JEE (Advanced) · Mathematics, Class 11
Statistics: JEE Advanced PYQ, 2007–2026
Statistics (Mathematics, Class 11) has been asked 3 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 0% of all Mathematics questions. It is the 19th most-asked of 19 Mathematics chapters. The two 2026 papers had 1 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 0in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 0in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 0in 2 papers |
| 2014 | 0in 2 papers |
| 2015 | 0in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 0in 2 papers |
| 2019 | 0in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 0in 2 papers |
| 2022 | 0in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 1in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Mean deviation2
- Standard deviation and variance1
Try it
10 Statistics questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Mathematics · Statistics
In a set of 100 observations, 25 are equal to 4 and the remaining 75 are equal to 12. The variance of the 100 observations is
Answer: (A)
Mean = (25 × 4 + 75 × 12)/100 = 10. Σfx²/N = (25 × 16 + 75 × 144)/100 = 112. Variance = 112 − 10² = 12. As a check, the deviations are −6 and +2, so the variance is (25 × 36 + 75 × 4)/100 = 12.
Q2 · Mathematics · Statistics
How many different sets of five positive integers (repetitions allowed, order irrelevant) have mean 12, median 15 and a unique mode equal to 18?
Answer: (A)
Write the data in ascending order as a ≤ b ≤ c ≤ d ≤ e. The median is c = 15. The sum is 5 × 12 = 60. A unique mode of 18 means 18 appears at least twice and more often than any other value. Since 18 > 15, the 18s must sit at d and e, so d = e = 18. Then a + b = 60 − 15 − 36 = 9. If a = b, 18 would not be the unique mode, but 9 is odd so a ≠ b anyway. With positive integers and a < b the pairs are (1, 8), (2, 7), (3, 6), (4, 5), all with b ≤ 15. Hence 4 sets.
Q3 · Mathematics · Statistics
Six observations have a mean deviation about their mean equal to 8. A seventh observation, equal to the mean of the first six, is added. What is the mean deviation of all seven observations about their mean?
Answer: (D)
The original sum of absolute deviations is 6 x 8 = 48. The new observation equals the mean, so the mean stays the same and the new observation contributes a deviation of 0. The sum is still 48, but there are now 7 observations. The new MD = 48/7.
Q4 · Mathematics · Statistics
For ten observations x₁, x₂, …, x₁₀ it is given that Σ(xᵢ − 3)² = 100 and Σ(xᵢ − 5)² = 60. What is the variance of the observations?
Answer: (A)
For any real a, Σ(xᵢ − a)² = Σ(xᵢ − x̄)² + n(x̄ − a)² = nσ² + n(x̄ − a)². Then 100 = 10σ² + 10(x̄ − 3)² and 60 = 10σ² + 10(x̄ − 5)². Subtracting gives 40 = 10[(x̄ − 3)² − (x̄ − 5)²] = 10(4x̄ − 16), so x̄ = 5. Substituting back, 60 = 10σ² + 0, so σ² = 6. Check: 10σ² + 10(5 − 3)² = 60 + 40 = 100.
Q5 · Mathematics · Statistics
Group A has 30 observations with mean 40 and standard deviation 3. Group B has 20 observations with mean 50 and standard deviation 7. What is the standard deviation of the combined 50 observations?
Answer: (A)
The combined mean is (30·40 + 20·50)/50 = 44. The deviations of the group means from it are d₁ = −4 and d₂ = 6. The combined variance is [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂) = [30(9 + 16) + 20(49 + 36)]/50 = (750 + 1700)/50 = 49. The standard deviation is 7. The combined spread includes the variation between the group means as well as the variation within each group.
Q6 · Mathematics · Statistics
A frequency distribution has class intervals 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 4, 8, x, 12, 6 respectively. Its mean is 27. What is its variance?
Answer: (D)
The class marks are 5, 15, 25, 35, 45. Σfx = 20 + 120 + 25x + 420 + 270 = 830 + 25x, and the mean condition gives 830 + 25x = 27(30 + x), so x = 10 and N = 40. The deviations about 27 are −22, −12, −2, 8, 18, so Σf(x − 27)² = 4(484) + 8(144) + 10(4) + 12(64) + 6(324) = 1936 + 1152 + 40 + 768 + 1944 = 5840. The variance is 5840/40 = 146.
Q7 · Mathematics · Statistics
The marks of 10 students are grouped into the classes 0–10, 10–20, 20–30 and 30–40 with frequencies 1, 2, 3 and 4 respectively. Using class marks as representatives of the classes, the variance of the marks is
Answer: (A)
Class marks are 5, 15, 25, 35 with frequencies 1, 2, 3, 4 and N = 10. Mean = (5 + 30 + 75 + 140)/10 = 25. Σfx² = 25 + 450 + 1875 + 4900 = 7250, so Σfx²/N = 725. Variance = 725 − 25² = 100.
Q8 · Mathematics · Statistics
A data set of ungrouped observations has mean 10, median 8 and mode 6. Every observation x is replaced by y = ax + b, where a and b are constants. The new data have mean 23 and median 19. What is the mode of the new data?
Answer: (A)
A linear change y = ax + b moves the mean, median and mode in the same way: each new value is a times the old value plus b. From the mean, 10a + b = 23. From the median, 8a + b = 19. Subtracting gives 2a = 4, so a = 2 and b = 3. The given mode is 6, so the new mode is 2(6) + 3 = 15. The empirical relation would give 11, but it is only approximate and the actual mode is stated.
Q9 · Mathematics · Statistics
Eleven observations form an arithmetic progression with first term 4 and common difference 3. What is their mean deviation about the mean?
Answer: (B)
The mean is the sixth term. The other terms lie 3, 6, 9, 12 and 15 away from it on each side, so the absolute deviations add up to 2 × 3 × (1 + 2 + 3 + 4 + 5) = 90. Mean deviation = 90/11.
Q10 · Mathematics · Statistics
In a grouped distribution with class width 10, the classes 30–40, 40–50 and 50–60 have frequencies 10, 24 and x respectively. The class 40–50 is the modal class and the mode of the distribution is 47. Find x.
Answer: (A)
Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h, with l = 40, f₁ = 24, f₀ = 10, f₂ = x and h = 10. So 47 = 40 + 140/(38 − x), giving 140/(38 − x) = 7, so 38 − x = 20 and x = 18. Check: 18 < 24, so 40–50 is indeed the modal class.
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All 3 real JEE Advanced questions from Statistics.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Statistics in JEE Advanced.
How many questions come from Statistics in JEE Advanced?
Statistics (Mathematics, Class 11) has been asked 3 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 0% of all Mathematics questions. It is the 19th most-asked of 19 Mathematics chapters. The two 2026 papers had 1 from it.
Which topics of Statistics are asked most in JEE Advanced?
Mean deviation (2), Standard deviation and variance (1), counted over the papers from 2007–2026.
Is Statistics important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 19 of 19 in Mathematics; use that to decide how much time it gets, not whether to study it.
Where can I solve Statistics JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Statistics with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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