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 $X$ whose values lie only between $-1$ and $3$. It is stated that the graph is symmetric about the line $x = 1$. Over the interval from $x = -1$ to $x = 3$, the graph is a quadratic curve. The random variable $S$ satisfies $E(S) = 2\text{E}(X)$ and $\operatorname{Var}(S) = \operatorname{Var}(X)$.
(a)[1]
On the grid below, draw a quadratic graph to represent the probability density function of $S$.
(b)[2]
The random variable $T$ satisfies $E(T) = \text{E}(X)$ and $\operatorname{Var}(T) = \frac{1}{4}\operatorname{Var}(X)$. On the grid below, sketch a quadratic graph for the probability density function of $T$.
(c)[3]
The pdf is now defined by $$f(x)=\begin{cases}\frac{3}{32}(3+2x-x^2), & -1 \le x \le 3, \\ 0, & \text{otherwise.}\end{cases}$$ 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: “The sketch has the correct shape, with its maximum at $(2,0.375)$” …