CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 13 Statistics
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Boards begin February 2027
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Statistics is in Statistics and Probability, which carries 11 of the 80 marks in the 2026-27 CBSE Maths paper
30 NCERT Exemplar questions from Chapter 13
Exercise 13.1
Exemplar Ex 13.1 Q41 mark
In the formula x̄ = a + h(Σfᵢuᵢ/Σfᵢ) for the mean of grouped frequency distribution, uᵢ equals
Step 1By definition in the step-deviation method, uᵢ = (xᵢ − a)/h, where xᵢ is the class mark, a is the assumed mean and h is the class size.
Step-by-step solution
By definition in the step-deviation method, uᵢ = (xᵢ − a)/h, where xᵢ is the class mark, a is the assumed mean and h is the class size.
In the formula x̄ = a + (Σfᵢdᵢ)/(Σfᵢ), for finding the mean of grouped data the dᵢ's are deviations from a of
Step 1[1 mark] In the assumed-mean method dᵢ = xᵢ − a, where xᵢ is the class mark (mid point) of the iᵗʰ class; so the deviations are taken from the mid points — option (C).
Step-by-step solution
[1 mark] In the assumed-mean method dᵢ = xᵢ − a, where xᵢ is the class mark (mid point) of the iᵗʰ class; so the deviations are taken from the mid points — option (C).
Consider the following frequency distribution: Class 0-5 → 13, 6-11 → 10, 12-17 → 15, 18-23 → 8, 24-29 → 11. The upper limit of the median class is
Step 1[1 mark] n = 57, so n/2 = 28.5; converting to continuous classes the cumulative frequencies are 13, 23, 38, 46, 57, so the median class is 12-17 (continuous 11.5-17.5), whose upper limit is 17.5 — option (B).
Step-by-step solution
[1 mark] n = 57, so n/2 = 28.5; converting to continuous classes the cumulative frequencies are 13, 23, 38, 46, 57, so the median class is 12-17 (continuous 11.5-17.5), whose upper limit is 17.5 — option (B).
Consider the data: Class 65-85 → 4, 85-105 → 5, 105-125 → 13, 125-145 → 20, 145-165 → 14, 165-185 → 7, 185-205 → 4. The difference of the upper limit of the median class and the lower limit of the modal class is
Step 1[1 mark] n = 67, n/2 = 33.5; cumulative frequencies 4, 9, 22, 42, ... give median class 125-145 (upper limit 145), and the maximum frequency 20 gives modal class 125-145 (lower limit 125), so the difference = 145 − 125 = 20 — option (C).
Step-by-step solution
[1 mark] n = 67, n/2 = 33.5; cumulative frequencies 4, 9, 22, 42, ... give median class 125-145 (upper limit 145), and the maximum frequency 20 gives modal class 125-145 (lower limit 125), so the difference = 145 − 125 = 20 — option (C).
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated: 13.8-14 → 2, 14-14.2 → 4, 14.2-14.4 → 5, 14.4-14.6 → 71, 14.6-14.8 → 48, 14.8-15 → 20. The number of athletes who completed the race in less than 14.6 seconds is
Step 1[1 mark] 'Less than 14.6 s' adds the frequencies of all classes up to 14.4-14.6: 2 + 4 + 5 + 71 = 82 — option (C).
Step-by-step solution
[1 mark] 'Less than 14.6 s' adds the frequencies of all classes up to 14.4-14.6: 2 + 4 + 5 + 71 = 82 — option (C).
Will the median class and modal class of a grouped data always be different? Justify your answer briefly.
Step 1No — they can coincide.
Step-by-step solution
No — they can coincide.
Example: a distribution with one class having both the maximum frequency and containing the (n/2)ᵗʰ observation will have the same class as median class and modal class.
The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.
Step 1[1 mark] No, it is not always correct.
Step-by-step solution
[1 mark] No, it is not always correct.
[1 mark] The grouped-median formula assumes the observations in each class are uniformly (equally) spaced within the class, which is generally only an approximation, so the two medians need not coincide.
Answer:Not always — the grouped-median formula assumes uniform distribution within classes, so the values can differ.
The following table gives the number of pages written by Sarika for completing her own book for 30 days: 16-18 → 1 day, 19-21 → 3, 22-24 → 4, 25-27 → 9, 28-30 → 13. Find the mean number of pages written per day.
Step 1[1 mark] Class marks (mid points) are 17, 20, 23, 26, 29 and Σfᵢ = 30.
Step-by-step solution
[1 mark] Class marks (mid points) are 17, 20, 23, 26, 29 and Σfᵢ = 30.
The daily income of a sample of 50 employees are tabulated as follows: 1-200 → 14, 201-400 → 15, 401-600 → 14, 601-800 → 7. Find the mean daily income of employees.
Step 1[1 mark] Converting to continuous classes, the class marks are 100.5, 300.5, 500.5, 700.5 and Σfᵢ = 50.
