Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 16

Because $X$ takes values only in $[0,1]$, its cumulative distribution function must satisfy $F(1)=1$. Thus

The full cumulative distribution function isThus the function graph is horizontal at zero for $-1\le x\le 0$, rises continuously from $(0,0)$ to $(1,1)$, and is horizontal at one for $1\le x\le 2$. Its derivative $12x^2(1-x)$ is positive on $(0,1)$ and zero at both endpoints. The second derivative $12x(2-3x)$ changes sign at $x=2/3$, giving an inflection point there.

The probability density function is the derivative of the cumulative distribution function:It is nonnegative and has integral $F(1)-F(0)=1$. This is a beta distribution with parameters $3$ and $2$.

The probability density function is zero outside $[0,1]$. Inside that interval, it rises from $(0,0)$, reaches its only interior maximum at $x=2/3$, and falls to $(1,0)$. The derivative ispositive for $0<x<2/3$ and negative for $2/3<x<1$. Its maximum height is $f(2/3)=16/9$. The second panel of the figure in part (b) shows the function graph on the requested full range, with this mode marked.

The interior stationary point of the probability density function solves $12x(2-3x)=0$ with $0<x<1$, giving $x=2/3$. The sign change of $f^{\prime }$ established in part (d), together with zero endpoint values, makes this the global maximum. Thus

At the mode, the cumulative distribution function has valueSince $F$ is strictly increasing on $(0,1)$, its unique crossing of $1/2$ occurs earlier. Therefore the median is less than the mode.

The expected value isThe second moment is $\mathbb{E}[X^2]=12(1/5-1/6)=2/5$. Hence the variance and standard deviation are

The requested bounds are $\mathbb{E}[X]-\sigma =2/5$ and $\mathbb{E}[X]+\sigma =4/5$, both within the support. Since the probability distribution is continuous, endpoints have zero probability, and subtraction of the cumulative distribution function givesThis leaves the answer in rational fractions as requested.