Mean Deviation from Mean and Variance – Statistics JEE Main 2026 PYQ
Quick Summary
- Question Type:
- Statistics (back-calculation from moments)
- Chapter:
- Statistics
- Difficulty:
- ⭐⭐⭐ Medium
- Time to Solve:
- 2 minutes
- Key Concept:
- Σx = 20 and Σx² = 4(4 + 25) = 116; the unknown pair satisfies u + v = 3 and u² + v² = 7 in deviations, hence |u| + |v| = 3 and MD = (3 + 0 + 3)/4 = 1.50
- Correct Answer:
- Option (2) – 1.50
- Why:
- The two unknown observations have deviations +((3+√5)/2) and +((3−√5)/2), both positive, so their absolute deviations sum to exactly 3
The Question
JEE Main 2026 (8 April – Evening Shift) – Statistics
A set of four observations has mean equal to 5 and variance equal to
4. If two of the observations are 2 and 5, then the mean
deviation about the mean of all four observations is:
Quick Answer
Correct Option: (2) – 1.50
Reasoning: work in deviations from the mean and the unknown pair never needs to be solved explicitly:
- Σx = 4 × 5 = 20 and Σx² = n(σ² + x̄²) = 4(4 + 25) = 116
- Unknown pair: a + b = 20 − 7 = 13, a² + b² = 116 − 29 = 87
- In deviations u = a − 5, v = b − 5: u + v = 3 and u² + v² = 87 − 10·13 + 50 = 7 → uv = (9 − 7)/2 = 1 > 0, so u, v are both positive (their sum is 3)
- MD = (|2−5| + |5−5| + |u| + |v|)/4 = (3 + 0 + 3)/4 = 1.50 ✓
Understanding the Concept
Rebuilding the Data from Moments
Mean and variance are the first two moments of the data, and together they pin down two
aggregate facts: Σx = n·x̄ = 20 and Σx² = n(σ² + x̄²) = 4(4 + 25) = 116. The identity
σ² = (Σx²/n) − x̄² rearranged as Σx² = n(σ² + x̄²) is the workhorse — students who remember
only σ² = Σx²/n (dropping the mean term) get Σx² = 16 and the question collapses into
nonsense. Once the two totals are known, the two unobserved observations a and b satisfy a
pair of symmetric equations: a + b = 13 and a² + b² = 87.
The Deviation Trick — Solve Without Solving
The goal is the mean deviation MD = Σ|x − x̄|/4, which lives entirely in deviations
u = a − 5, v = b − 5. Translating the pair of equations into deviations: u + v = 3 and
u² + v² = a² + b² − 10(a + b) + 50 = 87 − 130 + 50 = 7. Then uv = ((u + v)² − (u² + v²))/2 =
(9 − 7)/2 = 1. Now the sign question resolves itself: uv > 0 means u and v share a sign, and
u + v = 3 > 0 forces that shared sign to be positive. Therefore |u| + |v| = u + v = 3
— the absolute values evaporate without ever computing a and b. The mean deviation is
(|−3| + |0| + 3)/4 = 6/4 = 1.50. (For the record, the pair is a, b = 5 + (3 ± √5)/2 =
(13 ± √5)/2 — irrational, as the discriminant 5 confirms — which is exactly why solving
explicitly is wasted effort.)
Why the Answer Is Immune to the Actual Pair
Mean deviation depends on the observations only through |deviation| sums. Since the unknown
deviations are guaranteed positive with a fixed sum 3, their individual values —
(3 + √5)/2 ≈ 2.618 and (3 − √5)/2 ≈ 0.382 — are irrelevant to the answer. This is a designed
elegance: the examiner chose data whose absolute-value signs all cooperate, so the answer is a
clean 1.50. Had uv been negative, one deviation would be negative and |u| + |v| would exceed
|u + v| — then the actual (irrational) values would matter, and the question would turn
considerably uglier. Checking the sign of uv before dropping the modulus bars is the
professional move.
