| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2002 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Binomial from normal with unknown parameter |
| Difficulty | Standard +0.3 This is a standard S2 normal distribution question requiring routine application of inverse normal tables (part a), binomial probability (part b), and normal approximation to binomial with continuity correction (part c). All techniques are textbook exercises with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = 150.49347 = 150.5\) | M1 A1, B1; A1 (4 marks) | Marks shown for: M1 (setting up), A1 (finding z-value), B1 (diagram), A1 (final answer) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \leq 2) = 0.9885\) | B1; M1 A1 (3 marks) | Marks shown for: B1 (distribution), M1 A1 (calculation) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 35) \approx P\left(Z < \frac{34.5-25}{\sqrt{23.75 \text{ or } 25}}\right)\) | B1, B1; M1, M1 | \(\pm 0.5\) continuity correction required; standardise |
| Answer | Marks |
|---|---|
| \(\approx 0.9744\) or \(0.9713\) | A1; A1 (6 marks) |
## Part (a)
$L \sim N(\mu, 0.3^2)$, $P(L < 150) = 0.05 \Rightarrow P\left(Z < \frac{150-\mu}{0.3}\right) = 0.05$
$\Rightarrow \frac{150-\mu}{0.3} = -1.6449$
$\mu = 150.49347 = 150.5$ | M1 A1, B1; A1 (4 marks) | Marks shown for: M1 (setting up), A1 (finding z-value), B1 (diagram), A1 (final answer)
## Part (b)
$X$ represents number less than $150$ cm. $X \sim B(10, 0.05)$
$P(X \leq 2) = 0.9885$ | B1; M1 A1 (3 marks) | Marks shown for: B1 (distribution), M1 A1 (calculation)
## Part (c)
Normal approximation: $\mu = 500 \times 0.05 = 25$, $\sigma^2 = 23.75$ or $25$
$P(X < 35) \approx P\left(Z < \frac{34.5-25}{\sqrt{23.75 \text{ or } 25}}\right)$ | B1, B1; M1, M1 | $\pm 0.5$ continuity correction required; standardise
$\approx P(Z < 1.95 \text{ or } 1.9)$
$\approx 0.9744$ or $0.9713$ | A1; A1 (6 marks)A garden centre sells canes of nominal length 150 cm. The canes are bought from a supplier who uses a machine to cut canes of length $L$ where $L \sim \mathrm{N}(\mu, 0.3^2)$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\mu$, to the nearest 0.1 cm, such that there is only a 5\% chance that a cane supplied to the garden centre will have length less than 150 cm. [4]
\end{enumerate}
A customer buys 10 of these canes from the garden centre.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the probability that at most 2 of the canes have length less than 150 cm. [3]
\end{enumerate}
Another customer buys 500 canes.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Using a suitable approximation, find the probability that fewer than 35 of the canes will have length less than 150 cm. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2002 Q5 [13]}}