| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Symmetric probability given |
| Difficulty | Moderate -0.8 This is a straightforward normal distribution question requiring basic understanding of symmetry (median = mean), using inverse normal tables to find standard deviation from a given probability, and then applying normal tables again. All parts are routine S1 techniques with no problem-solving insight needed, making it easier than average but not trivial since it requires correct table usage. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| 200 or 200g | B1 | "mean = 200g" is B0 but "median = 200" or just "200" alone is B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(190 < X < 210) = 0.6\) or \(P(X < 210) = 0.8\) or \(P(X > 210) = 0.2\) or diagram | M1 | For a correct probability statement; shaded diagram must have values on z-axis and probability areas shown |
| Correct use of 0.8 or 0.2 | A1 | For correct use of 0.8 or \(p = 0.2\); may be implied by suitable z value (e.g. \(z = 0.84\)) |
| \(Z = (\pm)\dfrac{210 - 200}{\sigma}\) | M1 | For attempting to standardise; values for \(x\) and \(\mu\) used in formula. Don't need \(z\) for this M1 |
| \(\dfrac{10}{\sigma} = 0.8416\) | B1 | For \(z = 0.8416\) or better [\(z = 0.84\) usually just loses this mark] |
| \(\sigma = 11.882129\ldots\) | A1 | AWRT 11.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 180) = P\!\left(Z < \dfrac{180 - 200}{\sigma}\right)\) | M1 | For attempting to standardise with 200 and their sd\((>0)\), e.g. \((\pm)\dfrac{180-200}{\text{their }\sigma}\) |
| \(= P(Z < -1.6832)\) \(= 1 - 0.9535\) | M1 | NB on open this is an A mark, ignore and treat as \(2^{\text{nd}}\) M1; for \(1 -\) a probability from tables compatible with their probability statement |
| \(= 0.0465\) or AWRT \(0.046\) | A1 | For 0.0465 or AWRT 0.046 (dependent on both Ms in part (c)) |
| (3 marks) | Total 9 marks |
# Question 6:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| 200 or 200g | B1 | "mean = 200g" is B0 but "median = 200" or just "200" alone is B1 |
**(1 mark)**
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(190 < X < 210) = 0.6$ or $P(X < 210) = 0.8$ or $P(X > 210) = 0.2$ or diagram | M1 | For a correct probability statement; shaded diagram must have values on z-axis and probability areas shown |
| Correct use of 0.8 or 0.2 | A1 | For correct use of 0.8 or $p = 0.2$; may be implied by suitable z value (e.g. $z = 0.84$) |
| $Z = (\pm)\dfrac{210 - 200}{\sigma}$ | M1 | For attempting to standardise; values for $x$ and $\mu$ used in formula. Don't need $z$ for this M1 |
| $\dfrac{10}{\sigma} = 0.8416$ | B1 | For $z = 0.8416$ or better [$z = 0.84$ usually just loses this mark] |
| $\sigma = 11.882129\ldots$ | A1 | AWRT 11.9 |
**(5 marks)**
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 180) = P\!\left(Z < \dfrac{180 - 200}{\sigma}\right)$ | M1 | For attempting to standardise with 200 and their sd$(>0)$, e.g. $(\pm)\dfrac{180-200}{\text{their }\sigma}$ |
| $= P(Z < -1.6832)$ $= 1 - 0.9535$ | M1 | NB on open this is an A mark, ignore and treat as $2^{\text{nd}}$ M1; for $1 -$ a probability from tables compatible with their probability statement |
| $= 0.0465$ or AWRT $0.046$ | A1 | For 0.0465 or AWRT 0.046 (dependent on both Ms in part (c)) |
**(3 marks) | Total 9 marks**
---6. The weights of bags of popcorn are normally distributed with mean of 200 g and $60 \%$ of all bags weighing between 190 g and 210 g .
\begin{enumerate}[label=(\alph*)]
\item Write down the median weight of the bags of popcorn.
\item Find the standard deviation of the weights of the bags of popcorn.
A shopkeeper finds that customers will complain if their bag of popcorn weighs less than 180 g .
\item Find the probability that a customer will complain.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q6 [9]}}