Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2017 p62 q3
2915
In a probability distribution the random variable \(X\) takes the value \(x\) with probability \(kx^2\), where \(k\) is a constant and \(x\) takes values \(-2, -1, 2, 4\) only.
Show that \(P(X = -2)\) has the same value as \(P(X = 2)\).
Draw up the probability distribution table for \(X\), in terms of \(k\), and find the value of \(k\).
Find \(E(X)\).
Solution
(i) The probability \(P(X = -2) = k(-2)^2 = 4k\) and \(P(X = 2) = k(2)^2 = 4k\). Thus, \(P(X = -2) = P(X = 2)\).