Mathematics 9709 · AS & A Level · Continuous random variables

Continuous random variables — practice question

A random variable $X$ is defined by the probability density function given by $$f(x)=\begin{cases}\frac{1}{18}(9-x^2), & 0 \le x \le 3,\\ 0, & \text{otherwise}.\end{cases}$$
(a)[3]

Find the value of $P(X < 1.2)$.

(b)[3]

Find the value of $\text{E}(X)$.

(c)[3]

The median of $X$ is $m$. Show that $m^3 - 27m + 27 = 0$.

Worked solution & mark scheme

This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: Set up the integral $\frac{1}{18}\int_0^{1.2}(9-x^2)\,dx$

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