- 2072 Chaitra · 5 marks
Expectation of discrete variable; P(X≥4)
Short answer
Expectation is the probability-weighted average. From the table: P(X≥4)=0.36, P(0<X<4)=0.38, mean=2.08, variance=3.2936.
Define mathematical expectation of a discrete random variable. A probability distribution is given.
| X=x | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| p(X=x) | 0.26 | 0.25 | 0.11 | 0.02 | 0.25 | 0.11 |
| Find (a) P(X ≥ 4); (b) P(0 < X < 4); (c) mean and variance of X. |
Mathematical expectation
For a discrete random variable X with pmf p(x), the mathematical expectation (mean) is the probability-weighted average of its values:
E(X) = Σ x · p(x)
and E[g(X)] = Σ g(x) p(x). It is the long-run average value of X.
The given probabilities add to 0.26 + 0.25 + 0.11 + 0.02 + 0.25 + 0.11 = 1, so it is a valid distribution.
(a) P(X ≥ 4)
P(X ≥ 4) = P(4) + P(5) = 0.25 + 0.11 = 0.36
(b) P(0 < X < 4)
P(1) + P(2) + P(3) = 0.25 + 0.11 + 0.02 = 0.38
(c) Mean and variance
x p(x) x·p(x) x²·p(x)
0 0.26 0.00 0.00
1 0.25 0.25 0.25
2 0.11 0.22 0.44
3 0.02 0.06 0.18
4 0.25 1.00 4.00
5 0.11 0.55 2.75
Sum 1.00 2.08 7.62
Mean = E(X) = 2.08
Variance = E(X²) − [E(X)]² = 7.62 − (2.08)² = 7.62 − 4.3264 = 3.2936
Answer: (a) 0.36, (b) 0.38, (c) mean = 2.08, variance = 3.2936
✓ Answer checked by Claude Sonnet 5.5
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