Mathematics 9709 · AS & A Level · Continuous random variables
Continuous random variables — practice question
The continuous random variable $X$ can take values only in the interval from $0$ to $6$, and its probability distribution is symmetrical. For two values, $a$ and $b$, of $X$, $\mathrm{P}(a < X < b) = p$ and $\mathrm{P}(b < X < 3) = \frac{13}{10}p$, where $p$ is positive.
(a)[1]
Show that the inequality $p \leq \frac{5}{23}$ holds.
(b)[2]
Find $\mathrm{P}(b < X < 6 - a)$ in terms of $p$.
(c)[5]
The probability density function of $X$ is now given as $f$, where
$f(x) = \begin{cases}\frac{1}{36}(6x - x^2), & 0 \leq x \leq 6, \\ 0, & \text{otherwise}.\end{cases}$
With $b = 2$ and $p = \frac{5}{27}$ given, determine the value of $a$.
Worked solution & mark scheme
This 8-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Solve $p+\frac{13}{10}p\le\frac12$ to arrive at $p\le\frac{5}{23}$.” …