Feb/Mar 2021 p52 q4
2891
The random variable X takes the values 1, 2, 3, 4 only. The probability that X takes the value x is k(5 − x), where k is a constant.
(a) Draw up the probability distribution table for X, in terms of k.
\((b) Show that Var(X) = 1.05.\)
Solution
(a) The probability distribution table for X is:
To find k, use the fact that the sum of probabilities is 1:
\(4k + 6k + 6k + 4k = 1\)
\(20k = 1\)
\(k = \frac{1}{20}\)
(b) Calculate the expected value \(E(X)\):
\(E(X) = 1 \times \frac{4}{20} + 2 \times \frac{6}{20} + 3 \times \frac{6}{20} + 4 \times \frac{4}{20}\)
\(= \frac{4}{20} + \frac{12}{20} + \frac{18}{20} + \frac{16}{20}\)
\(= \frac{50}{20} = 2.5\)
Calculate the variance \(\text{Var}(X)\):
\(\text{Var}(X) = 1^2 \times \frac{4}{20} + 2^2 \times \frac{6}{20} + 3^2 \times \frac{6}{20} + 4^2 \times \frac{4}{20} - (2.5)^2\)
\(= \left(\frac{4}{20} + \frac{24}{20} + \frac{54}{20} + \frac{64}{20}\right) - 6.25\)
\(= \frac{146}{20} - 6.25\)
\(= 7.3 - 6.25\)
\(= 1.05\)
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