CBSE Class 10 Mathematics Statistics MCQs Set 13

Practice MCQs for Class 10 Mathematics Chapter 13 Statistics

Access targeted multiple-choice questions for Chapter 13 Statistics designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Chapter 13 Statistics Questions and Solutions

Access the complete set of multiple-choice questions for Chapter 13 Statistics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: The weights of 12 students: 67 kg (4 students), 70 kg (3), 72 kg (2), 73 kg (2), 75 kg (1). The mean weight is:
(a) 70.25 kg
(b) 71.5 kg
(c) 69.8 kg
(d) 72 kg
Show Answer & Explanation

Answer: (a) 70.25 kg

Explanation:
1. Σfx = 268 + 210 + 144 + 146 + 75 = 843. Mean = 843/12 = 70.25 kg.

Teacher's Note:
1. Use the assumed mean 72: Σfd = −21.

Question 2: The mean of the classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 7, 10, 15, 8, 10 is:
(a) 25
(b) 26.5
(c) 24.2
(d) 25.8
Show Answer & Explanation

Answer: (d) 25.8

Explanation:
1. Class marks 5, 15, 25, 35, 45. Σfx = 35 + 150 + 375 + 280 + 450 = 1290. Mean = 1290/50 = 25.8.

Teacher's Note:
1. Use class marks.

Question 3: The marks 'below 10, 20, 30, 40, 50, 60' have cumulative frequencies 6, 15, 29, 41, 60, 70. The median mark is:
(a) 30
(b) 32.5
(c) 35
(d) 40
Show Answer & Explanation

Answer: (c) 35

Explanation:
1. Class frequencies 6, 9, 14, 12, 19, 10. N/2 = 35, so the median class is 30-40 (cumulative 41). Median = 30 + 10(35 − 29)/12 = 35.

Teacher's Note:
1. Median class is the first with cf above N/2.

Question 4: The mean of the distribution x: 3, 5, 7, 9, 11, 13 with frequencies 6, 8, 15, p, 8, 4 is 7.5. The value of p is:
(a) 2
(b) 3
(c) 4
(d) 5
Show Answer & Explanation

Answer: (b) 3

Explanation:
1. Σf = 41 + p, Σfx = 303 + 9p. 7.5(41 + p) = 303 + 9p gives 307.5 + 7.5p = 303 + 9p, so 1.5p = 4.5 and p = 3.

Teacher's Note:
1. Write mean = Σfx / Σf and solve.

Question 5: The weekly wages of workers: 40-43 (31), 43-46 (58), 46-49 (60), 49-52 (missing), 52-55 (27). The mean is 47.2. The missing frequency is:
(a) 40
(b) 44
(c) 48
(d) 36
Show Answer & Explanation

Answer: (b) 44

Explanation:
1. With A = 47.5 and h = 3, Σfu = f − 66 and N = 176 + f. Mean = 47.5 + 3(f − 66)/(176 + f) = 47.2, so 3(f − 66)/(176 + f) = −0.3, that is f − 66 = −0.1(176 + f). Then 1.1f = 48.4 and f = 44.

Teacher's Note:
1. Use the step deviation formula with the unknown f.

Question 6: The mean of x: 2, 4, 6, 10, p + 5 with frequencies 3, 2, 3, 1, 2 is 6. The value of p is:
(a) 7
(b) 5
(c) 6
(d) 8
Show Answer & Explanation

Answer: (a) 7

Explanation:
1. Σf = 11. Σfx = 6 + 8 + 18 + 10 + 2(p + 5) = 52 + 2p. 52 + 2p = 66 gives p = 7.

Teacher's Note:
1. Mean = Σfx / Σf.

Question 7: The weekly cost of living index over 52 weeks: 140-150 (5), 150-160 (10), 160-170 (20), 170-180 (9), 180-190 (6), 190-200 (2). The mean index is about:
(a) 165.00
(b) 168.20
(c) 170.5
(d) 166.35
Show Answer & Explanation

Answer: (d) 166.35

Explanation:
1. Class marks 145, 155, ..., 195. Σfx = 725 + 1550 + 3300 + 1575 + 1110 + 390 = 8650. Mean = 8650/52 = 166.35.

Teacher's Note:
1. Use class marks.

Question 8: The marks distribution (0-10: 10, 10-20: b₁, 20-30: 25, 30-40: 30, 40-50: b₂, 50-60: 10) has N = 100 and median 32. The value of b₁ is:
(a) 16
(b) 10
(c) 9
(d) 12
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. b₁ + b₂ = 25. The median class is 30-40 with cf = 35 + b₁. 32 = 30 + 10(50 − 35 − b₁)/30, so 6 = 15 − b₁ and b₁ = 9 (then b₂ = 16).

