JEE Main 2024 · 9 April · evening · Single correct · +4 / −1
If the variance of the frequency distribution
is 160, then the value of c∈ N is
| x | c | 2c | 3c | 4c | 5c | 6c |
|---|---|---|---|---|---|---|
| f | 2 | 1 | 1 | 1 | 1 | 1 |
- (a) 5
- (b) 8
- (c) 6
- (d) 7
Answer and solution
Answer (d) 7
Solution
Maths sir · 4 steps
Pull c out of every value first, because variance scales by c². Then the arithmetic is easy.
- Factor out c. Every value is c times one of 1, 2, 3, 4, 5, 6, with frequencies 2, 1, 1, 1, 1, 1. Multiplying data by c multiplies the variance by c². So find the variance of u = 1, …, 6 and multiply it by c².
- Sums for u. N = 7. Σ fu = 2 + 2 + 3 + 4 + 5 + 6 = 22, so the mean is 22/7. Σ fu² = 2 + 4 + 9 + 16 + 25 + 36 = 92.
σᵤ² = 92/7 - (22/7)² = (644 - 484)/49 = 160/49
- Use the given variance. c² × 160/49 = 160, so c² = 49.
- So. Since c ∈ ℕ, c = 7, which is option (d).
The idea to keep. If every value is multiplied by k, the variance gets multiplied by k². Scale the data down to small numbers, find the variance there, then scale it back up.
Answer from the NTA final answer key. Solution by PrepWiser.