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Class 10 Maths Chapter 13: Statistics MCQs with answers

10 multiple-choice questions on Statistics, each with its answer and the reason, written from the NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
The class mark of the class 35-45 is
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Answer: (a) 40

Class mark = (35 + 45)/2 = 40. 10 is the class size; 80 forgets to halve.

Question 2
Marks in a class test at a school in Kanpur are grouped as 0-15, 15-30, 30-45, 45-60. Kavya scores exactly 30. Her score is counted in the class
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Answer: (b) 30-45

A score on the upper limit of a class goes into the next class: 30 is the upper limit of 15-30, so it is counted in 30-45.

Question 3
For a grouped distribution, Σfᵢ = 40 and Σfᵢxᵢ = 1260. The mean is
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Answer: (a) 31.5

x̄ = Σfᵢxᵢ ÷ Σfᵢ = 1260 ÷ 40 = 31.5.

Question 4
Using the assumed mean method with a = 45, a student finds Σfᵢ = 50 and Σfᵢdᵢ = −150. The mean is
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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 5
In the step deviation method, the assumed mean is 55 and the class size is 10. For the class with class mark 75, uᵢ is
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Answer: (a) 2

uᵢ = (xᵢ − a) ÷ h = (75 − 55) ÷ 10 = 2. 20 is dᵢ, not uᵢ.

Question 6
For a distribution with class size 20, the step deviation method gives a = 150, Σfᵢ = 40 and Σfᵢuᵢ = 12. The mean is
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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 7
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
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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 8
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
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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 9
The weights of 36 sacks of rice at a godown in Raipur are grouped below. The modal class is
Weight (kg)Number of sacks
45-505
50-559
55-6012
60-657
65-703
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Answer: (b) 55-60

The modal class is the class with the largest frequency, 12, which is 55-60.

Question 10
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
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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.

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