Teacher's Note:
1. Use the median formula.

Question 9: For the same distribution the value of b₂ is:
(a) 9
(b) 15
(c) 16
(d) 25
Show Answer & Explanation

Answer: (c) 16

Explanation:
1. b₁ + b₂ = 25 and b₁ = 9 so b₂ = 16.

Teacher's Note:
1. Total frequency is 100.

Question 10: The mode of the series 0-10 (8), 10-20 (10), 20-30 (missing), 30-40 (16), 40-50 (12), 50-60 (6), 60-70 (7) is 36. The missing frequency is:
(a) 12
(b) 10
(c) 8
(d) 14
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. The modal class is 30-40, f₁ = 16, f₀ = x, f₂ = 12. 36 = 30 + 10(16 − x)/(32 − x − 12), so 0.6(20 − x) = 16 − x, 12 − 0.6x = 16 − x, 0.4x = 4 and x = 10.

Teacher's Note:
1. Use the mode formula.

Question 11: The classes 0-10, 10-20, ..., 60-70 have frequencies 5, 15, 20, 23, 17, 11, 9. The median is about:
(a) 34.35
(b) 35
(c) 32.5
(d) 38
Show Answer & Explanation

Answer: (a) 34.35

Explanation:
1. N = 100, N/2 = 50. Cumulative: 5, 20, 40, 63 so the median class is 30-40: median = 30 + 10(50 − 40)/23 = 34.35. (An ogive gives about 35.)

Teacher's Note:
1. Median class is 30-40.

Question 12: The cumulative table 'below 10: 5, below 20: 9, below 30: 17, below 40: 29, below 50: 45, below 60: 60, below 70: 70, below 80: 78, below 90: 83, below 100: 85'. The mean mark is about:
(a) 45.6
(b) 50.2
(c) 52.0
(d) 48.41
Show Answer & Explanation

Answer: (d) 48.41

Explanation:
1. Class frequencies: 5, 4, 8, 12, 16, 15, 10, 8, 5, 2; N = 85. Σfx = 4115. Mean = 4115/85 = 48.41.

Teacher's Note:
1. Convert cumulative to class frequencies.

Question 13: Ages 16-18, 18-20, 20-22, 22-24, 24-26 have frequencies 50, 78, 46, 28, 23 (Group A) and 54, 89, 40, 25, 17 (Group B). The modal age of Group A is about:
(a) 18.83 years
(b) 19.2 years
(c) 18.5 years
(d) 18.93 years
Show Answer & Explanation

Answer: (d) 18.93 years

Explanation:
1. The modal class is 18-20 with f₁ = 78, f₀ = 50, f₂ = 46. Mode = 18 + 2(28)/(156 − 96) = 18.93.

Teacher's Note:
1. Use the mode formula.

Question 14: The median of the distribution 0-8 (8), 8-16 (10), 16-24 (x), 24-32 (24), 32-40 (y), 40-48 (7), with total frequency 80, is 26. The value of x is:
(a) 15
(b) 18
(c) 16
(d) 10
Show Answer & Explanation

Answer: (c) 16

Explanation:
1. x + y = 31. The median class is 24-32 with cf = 18 + x. 26 = 24 + 8(40 − 18 − x)/24, so 6 = 22 − x and x = 16 (then y = 15).

Teacher's Note:
1. Use the median formula.

Question 15: For the same distribution the value of y is:
(a) 16
(b) 15
(c) 14
(d) 21
Show Answer & Explanation

Answer: (b) 15

Explanation:
1. x = 16 and x + y = 31, so y = 15.

Teacher's Note:
1. Use the total frequency.

Question 16: For the same 25 households (4, 5, 12, 2, 2 over 100-150, ..., 300-350), the mean daily expenditure is:
(a) Rs. 211.00
(b) Rs. 220.59
(c) Rs. 215.50
(d) Rs. 205.00
Show Answer & Explanation

Answer: (a) Rs. 211.00

Explanation:
1. Class marks 125, 175, 225, 275, 325. Σfx = 500 + 875 + 2700 + 550 + 650 = 5275. Mean = 5275/25 = 211.

Teacher's Note:
1. Use class marks.

