JEE Main 2021 · 1 September · evening · Numerical answer · +4 / −0
Let X be a random variable with distribution.
If the mean of X is 2.3 and variance of X is σ², then 100 σ² is equal to :
| x | -2 | -1 | 3 | 4 | 6 |
|---|---|---|---|---|---|
| P(X = x) | 1/5 | a | 1/3 | 1/5 | b |
Answer and solution
Answer 781
Solution
Maths sir · 4 steps
Two unknowns need two facts: the probabilities add to 1, and the mean is 2.3. Find a and b first.
- Total probability. All probabilities add to 1, so 1/5 + a + 1/3 + 1/5 + b = 1. Since 1/5+1/3+1/5 = 11/15, this gives a + b = 4/15.
- Use the mean. E(X) = (-2)1/5 + (-1)a + 3 × 1/3 + 4 × 1/5 + 6b = 1.4 - a + 6b = 2.3, so -a + 6b = 9/10. Adding this to a + b = 4/15 gives 7b = 7/6, so b = 1/6 and a = 1/10.
σ² = E(X²) - (E(X))²
- Find E(X²). E(X²) = 4 × 1/5 + 1 × 1/10 + 9 × 1/3 + 16 × 1/5 + 36 × 1/6 = 0.8 + 0.1 + 3 + 3.2 + 6 = 13.1.
- So. σ² = 13.1 - (2.3)² = 13.1 - 5.29 = 7.81, so 100σ² = 781.
The common slip: 1310. Stopping at E(X²) and forgetting to subtract (2.3)² gives 1310, not 781.
The idea to keep. Missing probabilities in a table come from two equations: the probabilities sum to 1, and the given mean. Then variance is the mean of the squares minus the square of the mean.
Answer from the NTA final answer key. Solution by PrepWiser.