CBSE Class 10 Mathematics Statistics MCQs Set 15

Download CBSE MCQs for Class 10 Mathematics: Chapter 13 Statistics

Review structured MCQ sets for Class 10 Mathematics Chapter 13 Statistics. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chapter 13 Statistics

Navigate directly to the 50 objective questions for Chapter 13 Statistics using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The mean marks of 100 students were found to be 40. Later it was discovered that a score of 56 had been misread as 83. The correct mean is:
(a) 39.73
(b) 39.70
(c) 40.27
(d) 40.00
Show Answer & Explanation

Answer: (a) 39.73

Explanation:
1. Wrong total = 100 x 40 = 4000. Correct total = 4000 - 83 + 56 = 3973. Correct mean = 3973/100 = 39.73.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 2: The mean of the distribution x: 5, 8, 10, 12, p, 20, 25 with frequencies 2, 5, 8, 22, 7, 4, 2 is 12.58. The value of p is:
(a) 12
(b) 14
(c) 18
(d) 15
Show Answer & Explanation

Answer: (d) 15

Explanation:
1. Σf = 50, so Σfx = 12.58 x 50 = 629. Known part: 10 + 40 + 80 + 264 + 80 + 50 = 524. Then 7p = 105 and p = 15.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 3: Class 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, 70-80 have frequencies 6, 8, 13, 7, 3, 2, 1. The mean is:
(a) 34.25
(b) 36.50
(c) 35.75
(d) 38.75
Show Answer & Explanation

Answer: (c) 35.75

Explanation:
1. Class marks 15, 25, ..., 75 give Σfx = 90 + 200 + 455 + 315 + 165 + 130 + 75 = 1430. Σf = 40. Mean = 1430/40 = 35.75.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 4: For the data x = 15, 20, 22, 24, 25, 30, 33, 38 with frequencies 5, 8, 11, 20, 23, 18, 13, 2, taking an assumed mean of 25, the mean is:
(a) 24.23
(b) 25.77
(c) 26.77
(d) 25.00
Show Answer & Explanation

Answer: (b) 25.77

Explanation:
1. Deviations from 25: -10, -5, -3, -1, 0, 5, 8, 13. Σfd = -50 - 40 - 33 - 20 + 0 + 90 + 104 + 26 = 77, Σf = 100. Mean = 25 + 77/100 = 25.77.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 5: Classes 10-15, 15-20, 20-25, 25-30, 30-35, 35-40 have frequencies 5, 6, 8, 12, 6, 3. By the step-deviation method with a = 27.5 and h = 5, the mean is:
(a) 26.125
(b) 24.625
(c) 30.375
(d) 23.125
Show Answer & Explanation

Answer: (b) 24.625

Explanation:
1. u = -3, -2, -1, 0, 1, 2. Σfu = -15 - 12 - 8 + 0 + 6 + 6 = -23, Σf = 40. Mean = 27.5 + 5(-23/40) = 27.5 - 2.875 = 24.625.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 6: The mean of the frequency distribution 0-20, 20-40, 40-60, 60-80, 80-100, 100-120 with frequencies 5, f1, 10, f2, 7, 8 is 62.8 and the total frequency is 50. The values of f1 and f2 are:
(a) f1 = 8, f2 = 12
(b) f1 = 12, f2 = 8
(c) f1 = 10, f2 = 10
(d) f1 = 6, f2 = 14
Show Answer & Explanation

Answer: (a) f1 = 8, f2 = 12

Explanation:
1. f1 + f2 = 50 - 30 = 20. Using class marks 10, 30, 50, 70, 90, 110: 50 + 30f1 + 500 + 70f2 + 630 + 880 = 62.8 x 50 = 3140, so 30f1 + 70f2 = 1080 and 3f1 + 7f2 = 108. With f1 = 20 - f2: 60 + 4f2 = 108, so f2 = 12 and f1 = 8.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 7: The number of students scoring below 10, 20, ..., 100 marks is 5, 9, 17, 29, 45, 60, 70, 78, 83, 85 respectively. The mean marks are:
(a) 45.00
(b) 47.41
(c) 50.12
(d) 48.41
Show Answer & Explanation

Answer: (d) 48.41

Explanation:
1. The class frequencies are 5, 4, 8, 12, 16, 15, 10, 8, 5, 2. With a = 45 and h = 10, Σfu = 29 and Σf = 85. Mean = 45 + 10(29/85) = 48.41.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 8: The median of the lives (in hours) of 15 components: 715, 724, 725, 710, 729, 745, 649, 699, 696, 712, 734, 728, 716, 705, 719 is:
(a) 715
(b) 719
(c) 716
(d) 712
Show Answer & Explanation

Answer: (c) 716

Explanation:
1. Sorted: 649, 696, 699, 705, 710, 712, 715, 716, 719, 724, 725, 728, 729, 734, 745. N = 15, so the median is the 8th value = 716.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 9: The daily wages (Rs.) 125, 130, 135, 140, 145, 150, 160, 180 are earned by 6, 20, 24, 28, 15, 4, 2, 1 workers. The median wage is:
(a) 135.00 Rs.
(b) 140.00 Rs.
(c) 137.50 Rs.
(d) 138.50 Rs.
Show Answer & Explanation

Answer: (c) 137.50 Rs.

