| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution parameters found then approximation applied |
| Difficulty | Standard +0.3 This is a straightforward S2 question combining normal distribution with binomial probability. Part (a) requires inverse normal calculation using tables/calculator (routine), part (b) is direct binomial probability calculation with n=8, and part (c) applies normal approximation to binomial—all standard techniques with no novel insight required. Slightly easier than average due to clear structure and standard methods. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pm\frac{200-\mu}{\sqrt{0.04}} = \pm z\) value, \( | z | >1\) |
| \(\frac{200-\mu}{0.2} = -1.6449\) | A1 | Correct equation with compatible signs and \(z = 1.6449\) or better |
| \(\mu = 200.3\) | A1 | awrt 200.3 (condone awrt 200.5); Note: M1A0A1 is possible |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim B(8, 0.05)\) | B1 | Writing or using \(B(8, 0.05)\) |
| \(P(X \geq 3) = 1 - P(X \leq 2)\) | M1 | Writing or using \(1 - P(X \leq 2)\) |
| \(= 1 - 0.9942\) | ||
| \(= 0.0058\) | A1 | awrt 0.0058 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim Po(3)\) | B1 | Writing or using \(Po(3)\) |
| \(P(Y > 5) = 1 - P(Y \leq 5)\) | M1 | Writing or using \(1 - P(Y \leq 5)\) |
| \(= 1 - 0.9161\) | ||
| \(= 0.0839\) | A1 | awrt 0.0839 |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pm\frac{200-\mu}{\sqrt{0.04}} = \pm z$ value, $|z|>1$ | M1 | Method for standardising |
| $\frac{200-\mu}{0.2} = -1.6449$ | A1 | Correct equation with compatible signs and $z = 1.6449$ or better |
| $\mu = 200.3$ | A1 | awrt **200.3** (condone awrt 200.5); Note: M1A0A1 is possible |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(8, 0.05)$ | B1 | Writing or using $B(8, 0.05)$ |
| $P(X \geq 3) = 1 - P(X \leq 2)$ | M1 | Writing or using $1 - P(X \leq 2)$ |
| $= 1 - 0.9942$ | | |
| $= 0.0058$ | A1 | awrt **0.0058** |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim Po(3)$ | B1 | Writing or using $Po(3)$ |
| $P(Y > 5) = 1 - P(Y \leq 5)$ | M1 | Writing or using $1 - P(Y \leq 5)$ |
| $= 1 - 0.9161$ | | |
| $= 0.0839$ | A1 | awrt **0.0839** |
**Total: 9 marks**\begin{enumerate}
\item A shop sells rods of nominal length 200 cm . The rods are bought from a manufacturer who uses a machine to cut rods of length $L \mathrm {~cm}$, where $L \sim \mathrm {~N} \left( \mu , 0.2 ^ { 2 } \right)$
\end{enumerate}
The value of $\mu$ is such that there is only a $5 \%$ chance that a rod, selected at random from those supplied to the shop, will have length less than 200 cm .\\
(a) Find the value of $\mu$ to one decimal place.
A customer buys a random sample of 8 of these rods.\\
(b) Find the probability that at least 3 of these rods will have length less than 200 cm .
Another customer buys a random sample of 60 of these rods.\\
(c) Using a suitable approximation, find the probability that more than 5 of these rods will have length less than 200 cm .\\
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\hfill \mbox{\textit{Edexcel S2 2017 Q1 [9]}}