JEE Main 2021 · 16 March · evening · Numerical answer · +4 / −0
Consider the statistics of two sets of observations as follows :
If the variance of the combined set of these two observations is 17/9, then the value of n is equal to ___________.
| Size | Mean | Variance | |
|---|---|---|---|
| Observation I | 10 | 2 | 2 |
| Observation II | n | 3 | 1 |
Answer and solution
Answer 5
Solution
Maths sir · 5 steps
Pooling two groups adds two things: their average spread, plus extra spread because their means differ.
- Combined variance formula. For two groups of sizes n₁, n₂, means x̄₁, x̄₂ and variances σ₁², σ₂², the combined variance is the weighted average of the variances plus a term for the gap between the means.
σ² = (n₁σ₁² + n₂σ₂²)/(n₁+n₂) + (n₁ n₂(x̄₁-x̄₂)²)/(n₁+n₂)²
- Put in numbers. Here n₁ = 10, n₂ = n, σ₁² = 2, σ₂² = 1 and x̄₁ - x̄₂ = -1. So (20+n)/(10+n) + 10n/(10+n)² = 17/9.
- Clear denominators. Multiply by 9(10+n)²: 9(n² + 40n + 200) = 17(n² + 20n + 100). This simplifies to 8n² - 20n - 100 = 0, that is 2n² - 5n - 25 = 0.
- Factorise. (2n+5)(n-5) = 0, so n = 5 or n = -5/2. A group size must be a positive integer, so reject -5/2.
- So. n = 5. Check: the combined mean is 35/15 = 7/3 and the mean of squares is (60+50)/15 = 22/3, so the variance is 66/9 - 49/9 = 17/9.
The common slip: 1.25. Dropping the mean-difference term and using only the weighted average of the variances gives n = 1.25, which is not even a whole number.
The idea to keep. Combined variance is never just the weighted average of the variances. Add (n₁ n₂ (x̄₁-x̄₂)²)/(n₁+n₂)², or work with Σ x² = n(σ² + x̄²) for each group.
Answer from the NTA final answer key. Solution by PrepWiser.