Explanation:
1. N = 100. Cumulative frequencies: 6, 26, 50, 78, ... The 50th value is 135 and the 51st is 140. Median = (135 + 140)/2 = 137.50.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 10: Classes 0-10, 10-20, ..., 50-60 have frequencies 5, 10, 20, 7, 8, 5. The median is:
(a) 25.00
(b) 26.25
(c) 27.50
(d) 28.75
Show Answer & Explanation

Answer: (b) 26.25

Explanation:
1. N = 55, N/2 = 27.5; cumulative frequencies 5, 15, 35, ... so the median class is 20-30. Median = 20 + ((27.5 - 15)/20) x 10 = 26.25.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 11: The mode of the data 25, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 16, 22, 16, 18, 20, 16, 19 is:
(a) 19
(b) 16
(c) 20
(d) 21
Show Answer & Explanation

Answer: (a) 19

Explanation:
1. 19 occurs 5 times, 16 occurs 4 times, 20 occurs 3 times and the rest fewer. The mode is 19.

Teacher's Note:
1. Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h, where f1 is the modal class frequency and f0, f2 are the frequencies of the classes before and after it.
2. Empirical relation: Mode = 3 Median - 2 Mean.

Question 12: Ages (years) 5-14, 15-24, 25-34, 35-44, 45-54, 55-64 have frequencies 6, 11, 21, 23, 14, 5. The mode, correct to 2 decimal places, is:
(a) 34.50 years
(b) 38.73 years
(c) 40.12 years
(d) 36.32 years
Show Answer & Explanation

Answer: (d) 36.32 years

Explanation:
1. Convert to continuous classes: modal class 34.5-44.5, f1 = 23, f0 = 21, f2 = 14, h = 10. Mode = 34.5 + (2/11) x 10 = 36.32.

Teacher's Note:
1. Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h, where f1 is the modal class frequency and f0, f2 are the frequencies of the classes before and after it.
2. Empirical relation: Mode = 3 Median - 2 Mean.

Question 13: Daily wages in classes 31-36, 37-42, 43-48, 49-54, 55-60, 61-66 have frequencies 6, 12, 20, 15, 9, 4. The mode is:
(a) 45.5
(b) 47.8
(c) 49.2
(d) 46.2
Show Answer & Explanation

Answer: (d) 46.2

Explanation:
1. Continuous classes: 30.5-36.5, ... The modal class is 42.5-48.5 with f1 = 20, f0 = 12, f2 = 15, h = 6. Mode = 42.5 + (8/13) x 6 = 46.19, about 46.2.

Teacher's Note:
1. Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h, where f1 is the modal class frequency and f0, f2 are the frequencies of the classes before and after it.
2. Empirical relation: Mode = 3 Median - 2 Mean.

Question 14: The marks of 400 examinees in classes 400-450, 450-500, 500-550, 550-600, 600-650, 650-700, 700-750, 750-800 have frequencies 30, 45, 60, 52, 54, 67, 45, 47. The number of examinees scoring 700 and above is:
(a) 308
(b) 67
(c) 92
(d) 45
Show Answer & Explanation

Answer: (c) 92

Explanation:
1. 45 + 47 = 92 (equal to 400 - 308, where 308 is the 'less than 700' cumulative frequency).

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 15: For the marks classes 400-450, ..., 750-800 with frequencies 30, 45, 60, 52, 54, 67, 45, 47, the number of examinees scoring less than 650 is:
(a) 187
(b) 241
(c) 308
(d) 159
Show Answer & Explanation

Answer: (b) 241

Explanation:
1. 30 + 45 + 60 + 52 + 54 = 241.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 16: Marks 0-10, 10-20, ..., 50-60 have frequencies 7, 10, 23, 51, 6, 3. The median marks (formula) are:
(a) 31.96
(b) 34.00
(c) 35.00
(d) 30.00
Show Answer & Explanation

Answer: (a) 31.96

Explanation:
1. N = 100, N/2 = 50; cumulative frequencies 7, 17, 40, 91, so the median class is 30-40. Median = 30 + ((50 - 40)/51) x 10 = 31.96.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 17: The median of the numbers 520, 20, 340, 190, 35, 800, 1210, 50, 80 is:
(a) 190
(b) 520
(c) 35
(d) 340
Show Answer & Explanation

Answer: (a) 190

Explanation:
1. Sorted: 20, 35, 50, 80, 190, 340, 520, 800, 1210. N = 9, so the median is the 5th value = 190.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 18: If the arithmetic mean of 5, 7, 9 and x is 9, then x is:
(a) 11
(b) 18
(c) 16
(d) 15
Show Answer & Explanation

Answer: (d) 15

Explanation:
1. Sum = 4 x 9 = 36. 5 + 7 + 9 + x = 36 gives x = 15.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 19: The mode of the numbers 3, 5, 7, 4, 2, 1, 4, 3, 4 is:
(a) 3
(b) 7
(c) 4
(d) 1
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. 4 appears three times, 3 appears twice and the others once. The mode is 4.

