NCERT Solutions · Class 10 · Chapter 13: Statistics

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.1

Exercise 13.1: Mean of grouped data. Use class mid-points \(x_i\). Direct method: \(\bar x = \dfrac{\sum f_ix_i}{\sum f_i}\). With large numbers use an assumed mean \(a\): \(\bar x = a + \dfrac{\sum f_id_i}{\sum f_i}\), or step deviations \(u_i = \dfrac{x_i - a}{h}\): \(\bar x = a + h\dfrac{\sum f_iu_i}{\sum f_i}\). All give the same answer.

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Exercise 13.1 questions and solutions

Exercise 13.1, Question 1

A survey of 20 houses records the number of plants in each. Find the mean number of plants per house, and say which method you used.

Class	Frequency
0–2	1
2–4	2
4–6	1
6–8	5
8–10	6
10–12	2
12–14	3

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
Mid-points \(1, 3, 5, \ldots, 13\): \(\sum f_ix_i = 162\), \(\sum f_i = 20\).
The direct method, because the numbers are small.

Answer: \(8.1\) plants

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Exercise 13.1, Question 2

Daily wages (₹) of 50 workers. Find the mean daily wage.

Class	Frequency
500–520	12
520–540	14
540–560	8
560–580	6
580–600	10

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
Assumed mean \(a = 550\), \(h = 20\): \(u_i = -2, -1, 0, 1, 2\), \[\begin{aligned}\sum f_iu_i &= -24 - 14 + 0 + 6 + 20 \\ &= -12\end{aligned}\]
\[\begin{aligned}\bar x &= 550 + 20 \times \dfrac{-12}{50} \\ &= 545.2\end{aligned}\]

Answer: ₹545.20

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Exercise 13.1, Question 3

Daily pocket allowance (₹) of the children of a locality; the mean is ₹18. Find the missing frequency \(f\).

Class	Frequency
11–13	7
13–15	6
15–17	9
17–19	13
19–21	f
21–23	5
23–25	4

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Mid-points \(12, 14, \ldots, 24\): \(\sum f_ix_i = 752 + 20f\), \(\sum f_i = 44 + f\).
\[\begin{aligned}&\dfrac{752 + 20f}{44 + f} = 18 \\ \Rightarrow\ &752 + 20f = 792 + 18f\end{aligned}\]

Answer: \(f = 20\)

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Exercise 13.1, Question 4

Heartbeats per minute of 30 women. Find the mean.

Class	Frequency
65–68	2
68–71	4
71–74	3
74–77	8
77–80	7
80–83	4
83–86	2

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
Mid-points \(66.5, 69.5, \ldots, 84.5\); with \(a = 75.5\), \(h = 3\): \(\sum f_iu_i = 4\).
\[\begin{aligned}\bar x &= 75.5 + 3 \times \tfrac{4}{30} \\ &= 75.9\end{aligned}\]

Answer: \(75.9\) beats per minute

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Exercise 13.1, Question 5

Number of mangoes in 400 boxes (classes 50–52, 53–55, …). Find the mean number per box. Which method did you choose?

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

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Mid-points \(51, 54, 57, 60, 63\) (the gaps between classes don't change mid-points). Step deviation with \(a = 57\), \(h = 3\): \[\begin{aligned}\sum f_iu_i &= -30 - 110 + 0 + 115 + 50 \\ &= 25\end{aligned}\]
\[\begin{aligned}\bar x &= 57 + 3 \times \tfrac{25}{400} \\ &= 57.1875\end{aligned}\]

Answer: About \(57.19\) mangoes (step-deviation method)

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Exercise 13.1, Question 6

Daily expenditure on food (₹) of 25 households. Find the mean.

Class	Frequency
100–150	4
150–200	5
200–250	12
250–300	2
300–350	2

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
\(a = 225\), \(h = 50\): \(u_i = -2, -1, 0, 1, 2\), \(\sum f_iu_i = -8 - 5 + 0 + 2 + 4 = -7\).
\[\begin{aligned}\bar x &= 225 + 50 \times \dfrac{-7}{25} \\ &= 211\end{aligned}\]

Answer: ₹211

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Exercise 13.1, Question 7

Concentration of SO₂ in the air (ppm) in 30 localities. Find the mean.

Class	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

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
Mid-points \(0.02, 0.06, \ldots, 0.22\): \(\sum f_ix_i = 2.96\).
\(\bar x = \dfrac{2.96}{30} \approx 0.099\).

Answer: About \(0.099\) ppm

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Exercise 13.1, Question 8

Days absent in a term for 40 students (unequal classes). Find the mean.

Class	Frequency
0–6	11
6–10	10
10–14	7
14–20	4
20–28	4
28–38	3
38–40	1

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
Mid-points \(3, 8, 12, 17, 24, 33, 39\): \(\sum f_ix_i = 499\).
\(\bar x = \dfrac{499}{40} = 12.475\). (Unequal widths: use the direct or assumed-mean method, not step deviation.)

Answer: About \(12.48\) days

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Exercise 13.1, Question 9

Literacy rate (%) of 35 cities. Find the mean.

Class	Frequency
45–55	3
55–65	10
65–75	11
75–85	8
85–95	3

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Mean \(\bar x = \dfrac{\sum f_i x_i}{\sum f_i}\), with \(x_i\) the class mid-points.
\(a = 70\), \(h = 10\): \(\sum f_iu_i = -6 - 10 + 0 + 8 + 6 = -2\).
\[\bar x = 70 + 10 \times \dfrac{-2}{35} \approx 69.43\]

Answer: About \(69.43\%\)

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