Mathematics 9709 · AS & A Level · Continuous random variables

Continuous random variables — practice question

A continuous random variable $X$ can take only the values from $0$ to $6$ and its probability distribution is symmetrical. The two values $a$ and $b$ of $X$ satisfy $\text{P}(a < X < b) = p$ and $\text{P}(b < X < 3) = \frac{13}{10}p$, where $p$ is a positive constant.
(a)[1]

Show that $p \leq \frac{5}{23}$ from the information given.

(b)[2]

Determine $\text{P}(b < X < 6 - a)$ in terms of $p$.

(c)[5]

The probability density function of $X$ is now given by $f(x) = \begin{cases} \frac{1}{36}(6x - x^2), & 0 \leq x \leq 6, \\ 0, & \text{otherwise}. \end{cases}$ If $b = 2$ and $p = \frac{5}{27}$ are 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: “Use $p+\frac{13}{10}p<\frac12$ and solve it to reach $p<\frac{5}{23}$.” …

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