Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 18

Under independent daily tosses, attending every lecture is the intersection of $24$ independent events of probability $3/4$. Hence

Attending none requires a tail on every independent toss. Thus

Write the count as a sum of $24$ Bernoulli random variables, one for each attendance indicator. Each indicator random variable has expected value $3/4$, so linearity of expectation givesEquivalently, $N$ has a binomial distribution with parameters $24$ and $3/4$.

Independence makes the variances of the attendance indicators additive. The binomial distribution therefore gives $\mathrm{Var}(N)=np(1-p)=24(3/4)(1/4)=9/2$. Its positive square root is the standard deviation:

Interpreting $f$ as the probability density function of the random variable, for $\alpha \le \beta$ one hasA single point has zero probability, so including or excluding either endpoint does not change this result.

When the relevant moments are finite, the mean and variance areEquivalently, ${\sigma }^2=\int x^{2}f(x)dx-{\mu }^2$. The mean requires an integrable absolute first moment; finite second moment guarantees both quantities are finite.

Normalize the probability density function on its nonnegative support:The integral follows from one integration by parts, with the boundary term zero since $\lambda >0$. Hence $A={\lambda }^2$. This is a gamma distribution of shape $2$ and rate $\lambda$.

Repeated integration by parts gives, for every nonnegative integer $j$,With $A={\lambda }^2$, the first two moments are $\mathbb{E}[X]={\lambda }^2(2!/{\lambda }^3)=2/\lambda$ and $\mathbb{E}[X^2]={\lambda }^2(3!/{\lambda }^4)=6/{\lambda }^2$. Subtracting the square of the mean givesIf $X$ carries units, $\lambda$ has inverse units of $X$, so the mean and standard deviation have the units of $X$ and the variance has their square.

For $t\ge 0$, evaluate the survival function directly:At $t=\mu +2\sigma =(2+2\sqrt{2})/\lambda$, this becomesThe parameter $\lambda$ cancels because the threshold scales with the distribution. Also $\mu -2\sigma <0$ lies outside the support, so in this particular case the two-sided event $|X-\mu |>2\sigma$ has the same probability as the stated upper-tail event.