| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2005 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Poisson approximation to binomial |
| Difficulty | Moderate -0.3 This is a straightforward application of standard approximations to the binomial distribution. Part (a) is direct binomial calculation, part (b) uses Poisson approximation (np=3.2, small and manageable), and part (c) uses normal approximation with continuity correction. All three parts follow textbook procedures with no conceptual challenges—students simply need to recognize which approximation to use and apply standard formulas. Slightly easier than average due to clear signposting ('using a suitable approximation') and routine calculations. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=2) = \binom{10}{2}(0.032)^2(1-0.032)^8\) | M1 | Use of \(^nC_r p^r q^{n-r}\) |
| \(= 0.0355274\ldots\) | A1 | All correct |
| A1 | AWRT 0.0355; (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Large \(n\), small \(p \Rightarrow\) Poisson approximation with \(\lambda = 100 \times 0.032 = 3.2\) | B1 | Seen or implied |
| \(P(X < 4) = P(X \leq 3) = P(0)+P(1)+P(2)+P(3)\) | M1 | \(P(X \leq 3)\) stated or implied |
| \(= e^{-3.2}\left\{1 + 3.2 + \frac{(3.2)^2}{2} + \frac{(3.2)^3}{6}\right\}\) | A1 | All correct |
| \(= 0.602519\ldots\) | A1 | AWRT 0.603; (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(np\) and \(nq\) both \(> 5 \Rightarrow\) Normal approximation with \(np = 32\) and \(npq = 30.976\) | M1, A1 | both |
| \(P(X > 42) \approx P\left(Y > 42.5\right)\) where \(Y \sim N(32, 30.976)\) | M1 | Standardise with \(np\), \(\sqrt{npq}\) |
| \(= P\left(Z > \frac{42.5 - 32}{\sqrt{30.976}}\right)\) | A1 | All correct |
| \(= P(Z > 1.8845\ldots)\) | A1 | AWRT 1.89 |
| \(= 0.0294\) | A1 | \(0.0294\)–\(0.0297\); (6 marks) |
## Question 5:
Let $X$ represent the number of defective articles, $X \sim B(10, 0.032)$
**(a)**
$P(X=2) = \binom{10}{2}(0.032)^2(1-0.032)^8$ | M1 | Use of $^nC_r p^r q^{n-r}$
$= 0.0355274\ldots$ | A1 | All correct
| A1 | AWRT 0.0355; (3 marks)
**(b)**
Large $n$, small $p \Rightarrow$ Poisson approximation with $\lambda = 100 \times 0.032 = 3.2$ | B1 | Seen or implied
$P(X < 4) = P(X \leq 3) = P(0)+P(1)+P(2)+P(3)$ | M1 | $P(X \leq 3)$ stated or implied
$= e^{-3.2}\left\{1 + 3.2 + \frac{(3.2)^2}{2} + \frac{(3.2)^3}{6}\right\}$ | A1 | All correct
$= 0.602519\ldots$ | A1 | AWRT 0.603; (4 marks)
[NB: Normal approx $\Rightarrow$ 0/4]
**(c)**
$np$ and $nq$ both $> 5 \Rightarrow$ Normal approximation with $np = 32$ and $npq = 30.976$ | M1, A1 | both
$P(X > 42) \approx P\left(Y > 42.5\right)$ where $Y \sim N(32, 30.976)$ | M1 | Standardise with $np$, $\sqrt{npq}$
$= P\left(Z > \frac{42.5 - 32}{\sqrt{30.976}}\right)$ | A1 | All correct
$= P(Z > 1.8845\ldots)$ | A1 | AWRT 1.89
$= 0.0294$ | A1 | $0.0294$–$0.0297$; (6 marks)
---5. From company records, a manager knows that the probability that a defective article is produced by a particular production line is 0.032 .
A random sample of 10 articles is selected from the production line.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that exactly 2 of them are defective.
On another occasion, a random sample of 100 articles is taken.
\item Using a suitable approximation, find the probability that fewer than 4 of them are defective.
At a later date, a random sample of 1000 is taken.
\item Using a suitable approximation, find the probability that more than 42 are defective.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2005 Q5 [13]}}