Step-by-step solution
[1 mark] Converting to continuous classes, the class marks are 100.5, 300.5, 500.5, 700.5 and Σfᵢ = 50.
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given: 100-104 → 15, 104-108 → 20, 108-112 → 32, 112-116 → 18, 116-120 → 15. Determine the mean number of seats occupied over the flights.
Step 1[1 mark] Class marks are 102, 106, 110, 114, 118 and Σfᵢ = 100.
Step-by-step solution
[1 mark] Class marks are 102, 106, 110, 114, 118 and Σfᵢ = 100.
The weights (in kg) of 50 wrestlers are recorded: 100-110 → 4, 110-120 → 14, 120-130 → 21, 130-140 → 8, 140-150 → 3. Find the mean weight of the wrestlers.
Step 1[1 mark] Class marks are 105, 115, 125, 135, 145 and Σfᵢ = 50.
Step-by-step solution
[1 mark] Class marks are 105, 115, 125, 135, 145 and Σfᵢ = 50.
The mileage (km per litre) of 50 cars of the same model was tested: 10-12 → 7, 12-14 → 12, 14-16 → 18, 16-18 → 13. Find the mean mileage. The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?
Step 1[1 mark] Class marks are 11, 13, 15, 17 and Σfᵢ = 50.
Step-by-step solution
[1 mark] Class marks are 11, 13, 15, 17 and Σfᵢ = 50.
The following table shows the cumulative frequency distribution of marks of 800 students: Below 10 → 10, Below 20 → 50, Below 30 → 130, Below 40 → 270, Below 50 → 440, Below 60 → 570, Below 70 → 670, Below 80 → 740, Below 90 → 780, Below 100 → 800. Construct a frequency distribution table for the data.
Step 1[1 mark] Each class frequency is the difference of successive 'below' cumulative frequencies.
Step-by-step solution
[1 mark] Each class frequency is the difference of successive 'below' cumulative frequencies.
Form the frequency distribution table from the following data: More than or equal to 0 → 34, ≥10 → 32, ≥20 → 30, ≥30 → 27, ≥40 → 23, ≥50 → 17, ≥60 → 11, ≥70 → 6, ≥80 → 4 (Marks out of 90).
Step 1[1 mark] Each class frequency is the difference of successive 'more-than-or-equal-to' cumulative frequencies.
Step-by-step solution
[1 mark] Each class frequency is the difference of successive 'more-than-or-equal-to' cumulative frequencies.
Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students (Total 50): 150-155 → freq 12, cf a; 155-160 → freq b, cf 25; 160-165 → freq 10, cf c; 165-170 → freq d, cf 43; 170-175 → freq e, cf 48; 175-180 → freq 2, cf f.
Step 1[1 mark] a = 12 (first cumulative frequency equals the first frequency); from a + b = 25, b = 13.
Step-by-step solution
[1 mark] a = 12 (first cumulative frequency equals the first frequency); from a + b = 25, b = 13.
[1 mark] c = 25 + 10 = 35; from c + d = 43, d = 8.
[1 mark] from 43 + e = 48, e = 5; and f = 48 + 2 = 50 (= total).
Answer:a = 12, b = 13, c = 35, d = 8, e = 5, f = 50.
Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class: Below 20 → 17, Below 40 → 22, Below 60 → 29, Below 80 → 37, Below 100 → 50. Form the frequency distribution table for the data.
Step 1[1 mark] Each class frequency is the difference of successive 'below' cumulative frequencies.
Step-by-step solution
[1 mark] Each class frequency is the difference of successive 'below' cumulative frequencies.
In the distribution (more-than-type): >₹10000 → 100, >₹13000 → 85, >₹16000 → 69, >₹19000 → 50, >₹22000 → 33, >₹25000 → 15, the number of families having income in the range ₹16000-₹19000 is
Step 1Number in 16000-19000 range = (>16000 count) − (>19000 count) = 69 − 50 = 19.
Step-by-step solution
Number in 16000-19000 range = (>16000 count) − (>19000 count) = 69 − 50 = 19.
Daily wages (₹) of 110 workers are tabulated: 100-120 → 10, 120-140 → 15, 140-160 → 20, 160-180 → 22, 180-200 → 18, 200-220 → 12, 220-240 → 13. Compute the mean daily wage.
Step 1[1 mark] Take the class mark (midpoint) of each class as xᵢ: 110, 130, 150, 170, 190, 210, 230; the frequencies fᵢ are 10, 15, 20, 22, 18, 12, 13 with Σfᵢ = 110.
Step-by-step solution
[1 mark] Take the class mark (midpoint) of each class as xᵢ: 110, 130, 150, 170, 190, 210, 230; the frequencies fᵢ are 10, 15, 20, 22, 18, 12, 13 with Σfᵢ = 110.
[1 mark] Direct-method formula: mean x̄ = Σfᵢxᵢ / Σfᵢ.
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Questions from the NCERT Class 10 Mathematics Exemplar Problems; the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.