In a probability distribution the random variable \(X\) takes the value \(x\) with probability \(kx\), where \(x\) takes values 1, 2, 3, 4, 5 only.
- Draw up a probability distribution table for \(X\), in terms of \(k\), and find the value of \(k\).
- Find \(E(X)\).
Solution
(i) The probability distribution table for \(X\) is:
| \(x\) | 1 | 2 | 3 | 4 | 5 |
| Prob | \(k\) | \(2k\) | \(3k\) | \(4k\) | \(5k\) |
Since the sum of all probabilities must equal 1, we have:
\(k + 2k + 3k + 4k + 5k = 1\)
\(15k = 1\)
\(k = \frac{1}{15}\) (0.0667)
(ii) To find \(E(X)\), we calculate:
\(E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 3k + 4 \cdot 4k + 5 \cdot 5k\)
\(= k + 4k + 9k + 16k + 25k\)
\(= 55k\)
Substituting \(k = \frac{1}{15}\), we get:
\(E(X) = 55 \times \frac{1}{15} = \frac{55}{15} = \frac{11}{3}\) (3.67)
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