Teacher's Note:
1. Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h, where f1 is the modal class frequency and f0, f2 are the frequencies of the classes before and after it.
2. Empirical relation: Mode = 3 Median - 2 Mean.

Question 20: Of the ten observations x1, ..., x10, the first five are each increased by 5 and the remaining five are each decreased by 5. The mean:
(a) increases by 5
(b) does not change
(c) decreases by 5
(d) increases by 25
Show Answer & Explanation

Answer: (b) does not change

Explanation:
1. The total changes by 5 x 5 - 5 x 5 = 0, so the mean is unchanged.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 21: The mean and median of a set of data are 8.9 and 9. Using the empirical relation, the mode is:
(a) 7.2
(b) 9.2
(c) 8.2
(d) 10.2
Show Answer & Explanation

Answer: (b) 9.2

Explanation:
1. Mode = 3 Median - 2 Mean = 27 - 17.8 = 9.2.

Teacher's Note:
1. Mode = l + ((f1 - f0)/(2f1 - f0 - f2)) x h, where f1 is the modal class frequency and f0, f2 are the frequencies of the classes before and after it.
2. Empirical relation: Mode = 3 Median - 2 Mean.

Question 22: The mean of the distribution x = 15, 17, 19, 20 + p, 23 with frequencies 2, 3, 4, 5p, 6 is 20. The (positive) value of p is:
(a) 1
(b) 2
(c) 3
(d) 5
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. Σf = 15 + 5p and Σfx = 295 + 100p + 5p². Setting Σfx = 20(15 + 5p) = 300 + 100p gives 5p² = 5, so p = 1 (p = -1 would give a negative frequency).

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 23: The workers' data 50-70, 70-90, 90-110, 110-130, 130-150, 150-170 have frequencies 18, 12, 13, 27, 8, 22. The mean is:
(a) 110.00
(b) 115.50
(c) 100.00
(d) 112.20
Show Answer & Explanation

Answer: (d) 112.20

Explanation:
1. Class marks 60, 80, ..., 160. Σfx = 1080 + 960 + 1300 + 3240 + 1120 + 3520 = 11220 and Σf = 100. Mean = 112.20.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 24: Monthly rents (Rs.) 15-25, 25-35, ..., 85-95 are paid by 8, 10, 15, 25, 40, 20, 15, 7 households. The median rent is:
(a) 55 Rs.
(b) 60 Rs.
(c) 58 Rs.
(d) 65 Rs.
Show Answer & Explanation

Answer: (c) 58 Rs.

Explanation:
1. N = 140, N/2 = 70. Cumulative frequencies: 8, 18, 33, 58, 98, so the median class is 55-65. Median = 55 + ((70 - 58)/40) x 10 = 58.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

Question 25: In a distribution of the number of accidents 0, 1, 2, 3, 4, 5 the frequencies (days) are 46, f1, f2, 25, 10, 5 with total 200 and mean 1.46. The values of f1 and f2 are:
(a) f1 = 38, f2 = 76
(b) f1 = 66, f2 = 48
(c) f1 = 76, f2 = 38
(d) f1 = 80, f2 = 34
Show Answer & Explanation

Answer: (c) f1 = 76, f2 = 38

Explanation:
1. f1 + f2 = 200 - 86 = 114. Σfx = f1 + 2f2 + 75 + 40 + 25 = 1.46 x 200 = 292, so f1 + 2f2 = 152. Subtracting: f2 = 38, f1 = 76.

Teacher's Note:
1. Mean = Σ f x / Σ f. For grouped data use the class mark x = (lower + upper)/2. Step-deviation: mean = a + h (Σ f u / Σ f), where u = (x - a)/h.
2. When one observation is corrected, correct the total first (total = mean x number of items) and then divide again.

Question 26: The classes 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 have frequencies 5, f1, 20, 15, f2, 5 with total 60 and median 28.5. The values of f1 and f2 are:
(a) f1 = 7, f2 = 8
(b) f1 = 8, f2 = 7
(c) f1 = 10, f2 = 5
(d) f1 = 5, f2 = 10
Show Answer & Explanation

Answer: (b) f1 = 8, f2 = 7

Explanation:
1. f1 + f2 = 15. The median class is 20-30: 28.5 = 20 + ((30 - 5 - f1)/20) x 10 gives (25 - f1)/2 = 8.5, so f1 = 8 and f2 = 7.

Teacher's Note:
1. Median = l + ((N/2 - cf)/f) x h, where l is the lower limit of the median class, cf the cumulative frequency before it, f its frequency and h its width.
2. The median class is the class that contains the (N/2)th observation.

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FAQs

Where can I access latest CBSE Class 10 Mathematics Statistics MCQs Set 15?

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