Moments Checklist for Every Such Question
Statistics back-calculation questions all run the same four-beat rhythm: (1) write Σx from the
mean; (2) write Σx² from the variance using n(σ² + x̄²); (3) subtract the known observations
to get symmetric equations for the unknowns; (4) express the requested statistic in terms of
the unknowns’ sums and products — usually via (a + b)² = a² + b² + 2ab — and finish. Whether
the ask is MD, the product ab, or the remaining pair themselves, the skeleton never changes.
Detailed Step-by-Step Solution
Step 1: Write the Two Moment Equations
- Σx = n·x̄ = 4 × 5 = 20
- Σx² = n(σ² + x̄²) = 4(4 + 25) = 116
Step 2: Reduce to the Unknown Pair
- Known observations 2 and 5 consume: sum 7, sum of squares 4 + 25 = 29
- Unknown pair: a + b = 13, a² + b² = 87
Step 3: Switch to Deviations and Fix the Signs
- u = a − 5, v = b − 5: u + v = 13 − 10 = 3; u² + v² = 87 − 130 + 50 = 7
- uv = ((u + v)² − (u² + v²))/2 = (9 − 7)/2 = 1 > 0 and u + v > 0 ⟹ u, v > 0
- So |u| + |v| = u + v = 3 (no need to solve for u, v)
Step 4: Assemble the Mean Deviation
- MD = (|2 − 5| + |5 − 5| + |u| + |v|)/4 = (3 + 0 + 3)/4 = 6/4
- MD = 1.50 — Answer: Option (2)
Final Answer
Correct Option: (2) – 1.50 ✓
The moment equations give the unknown pair a + b = 13, a² + b² = 87; in deviations from the
mean this is u + v = 3 with u² + v² = 7, so uv = 1 > 0 and both deviations are positive.
Hence |u| + |v| = 3 and MD = (3 + 0 + 3)/4 = 1.50 (Option 2).
Essential Facts for This Topic
Statistics Back-Calculation Toolkit
- Sum from mean: Σx = n·x̄
- Sum of squares from variance: Σx² = n(σ² + x̄²) — the single most forgotten +n·x̄² term in the chapter
- Symmetric pair identities: ab = ((a + b)² − (a² + b²))/2; a² + b² = (a + b)² − 2ab
- Sign resolution: uv > 0 with u + v > 0 ⟹ both positive ⟹ |u| + |v| = u + v
- Mean deviation about the mean: MD = (1/n) Σ |xᵢ − x̄| — absolute values demand the sign check first
- Reality check: for real observations the quadratic t² − (a+b)t + ab must have non-negative discriminant
Common Mistakes to Avoid
❌ Mistake 1: Using Σx² = n·σ² Without the Mean Term
Wrong Thinking: “Variance 4 with four observations means Σx² = 16.”
Correct Approach: Variance is Σ(x − x̄)²/n = Σx²/n − x̄². The identity is Σx² = n(σ² + x̄²) = 4(4 + 25) = 116, not 16. Dropping the +n·x̄² term is the single most common error in every mean-variance back-calculation question and makes the data impossible.
❌ Mistake 2: Assuming the Unknown Observations Are Nice Integers
Wrong Thinking: “The remaining two must be, say, 6 and 7 — their deviations are 1 and 2, so MD = (3 + 0 + 3)/4… wait, that still works?”
Correct Approach: Here the pair is (13 ± √5)/2 ≈ 7.62 and 5.38 — irrational, and the question never needs them. Forcing integer guesses risks wrong deviations; the deviation-equation route (u + v, u² + v²) is both faster and exact.
❌ Mistake 3: Skipping the Sign Analysis Before Dropping Modulus Bars
Wrong Thinking: “|u| + |v| = |u + v| = 3 always.”
Correct Approach: The identity |u| + |v| = |u + v| holds only when u and v share a sign. Here uv = 1 > 0 with u + v = 3 > 0 licenses it. If uv had been negative, the bars could not be dropped and the raw values would be needed — always certify the sign of uv first.
