The random variable X takes the values −2, 1, 2, 3. It is given that \(P(X = x) = kx^2\), where \(k\) is a constant.
- Draw up the probability distribution table for X, giving the probabilities as numerical fractions.
- Find \(E(X)\) and \(\text{Var}(X)\).
Solution
To find the probability distribution, we first determine \(k\) by ensuring the probabilities sum to 1:
\(k(-2)^2 + k(1)^2 + k(2)^2 + k(3)^2 = 1\)
\(4k + k + 4k + 9k = 18k = 1\)
\(k = \frac{1}{18}\)
Thus, the probability distribution table is:
| x | -2 | 1 | 2 | 3 |
|---|
| P(X=x) | \(\frac{4}{18}\) | \(\frac{1}{18}\) | \(\frac{4}{18}\) | \(\frac{9}{18}\) |
To find \(E(X)\):
\(E(X) = \frac{4 \times (-2) + 1 \times 1 + 4 \times 2 + 9 \times 3}{18}\)
\(= \frac{-8 + 1 + 8 + 27}{18}\)
\(= \frac{28}{18} = \frac{14}{9}\)
To find \(\text{Var}(X)\):
\(\text{Var}(X) = \frac{4 \times (-2)^2 + 1 \times 1^2 + 4 \times 2^2 + 9 \times 3^2}{18} - \left(\frac{14}{9}\right)^2\)
\(= \frac{16 + 1 + 16 + 81}{18} - \left(\frac{14}{9}\right)^2\)
\(= \frac{114}{18} - \frac{196}{81}\)
\(= \frac{317}{81}\)
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