Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 14 / b

Let $E$ be the finish-time offset from the start of its displayed minute. The elapsed time is $25+E-S$, so $T=E-S$. To obtain the stated density, use the intended model that $E$ and $S$ are independent random variables, both with the uniform distribution on $[0,1)$.

Their joint probability density function is one on the unit square. Equivalently, the convolution of probability densities givesThe integrand is one exactly where $s\in [0,1]$ and $s+t\in [0,1]$. The overlap has length $1+t$ for $-1\le t<0$, length $1-t$ for $0\le t<1$, and zero otherwise. ThereforeThis is a symmetric triangular distribution. The independence and finish-time uniformity are additional modelling assumptions: the two truncated clock readings alone do not imply them. For example, if $E=S$ then the elapsed time is exactly twenty-five minutes and $T=0$, even though each endpoint offset is individually uniform.

The probability density function graph joins $(-1,0)$ to $(0,1)$ to $(1,0)$ and is zero elsewhere, as shown in the right panel of the timing figure. Its area isThus the probability distribution is normalized.

Since $q$ is an even function, $tq(t)$ is an odd function and its integral over the symmetric interval vanishes. Thus $\mathbb{E}[T]=0$. Its variance isConsequentlyThe standard deviation is approximately $0.40825$ minute, or $24.49$ seconds.

An excess of thirty seconds is $T>1/2$ in minute units. ThereforeThis is the shaded triangular area under the probability density function.