Key Concept Summary
What You Must Remember
- Moments first: Σx = 20, Σx² = n(σ² + x̄²) = 116 — never omit the n·x̄² term
- Unknown pair in deviations: u + v = 3, u² + v² = 7 → uv = 1
- Sign certificate: uv > 0 and u + v > 0 ⟹ u, v both positive ⟹ |u| + |v| = 3
- MD = (3 + 0 + 3)/4 = 1.50 — the known observation 5 contributes a zero deviation
- No solving required: the pair (13 ± √5)/2 is irrational and irrelevant — the symmetric identities finish the job
The Golden Rule for Moment Back-Calculation
“Mean hands you Σx, variance hands you Σx² through n(σ² + x̄²) — subtract what you know, then let (a + b)² = a² + b² + 2ab and a sign check do the rest.”
Frequently Asked Questions
Q1: Why is Σx² equal to n(σ² + x̄²) and not just n·σ²?
A: Variance is the mean of squared deviations: σ² = Σ(xᵢ − x̄)²/n. Expanding (xᵢ − x̄)² = xᵢ² − 2x̄xᵢ + x̄² and summing gives σ² = Σx²/n − x̄², so Σx² = n(σ² + x̄²). The x̄² term is part of the data’s raw second moment and forgetting it is the classic error — here it is the difference between 116 and an impossible 16.
Q2: What are the two unknown observations?
A: They satisfy a + b = 13 and a² + b² = 87, giving ab = 41 and the quadratic t² − 13t + 41 = 0 with roots t = (13 ± √5)/2 ≈ 7.618 and 5.382. The question deliberately never requires them — the deviation trick extracts the mean deviation from their sums alone.
Q3: How do we know both unknown deviations are positive?
A: The deviations satisfy u + v = 3 and uv = 1. A positive product means the two share a sign; a positive sum then forces that shared sign to be positive. Hence u, v > 0, and the absolute values drop without ambiguity: |u| + |v| = u + v = 3.
Q4: What would happen if the two given observations made the data impossible?
A: The remaining pair must be real, so the quadratic t² − (sum)t + (product) needs a non-negative discriminant. For example, with given observations 2 and 4 instead, the pair equations become t² − 14t + 50 = 0 whose discriminant is 196 − 200 = −4 — no real observations exist. Whenever a mean-variance dataset produces a negative discriminant, the data is inconsistent and the numbers must be re-read.
Q5: Is mean deviation about the mean the same as about the median here?
A: No. Mean deviation depends on the point about which absolute deviations are taken. About the mean (5) the answer is 1.50; about the median of the four observations ((5 + (13 − √5)/2)/2 ≈ 5.19) the sum of absolute deviations — and hence the MD — would differ. JEE always specifies the point; read it carefully.
Prerequisites to Solve This Question
Before attempting this problem, you should be comfortable with:
- Mean and variance definitions: x̄ = Σx/n and σ² = Σ(x − x̄)²/n
- The expanded moment identity: Σx² = n(σ² + x̄²) and its rearrangements
- Symmetric functions of two variables: ab from (a + b)² and a² + b²; the quadratic with given sum and product
- Absolute-value logic: when |u| + |v| equals |u + v| and when it does not
- Mean deviation definition: MD about a point A = (1/n) Σ |xᵢ − A|
After Solving This, You Can:
- Reconstruct full datasets from mean + variance + partial observations in 2 minutes
- Solve the sibling asks: find ab, find the remaining pair, or find MD about the median
- Verify dataset consistency via the discriminant test before computing anything
- Extend the routine to five observations with three unknowns (two symmetric equations + one extra given)
- Move fluently between raw moments and central moments — the translation table of statistics
Study Tips for This Topic
For JEE Main:
- Memorise Σx² = n(σ² + x̄²): write it at the top of every statistics question — it is the engine of this entire family
- Work in deviations early: shifting to u = x − x̄ shrinks the numbers and connects directly to MD, which lives in deviations
- Budget 2 minutes: moments (30s) + unknown-pair equations (30s) + sign certificate (30s) + MD assembly (30s)
Common JEE Variants:
- “Find the remaining two observations” — solve the quadratic explicitly
- Five observations, three unknown — one more given (say Σx³ or an observation) closes the system
- “Find MD about the median” — reorder the four values first; the point changes to the median
- Variance given as sample variance (÷ (n−1)) — convert to population variance before using the identity
- Combined datasets: merge two groups with known means/variances and find the new MD structure
Difficulty Rating & Exam Frequency
Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Statistics – Moment Reconstruction and Mean Deviation