Feb/Mar 2018 p62 q4
2983
The discrete random variable X has the following probability distribution.
| x | -2 | 0 | 1 | 3 | 4 |
|---|
| P(X = x) | 0.2 | 0.1 | p | 0.1 | q |
(i) Given that \(E(X) = 1.7\), find the values of \(p\) and \(q\).
(ii) Find \(\text{Var}(X)\).
Solution
(i) The sum of probabilities must equal 1: \(0.2 + 0.1 + p + 0.1 + q = 1\). Simplifying gives \(p + q = 0.6\).
Given \(E(X) = 1.7\), we have \(\Sigma px = -0.4 + 0 + p + 0.3 + 4q = 1.7\). Simplifying gives \(p + 4q = 1.8\).
Solving the simultaneous equations \(p + q = 0.6\) and \(p + 4q = 1.8\), we find \(p = 0.2\) and \(q = 0.4\).
(ii) The variance is given by \(\text{Var}(X) = \Sigma x^2p - (E(X))^2\).
Calculate \(\Sigma x^2p = 4 \times 0.2 + 1 \times p + 9 \times 0.1 + 16 \times q\).
Substitute \(p = 0.2\) and \(q = 0.4\) to get \(4 \times 0.2 + 1 \times 0.2 + 9 \times 0.1 + 16 \times 0.4 = 8.3\).
Then \(\text{Var}(X) = 8.3 - 1.7^2 = 8.3 - 2.89 = 5.41\).
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