Subject: Applied Mathematics 4

Topic: Probability

Difficulty: Medium

Sum of all probabilities = 1

$ \therefore \sum P(x) = 1 \\ \therefore \frac{1}{8} + m + n + \frac{1}{4} + \frac{1}{12} = 1 \\ \frac{11}{24} + m + n = 1 \\ m + n = \frac{13}{24} \hspace{0.5cm} ...(1) $

Also mean = E(X)

$ = \sum xP(x) \\ = 8(\frac{1}{8}) + 12m + 16n + 20 (\frac{1}{4}) + 24 (\frac{1}{12}) \\ = 1 + 12m + 16n + 5 + 2 \\ = 12m + 16n + 8 $

Mean = 16

$ \therefore 12m + 16n + 8 = 16 \\ 12m + 16n = 8 \hspace{0.5cm} ...(2) $

Solving (1) and (2), we get, m = $\frac{1}{6}$ and n = $\frac{3}{8}$

$ E(X) = 12m + 16n + 8 = 12 (\frac{1}{6}) +16(\frac{3}{8}) + 8 \\ = 2 + 6 + 8 = 16 \\ E(X^2) = \sum x^2 P(x) \\ 8^2(\frac{1}{8}) + 12^2(m) + 16^2 (n) + 20^2 (\frac{1}{4}) + 24^2 (\frac{1}{12}) \\ = 8 + 144(\frac{1}{6}) + 16 \times 16 (\frac{3}{8}) + 5 \times 20 + 2 \times 24 \\ = 8 + 24 + 16 \times 6 + 100 + 48 \\ = 8 + 24 + 96 + 100 + 48 = 276 \\ Var(X) = E(X^2) - [E(X)]^2 = 276 - 256 = 20 $