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Nov 2023 p52 q1
2963
A competitor in a throwing event has three attempts to throw a ball as far as possible. The random variable \(X\) denotes the number of throws that exceed 30 metres. The probability distribution table for \(X\) is shown below.
\(x\)
0
1
2
3
\(P(X = x)\)
0.4
\(p\)
\(r\)
0.15
Given that \(E(X) = 1.1\), find the value of \(p\) and the value of \(r\). [3]
Find the numerical value of \(\text{Var}(X)\). [2]
Solution
(a) The sum of probabilities must equal 1: \(p + r + 0.4 + 0.15 = 1\). Simplifying gives \(p + r + 0.55 = 1\), so \(p + r = 0.45\).
Given \(E(X) = 1.1\), we have \(0 \times 0.4 + 1 \times p + 2 \times r + 3 \times 0.15 = 1.1\). Simplifying gives \(p + 2r + 0.45 = 1.1\), so \(p + 2r = 0.65\).
Solving the equations \(p + r = 0.45\) and \(p + 2r = 0.65\) simultaneously, we subtract the first from the second: \((p + 2r) - (p + r) = 0.65 - 0.45\), giving \(r = 0.2\).
Substituting \(r = 0.2\) into \(p + r = 0.45\), we get \(p + 0.2 = 0.45\), so \(p = 0.25\).
(b) The variance \(\text{Var}(X)\) is calculated as \(E(X^2) - [E(X)]^2\).
