| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard normal distribution techniques: z-score calculations, independence of events, binomial distribution recognition, and normal approximation to binomial. Part (iv)(A) requires working backwards from a probability to find σ using inverse normal tables, which is slightly less routine. However, all parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{apple} \geq 220\text{g}) = P\!\left(Z > \frac{220 - 210.5}{15.2}\right)\) | M1 | For standardising; condone numerator reversed but penalise continuity corrections |
| \(= P(Z > 0.625) = 1 - \Phi(0.625) = 1 - 0.7340\) | M1 | For correct structure |
| \(= 0.2660\) | A1 CAO | i.e. \(1 - \Phi(\text{positive } z \text{ value})\); allow 0.266 but not 0.27 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{all 3 weigh} \geq 220\text{g}) = 0.2660^3 = 0.0188\) | M1 | M1 for their answer to part (i) cubed |
| A1 FT | Allow 0.019 and 0.01882 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (A) Binomial \((100, 0.0188)\) | B1 | For binomial |
| B1 | For parameters; second B1 dependent on first B1; FT their answer to part (ii) for second B1 | |
| (B) Use Poisson with \(\lambda = 100 \times 0.0188 = 1.88\) | B1 | For Poisson SOI; consistent with \(p < 0.1\) from part (iii)(A) |
| \(P(\text{at most one}) = e^{-1.88}\frac{1.88^0}{0!} + e^{-1.88}\frac{1.88^1}{1!}\) | B1 | For Poisson mean; FT answer to part (ii) with \(p < 0.1\) |
| \(= 0.1525 + 0.2869 = 0.4394\) | M1 | For either probability |
| M1 | For sum of both | |
| A1 CAO | For 0.44 or better; allow 0.4395 but not 0.4337 | |
| (C) \(n\) is large and \(p\) is small | B1 | Allow '\(n\) is large and \(np < 10\)' and '\(n\) is large and \(np \approx npq\)' |
| Total: 5+1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Phi^{-1}(0.1) = -1.282\) | B1 | For \(\pm 1.282\) |
| \(\frac{170 - 185}{\sigma} = -1.282\) | M1 | For correct equation as written o.e.; allow M1 if different \(z\)-value used |
| \(1.282\sigma = 15 \Rightarrow \sigma = 11.70\) | A1 CAO | Without incorrect working seen; allow 11.7 |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Two Normal curves drawn, intersecting | G1 | For shape; ignore labelling of vertical axis |
| Means shown at 185 and 210.5 | G1 | For means shown explicitly or by scale |
| Cox's curve has lower maximum height | G1 | For lower max height for Braeburns |
| Braeburns curve has greater width | G1 | For greater width (variance) for Braeburns |
| Total: 4 |
# Question 3:
## Part 3(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{apple} \geq 220\text{g}) = P\!\left(Z > \frac{220 - 210.5}{15.2}\right)$ | M1 | For standardising; condone numerator reversed but penalise continuity corrections |
| $= P(Z > 0.625) = 1 - \Phi(0.625) = 1 - 0.7340$ | M1 | For correct structure |
| $= 0.2660$ | A1 CAO | i.e. $1 - \Phi(\text{positive } z \text{ value})$; allow 0.266 but not 0.27 |
| **Total: 3** | | |
## Part 3(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{all 3 weigh} \geq 220\text{g}) = 0.2660^3 = 0.0188$ | M1 | M1 for their answer to part (i) cubed |
| | A1 FT | Allow 0.019 and 0.01882 |
| **Total: 2** | | |
## Part 3(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (A) Binomial $(100, 0.0188)$ | B1 | For binomial |
| | B1 | For parameters; second B1 dependent on first B1; FT their answer to part (ii) for second B1 |
| (B) Use Poisson with $\lambda = 100 \times 0.0188 = 1.88$ | B1 | For Poisson SOI; consistent with $p < 0.1$ from part (iii)(A) |
| $P(\text{at most one}) = e^{-1.88}\frac{1.88^0}{0!} + e^{-1.88}\frac{1.88^1}{1!}$ | B1 | For Poisson mean; FT answer to part (ii) with $p < 0.1$ |
| $= 0.1525 + 0.2869 = 0.4394$ | M1 | For either probability |
| | M1 | For sum of both |
| | A1 CAO | For 0.44 or better; allow 0.4395 but not 0.4337 |
| (C) $n$ is large and $p$ is small | B1 | Allow '$n$ is large and $np < 10$' and '$n$ is large and $np \approx npq$' |
| **Total: 5+1** | | |
## Part 3(iv)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Phi^{-1}(0.1) = -1.282$ | B1 | For $\pm 1.282$ |
| $\frac{170 - 185}{\sigma} = -1.282$ | M1 | For correct equation as written o.e.; allow M1 if different $z$-value used |
| $1.282\sigma = 15 \Rightarrow \sigma = 11.70$ | A1 CAO | Without incorrect working seen; allow 11.7 |
| **Total: 3** | | |
## Part 3(iv)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Two Normal curves drawn, intersecting | G1 | For shape; ignore labelling of vertical axis |
| Means shown at 185 and 210.5 | G1 | For means shown explicitly or by scale |
| Cox's curve has lower maximum height | G1 | For lower max height for Braeburns |
| Braeburns curve has greater width | G1 | For greater width (variance) for Braeburns |
| **Total: 4** | | |
---
3 The weights of Braeburn apples on display in a supermarket, measured in grams, are Normally distributed with mean 210.5 and standard deviation 15.2.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that a randomly selected apple weighs at least 220 grams.
\item These apples are sold in packs of 3. You may assume that the weights of apples in each pack are independent. Find the probability that all 3 of the apples in a randomly selected pack weigh at least 220 grams.
\item 100 packs are selected at random.\\
(A) State the exact distribution of the number of these 100 packs in which all 3 apples weigh at least 220 grams.\\
(B) Use a suitable approximating distribution to find the probability that in at most one of these packs all 3 apples weigh at least 220 grams.\\
(C) Explain why this approximating distribution is suitable.
\item The supermarket also sells Cox's Orange Pippin apples. The weights of these apples, measured in grams, are Normally distributed with mean 185 and standard deviation $\sigma$.\\
(A) Given that $10 \%$ of randomly selected Cox's Orange Pippin apples weigh less than 170 grams, calculate the value of $\sigma$.\\
(B) Sketch the distributions of the weights of both types of apple on a single diagram.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2011 Q3 [20]}}