Class 10 Maths Chapter 13 Statistics Worksheet with Answers, free PDF (based on NCERT)

A 20-mark worksheet on Statistics for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

For a distribution with class size 10, the modal class is 40-50 with frequency 20; the class before it has frequency 12 and the class after it has frequency 8. The mode is

(a) 46(b) 44(c) 42(d) 48

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Answer: (b) 44

Mode = l + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h = 40 + (8 ÷ 20) × 10 = 44. 46 uses f₁ − f₂ on top; 42 leaves f₀ and f₂ out of the denominator.

Question 2

For a distribution with class size 20, the step deviation method gives a = 150, Σfᵢ = 40 and Σfᵢuᵢ = 12. The mean is

(a) 144(b) 150.3(c) 162(d) 156

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Answer: (d) 156

x̄ = a + (Σfᵢuᵢ ÷ Σfᵢ) × h = 150 + (12 ÷ 40) × 20 = 150 + 6 = 156. 150.3 forgets to multiply by h.

Question 3

Using the assumed mean method with a = 45, a student finds Σfᵢ = 50 and Σfᵢdᵢ = −150. The mean is

(a) −3(b) 48(c) 45(d) 42

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Answer: (d) 42

x̄ = a + Σfᵢdᵢ ÷ Σfᵢ = 45 + (−150 ÷ 50) = 45 − 3 = 42. 48 adds the deviation with the wrong sign; −3 forgets to add a.

Question 4

Asha finds the mean of a grouped distribution of 50 observations by the direct method and gets 41.6. Ravi uses the step deviation method on the same data with a = 45 and h = 10. His Σfᵢuᵢ must be

(a) −3.4(b) 17(c) −17(d) −34

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Answer: (c) −17

The methods give the same mean, so 41.6 = 45 + (Σfᵢuᵢ ÷ 50) × 10. Then Σfᵢuᵢ ÷ 5 = −3.4, so Σfᵢuᵢ = −17. −3.4 stops one step short.

Question 5

The class marks of a distribution are 1250, 1350, 1450, 1550 and 1650, and the frequencies run into the hundreds. The way of finding the mean that keeps the arithmetic smallest is

(a) the direct method(b) the step deviation method(c) adding the class marks and dividing by 5(d) the mode formula

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Answer: (b) the step deviation method

With large xᵢ and fᵢ, taking a = 1450 and dividing every dᵢ by h = 100 leaves uᵢ = −2, −1, 0, 1, 2. The direct method multiplies the large numbers as they stand; adding class marks ignores the frequencies.

Question 6 · 2 marks

The weekly hours spent on homework by 40 students of a school in Pune are given below. Find the mean number of hours by the direct method.

Hours	Number of students
0-4	5
4-8	8
8-12	12
12-16	9
16-20	6

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Class marks 2, 6, 10, 14, 18. fᵢxᵢ: 10, 48, 120, 126, 108; Σfᵢxᵢ = 412. x̄ = 412 ÷ 40 = 10.3 hours.

class marks and the fᵢxᵢ column, Σfᵢxᵢ = 412 (1)
mean = 412 ÷ 40 = 10.3 hours (1)

Question 7 · 2 marks

The table gives the number of runs made by 47 batters in a district cricket camp in Ranchi, grouped as below. Find n/2 and the median class.

Runs	Number of batters
0-10	5
10-20	9
20-30	14
30-40	12
40-50	7

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n = 47, so n/2 = 23.5 (the grouped formula halves n whatever its parity). Cumulative frequencies: 5, 14, 28, 40, 47. The first to exceed 23.5 is 28, so the median class is 20-30.

n/2 = 23.5 and cumulative frequencies 5, 14, 28, ... (1)
median class 20-30 (1)

Question 8 · 2 marks

The mean and median of the monthly rents of shops in a market in Indore are ₹52.4 thousand and ₹50 thousand. Estimate the mode, and say what the relationship you used is for.

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3 Median = Mode + 2 Mean, so Mode = 3(50) − 2(52.4) = 150 − 104.8 = ₹45.2 thousand. The relationship is empirical: it estimates the third measure when two are known.

mode = 3 Median − 2 Mean = ₹45.2 thousand (1)
states it is an empirical estimate from the other two measures (1)

Question 9 · 3 marks

The marks of 45 students in a unit test at a school in Jaipur are given below. Find the mode.

Marks	Number of students
1-10	5
11-20	9
21-30	15
31-40	11
41-50	5

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The classes have gaps, so make them continuous: 0.5-10.5, 10.5-20.5, 20.5-30.5, 30.5-40.5, 40.5-50.5. Modal class 20.5-30.5; l = 20.5, f₁ = 15, f₀ = 9, f₂ = 11, h = 10. Mode = 20.5 + ((15 − 9) ÷ (30 − 9 − 11)) × 10 = 20.5 + 6 = 26.5.

classes made continuous, l = 20.5 (1)
f₁, f₀, f₂, h substituted (1)
mode = 26.5 (1)

Question 10 · 3 marks

The monthly mobile data (GB) used by the members of a youth club in Bhopal is given below. The mean is 25.75 GB. Find k.

Data used (GB)	Number of members
0-10	5
10-20	8
20-30	k
30-40	9
40-50	6

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Class marks 5, 15, 25, 35, 45. Σfᵢ = 28 + k; Σfᵢxᵢ = 25 + 120 + 25k + 315 + 270 = 730 + 25k. Then (730 + 25k) ÷ (28 + k) = 25.75, so 730 + 25k = 721 + 25.75k, giving 0.75k = 9 and k = 12.

Σfᵢ = 28 + k and Σfᵢxᵢ = 730 + 25k (1)
equation (730 + 25k) = 25.75(28 + k) (1)
k = 12 (1)

Question 11 · 3 marks

The time taken by 50 students of a school in Lucknow to reach school is given below. Find the median time.

Time (minutes)	Number of students
0-10	7
10-20	12
20-30	15
30-40	10
40-50	6

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Cumulative frequencies: 7, 19, 34, 44, 50. n/2 = 25; the first cumulative frequency above 25 is 34, so the median class is 20-30. l = 20, cf = 19, f = 15, h = 10. Median = 20 + ((25 − 19) ÷ 15) × 10 = 24 minutes.

cumulative frequency column (1)
median class 20-30 with l, cf, f, h (1)
median = 24 minutes (1)

Class 10 Maths Chapter 13: Statistics worksheet with answer key