of days.

Using Assumed Mean Method

Note: Unequal class sizes. Use Assumed Mean $a=17$.

Days	$f_i$	$x_i$	$d_i=x_i-17$	$f_i d_i$
0-6	11	3	-14	-154
6-10	10	8	-9	-90
10-14	7	12	-5	-35
14-20	4	17	0	0
20-28	4	24	7	28
28-38	3	33	16	48
38-40	1	39	22	22
Total	40			-181

$$\bar{x} = 17 + \frac{-181}{40} = 17 – 4.525 = 12.475$$

✔️ Mean days = 12.48 days

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Using Step Deviation Method

$a = 70, h = 10$.

Literacy Rate	Cities ($f_i$)	$x_i$	$u_i$	$f_i u_i$
45-55	3	50	-2	-6
55-65	10	60	-1	-10
65-75	11	70	0	0
75-85	8	80	1	8
85-95	3	90	2	6
Total	35			-2

$$\bar{x} = 70 + \left(\frac{-2}{35}\right) \times 10 = 70 – 0.57$$

✔️ Mean literacy rate = 69.43%

🎉 Exercise 13.1 Completed | Chapter 13 Statistics