Mathematics · Statistics · NCERT Exercises

Exercise 13.1

Complete, independently verified solutions for NCERT Exercise 13.1.

ChapterResourceExercise / Example

Mathematics · Chapter 13 · NCERT Exercises

Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

Question

Question 1

mean plants · direct method

A survey recorded the number of plants in 20 houses. Find the mean number of plants per house and state a suitable method.

Plants	Houses (fᵢ)
0–2	1
2–4	2
4–6	1
6–8	5
8–10	6
10–12	2
12–14	3

Solution

Board-exam working:

The class marks and frequencies are small, so use the direct method.

Class	fᵢ	xᵢ	fᵢxᵢ
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

\sum f_i=20,\qquad \sum f_ix_i=162

\bar{x}=\frac{162}{20}=8.1

Final answer

\boxed{\bar{x}=8.1\text{ plants per house}}

Question

Question 2

mean wages · assumed mean

The daily wages of 50 factory workers are distributed as follows. Find the mean daily wage using an appropriate method.

Daily wages (₹)	Workers (fᵢ)
500–520	12
520–540	14
540–560	8
560–580	6
580–600	10

Solution

Board-exam working:

Take assumed mean 550 and class size 20.

Class	fᵢ	xᵢ	uᵢ = (xᵢ−550)/20	fᵢuᵢ
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

\sum f_i=50,\qquad \sum f_iu_i=-12

\bar{x}=550+20\left(\frac{-12}{50}\right)

=550-4.8=545.2

Final answer

\boxed{\text{Mean daily wage}=\text{Rs. }545.20}

Question

Question 3

mean · missing frequency · pocket allowance

The mean daily pocket allowance is ₹18. Find the missing frequency f.

Allowance (₹)	Children
11–13	7
13–15	6
15–17	9
17–19	13
19–21	f
21–23	5
23–25	4

Solution

Board-exam working:

Class mark (xᵢ)	Frequency (fᵢ)	fᵢxᵢ
12	7	84
14	6	84
16	9	144
18	13	234
20	f	20f
22	5	110
24	4	96

\sum f_i=44+f,\qquad \sum f_ix_i=752+20f

Use the given mean.

18=\frac{752+20f}{44+f}

792+18f=752+20f

40=2f\Rightarrow f=20

Final answer

\boxed{f=20}

Question

Question 4

mean heartbeats · step deviation

Thirty women were examined in a hospital. Find their mean heartbeats per minute from the distribution below.

Heartbeats/min	Women
65–68	2
68–71	4
71–74	3
74–77	8
77–80	7
80–83	4
83–86	2

Solution

Board-exam working:

Take assumed mean 75.5 and class size 3.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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

\sum f_i=30,\qquad \sum f_iu_i=4

\bar{x}=75.5+3\left(\frac4{30}\right)=75.9

Final answer

\boxed{\bar{x}=75.9\text{ heartbeats per minute}}

Question

Question 5

mean mangoes · step deviation

Packing boxes contain varying numbers of mangoes. Find the mean number of mangoes per box and state a suitable method.

Mangoes	Boxes
50–52	15
53–55	110
56–58	135
59–61	115
62–64	25

Solution

Board-exam working:

The inclusive classes have class marks 51, 54, 57, 60 and 63. Take assumed mean 57 and step 3.

Class	fᵢ	xᵢ	uᵢ	fᵢuᵢ
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

\sum f_i=400,\qquad \sum f_iu_i=25

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

=57+0.1875=57.1875

Final answer

\boxed{\bar{x}\approx57.19\text{ mangoes per box}}

Question

Question 6

mean expenditure · direct method

The table shows the daily food expenditure of 25 households. Find the mean daily expenditure.

Expenditure (₹)	Households
100–150	4
150–200	5
200–250	12
250–300	2
300–350	2

Solution

Board-exam working:

Class	fᵢ	xᵢ	fᵢxᵢ
100–150	4	125	500
150–200	5	175	875
200–250	12	225	2700
250–300	2	275	550
300–350	2	325	650

\sum f_i=25,\qquad \sum f_ix_i=5275

\bar{x}=\frac{5275}{25}=211

Final answer

\boxed{\text{Mean daily expenditure}=\text{Rs. }211}

Question

Question 7

mean sulphur dioxide · ppm

The concentration of sulphur dioxide in the air was measured in 30 localities. Find the mean concentration.

SO₂ concentration (ppm)	Frequency
0.00–0.04	4
0.04–0.08	9
0.08–0.12	9
0.12–0.16	2
0.16–0.20	4
0.20–0.24	2

Solution

Board-exam working:

Class	fᵢ	xᵢ	fᵢxᵢ
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

\sum f_i=30,\qquad \sum f_ix_i=2.96

\bar{x}=\frac{2.96}{30}=0.098666\ldots

Final answer

\boxed{\bar{x}\approx0.099\text{ ppm}}

Question

Question 8

mean absence · unequal intervals

The absentee record of 40 students for a term is given below. Find the mean number of days absent.

Days absent	Students
0–6	11
6–10	10
10–14	7
14–20	4
20–28	4
28–38	3
38–40	1

Solution

Board-exam working:

Since the class widths differ, use the direct method with class marks.

Class	fᵢ	xᵢ	fᵢxᵢ
0–6	11	3	33
6–10	10	8	80
10–14	7	12	84
14–20	4	17	68
20–28	4	24	96
28–38	3	33	99
38–40	1	39	39

\sum f_i=40,\qquad \sum f_ix_i=499

\bar{x}=\frac{499}{40}=12.475

Final answer

\boxed{\bar{x}\approx12.48\text{ days}}

Question

Question 9

mean literacy rate · direct method

The literacy rates of 35 cities are grouped below. Find the mean literacy rate.

Literacy rate (%)	Cities
45–55	3
55–65	10
65–75	11
75–85	8
85–95	3

Solution

Board-exam working:

Class	fᵢ	xᵢ	fᵢxᵢ
45–55	3	50	150
55–65	10	60	600
65–75	11	70	770
75–85	8	80	640
85–95	3	90	270

\sum f_i=35,\qquad \sum f_ix_i=2430

\bar{x}=\frac{2430}{35}=69.4285\ldots

Final answer

\boxed{\bar{x}\approx69.43\%}