Problem Statement
A continuous random variable $X$ has probability density function given by
$$f(x) = \begin{cases} k x (1 - x)^2, & 0 \le x \le 1 \\ 0, & \text{otherwise} \end{cases}$$
(a) Show that $k = 12$. [3 marks]
(b) Find $\text{E}(X)$ and $\text{Var}(X)$ in exact fractional form. [4 marks]
Verified Solution & Marking Scheme
Part (a): Total probability integral condition
$$\int_{0}^{1} f(x) \, dx = 1 \implies k \int_{0}^{1} x(1 - 2x + x^2) \, dx = 1$$
$$k \int_{0}^{1} (x - 2x^2 + x^3) \, dx = k \left[ \frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4} \right]_{0}^{1} = 1$$
$$k \left( \frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) = k \left( \frac{6 - 8 + 3}{12} \right) = k \left( \frac{1}{12} \right) = 1 \implies k = 12 \quad \text{(AG)}$$
Part (b): Calculate E(X)
$$\text{E}(X) = \int_{0}^{1} x f(x) \, dx = 12 \int_{0}^{1} (x^2 - 2x^3 + x^4) \, dx$$
$$= 12 \left[ \frac{x^3}{3} - \frac{2x^4}{4} + \frac{x^5}{5} \right]_{0}^{1} = 12 \left( \frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = 12 \left( \frac{10 - 15 + 6}{30} \right) = 12 \left( \frac{1}{30} \right) = \frac{2}{5} = 0.4$$
Calculate E(X²) and Variance
$$\text{E}(X^2) = \int_{0}^{1} x^2 f(x) \, dx = 12 \int_{0}^{1} (x^3 - 2x^4 + x^5) \, dx$$
$$= 12 \left[ \frac{1}{4} - \frac{2}{5} + \frac{1}{6} \right] = 12 \left( \frac{15 - 24 + 10}{60} \right) = 12 \left( \frac{1}{60} \right) = \frac{1}{5}$$
$$\text{Var}(X) = \text{E}(X^2) - [\text{E}(X)]^2 = \frac{1}{5} - \left(\frac{2}{5}\right)^2 = \frac{1}{5} - \frac{4}{25} = \frac{5 - 4}{25} = \frac{1}{25} = 0.04$$