Mathematics 9709 · AS & A Level · Continuous random variables
Continuous random variables — practice question
The diagram presents the graph of the probability density function for a random variable $X$ that can take only values from $-1$ to $3$. It is stated that the graph is symmetric about the line $x = 1$. For $x=-1$ to $x=3$ the graph is a quadratic curve. The random variable $S$ is defined so that $E(S)=2\text{E}(X)$ and $\mathrm{Var}(S)=\mathrm{Var}(X)$.
(a)[1]
On the grid below, sketch the quadratic graph for the probability density function of $S$.
(b)[2]
The random variable $T$ is defined so that $E(T)=\text{E}(X)$ and $\mathrm{Var}(T)=\tfrac{1}{4}\mathrm{Var}(X)$. On the grid below, sketch the quadratic graph for the probability density function of $T$.
(c)[3]
It is now given that
$f(x)=\begin{cases} \tfrac{3}{32}(3+2x-x^2), & -1 \le x \le 3, \\\ 0, & \text{otherwise}. \end{cases}$
Given that $P(1-a < X < 1+a)=0.5$, show that $a^3-12a+8=0$.
(d)[1]
Hence confirm that $0.69 \le a \le 0.70$.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Accurate sketch on the interval from $x=0$ to $x=4$ with highest point at $(2,0.375)$” …