Mathematics 9709/62 — October/November 2023

Walkthrough

We are told $X$ follows a Poisson distribution with mean $25$. For a Poisson distribution, mean = variance = $\lambda$, so both are $25$. When $\lambda$ is large, the distribution is approximately normal with the same mean and variance, so we replace $X$ by a normal random variable $\text{N}(25, 25)$.

Because $X$ is discrete and the normal distribution is continuous, we need a continuity correction. The event $X > 30$ means $X$ can be $31, 32, \ldots$. In the continuous approximation, the boundary between $30$ and $31$ is $30.5$, so we compute $\text{P}(X > 30.5)$ in the normal model. We standardise by subtracting the mean $25$ and dividing by the standard deviation $\sqrt{25} = 5$. This gives $z = \frac{30.5 - 25}{5} = 1.1$. We want the probability above $1.1$, so it is $1 - \Phi(1.1)$. From tables, $\Phi(1.1) = 0.8643$, so the answer is $0.1357$, which rounds to $0.136$.

Key Takeaways

• For a Poisson distribution, mean = variance = $\lambda$.
• The normal approximation to a Poisson distribution is $\text{N}(\lambda, \lambda)$.
• A continuity correction is needed when approximating a discrete distribution by a continuous one.
• For $\text{P}(X > k)$, use $k + 0.5$ as the boundary; for $\text{P}(X < k)$, use $k - 0.5$.

Common Mistakes

• Forgetting the continuity correction and using $30$ instead of $30.5$. The mark scheme allows this in the standardising M1, but the final answer would be different.
• Using the variance $25$ as the standard deviation instead of $\sqrt{25} = 5$.
• Forgetting to subtract $\Phi(1.1)$ from $1$, giving $0.864$ instead of $0.136$.
• Rounding too early.

Things to Be Careful About

• The normal approximation is stated as $\text{N}(25, 25)$: the second parameter is the variance, not the standard deviation.
• The continuity correction direction: $X > 30$ becomes $X > 30.5$.
• Give the final answer to 3 significant figures as requested: $0.136$.
• The mark scheme awards B1 for stating the normal approximation, M1 for standardising, M1 for the tail area, and A1 for the final value.

Walkthrough

$Y$ is binomial with $n = 100$ and $p < 0.05$. The rule of thumb for the Poisson approximation to the binomial is that $n$ is large and $p$ is small, with $np$ moderate. Here $np = 100p$, so we approximate $Y$ by $\text{Po}(100p)$.

We need $\text{P}(Y < 3)$. Since $Y$ is discrete and takes values $0, 1, 2, \ldots$, $Y < 3$ means $Y = 0, 1, 2$. For a Poisson distribution, $\text{P}(Y=r) = \mathrm{e}^{-\lambda} \frac{\lambda^r}{r!}$. With $\lambda = 100p$, add the three probabilities:

Factoring out $\mathrm{e}^{-100p}$ gives the required expression. Since $\frac{(100p)^2}{2!} = 5000p^2$, an equivalent simplified form is $\mathrm{e}^{-100p}(1 + 100p + 5000p^2)$.

Key Takeaways

• Poisson approximation to binomial: $\text{B}(n,p) \approx \text{Po}(np)$ when $n$ is large and $p$ is small.
• For $\text{P}(Y < 3)$, include $r = 0, 1, 2$ only; do not include $r = 3$.
• The Poisson probability formula is $\mathrm{e}^{-\lambda} \frac{\lambda^r}{r!}$.

Common Mistakes

• Using $\lambda = p$ instead of $\lambda = 100p$.
• Including the $r = 3$ term or omitting one of $0, 1, 2$.
• Forgetting the factorial in the $r = 2$ term.
• Omitting brackets, e.g. writing $\mathrm{e}^{-100p} + 100p + \cdots$ instead of $\mathrm{e}^{-100p}(1 + 100p + \cdots)$. The mark scheme requires brackets for the final A1.
• Using $p^0/0!$ in the final answer; the mark scheme disallows $0!$ and $p^0$ in the final simplified expression.

Things to Be Careful About

• The condition $p < 0.05$ justifies the approximation; state $\lambda = 100p$.
• $Y < 3$ is strict, so $3$ is not included.
• The final answer must have brackets; an unsimplified form with a factorial is acceptable.
• The mark scheme gives M1 for the Poisson expression and A1 for a fully correct bracketed expression; once correct, ignore subsequent working (ISW).