Question 17: The numbers of students with marks less than 10, 20, 30, 40, 50 are 7, 19, 32, 42, 50. The mean mark is:
(a) 25
(b) 24
(c) 26.5
(d) 22.5
Show Answer & Explanation

Answer: (a) 25

Explanation:
1. Class frequencies: 7, 12, 13, 10, 8. Σfx = 35 + 180 + 325 + 350 + 360 = 1250. Mean = 1250/50 = 25.

Teacher's Note:
1. Convert cumulative to class frequencies.

Question 18: For the same 200 students, the weight above which the heaviest 30% of the students fall is:
(a) 60 kg
(b) 62.5 kg
(c) 70 kg
(d) 65 kg
Show Answer & Explanation

Answer: (d) 65 kg

Explanation:
1. 30% of 200 = 60 students. From the top: 9 + 20 + 31 = 60 students weigh 65 kg or more.

Teacher's Note:
1. Count from the top.

Question 19: Number of accidents x = 0, 1, 2, 3, 4, 5 with frequencies 46, f₁, f₂, 25, 10, 5 (total 200) and mean 1.46. The value of f₂ is:
(a) 76
(b) 30
(c) 38
(d) 45
Show Answer & Explanation

Answer: (c) 38

Explanation:
1. f₁ + f₂ = 114. Σfx = f₁ + 2f₂ + 75 + 40 + 25 = 292, so f₁ + 2f₂ = 152 and f₂ = 38 (f₁ = 76).

Teacher's Note:
1. Two equations.

Question 20: If a set of data contains zero as an observation, which measure of central tendency is NOT appropriate?
(a) Arithmetic mean
(b) Geometric mean
(c) Median
(d) Mode
Show Answer & Explanation

Answer: (b) Geometric mean

Explanation:
1. The geometric mean is the nth root of the product of the observations, which is 0 when any observation is 0. It then gives no useful information. The other three measures work normally.

Teacher's Note:
1. Think of the product.

Question 21: A: 2, 3, 7, 1, 3, 2, 3. B: 7, 5, 9, 12, 5, 3, 8. C: 4, 4, 11, 7, 2, 3, 4. Which statement is correct?
(a) Mean of A equals mode of C
(b) Mean, median and mode of A are the same
(c) Mean of C equals median of B
(d) Median of B equals mode of A
Show Answer & Explanation

Answer: (b) Mean, median and mode of A are the same

Explanation:
1. A: mean = 21/7 = 3, sorted 1, 2, 2, 3, 3, 3, 7 so median 3 and mode 3. C: mean 5, mode 4. B: median 7. Only the last option is true.

Teacher's Note:
1. Compute mean, median and mode of each.

Question 22: If the variate is of discrete type, the frequency distribution can be represented by:
(a) a bar diagram
(b) a scatter diagram
(c) a histogram
(d) a pie diagram
Show Answer & Explanation

Answer: (a) a bar diagram

Explanation:
1. Bar diagrams show frequencies of separate discrete values; a histogram is for continuous classes.

Teacher's Note:
1. Discrete values need separate bars.

Question 23: A group has 40 observations. The average of the first 10 is 4.5 and the average of the remaining 30 is 3.5. The average of the whole group is:
(a) \( \frac15 \)
(b) 4
(c) 8
(d) \( \frac{15}{4} \)
Show Answer & Explanation

Answer: (d) \( \frac{15}{4} \)

Explanation:
1. Total = 45 + 105 = 150. Mean = 150/40 = 15/4.

Teacher's Note:
1. Add the totals.

Question 24: In a histogram the heights of the rectangles are:
(a) always proportional to the frequencies of the classes
(b) always proportional to the cumulative frequencies
(c) proportional to the frequency densities when the class sizes are unequal
(d) proportional to the cumulative frequencies only when classes are equal
Show Answer & Explanation

Answer: (c) proportional to the frequency densities when the class sizes are unequal

Explanation:
1. Area is proportional to frequency, so with unequal widths the height must be frequency divided by width (frequency density). With equal widths it is proportional to frequency, but 'always' is false.

Teacher's Note:
1. Area, not height, represents frequency.

Question 25: The arithmetic mean of n observations is m. If two observations 0 and m are added, the new mean is:
(a) m
(b) \( \frac{m}{n+1} \)
(c) \( \frac{(n+1)m}{n+2} \)
(d) \( \frac{nm}{n+1} \)
Show Answer & Explanation

Answer: (c) \( \frac{(n+1)m}{n+2} \)

Explanation:
1. New total = nm + 0 + m = (n + 1)m over n + 2 observations.

Teacher's Note:
1. Add to the total and count.

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Practice MCQs for Class 10 Mathematics Chapter 13 Statistics

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FAQs

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