Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 2 / 9F (Probability)

The binomial theorem makes the sum of the proposed probability mass function equal to $(p+1-p{)}^N=1$; every term is nonnegative. Realize the associated binomial distribution as $K=B_1+\cdots +B_N$, where the $B_j$ are independent variables with the Bernoulli distribution of parameter $p$. Each has expected value $p$ and variance $p(1-p)$. Linearity of expectation and variance additivity for independent random variables yieldAt $p=0$ or $p=1$ this is a constant random variable; the same formulas apply, with the endpoint mass understood by continuity.

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.

Let ${\sigma }_N=\sqrt{Np(1-p)}$ and use the floor function to chooseInterpret an empty integer interval as an empty sum. These bounds give exactly the event $a<(K_N-Np)/{\sigma }_N\le b$, where $K_N$ has the binomial distribution of parameters $N,p$. Since $0<p<1$, the central limit theorem for a sum of independent Bernoulli distribution variables gives convergence in distribution of that standardized count to the standard normal distribution. Its cumulative distribution function is continuous at both endpoints, soThe TeX's additional malformed inequality involving $p/k_a(N)$ and $p/k_b(N)$ is absent from the PDF and is not a condition on the requested integers.