| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question requiring standard z-score calculations and basic probability of independent events. Parts (a) and (b) are routine table lookups, while part (c) applies simple binomial reasoning (2 × p × (1-p)) with no conceptual challenges—slightly easier than average due to its mechanical nature. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(P(W < 224) = P\left(z < \frac{224 - 232}{5}\right)\) | M1 | M1 for standardising with 232 and 5. (i.e. not \(5^2\) or \(\sqrt{5}\)). Accept \(\pm \frac{w-232}{5}\). |
| \(= P(z < -1.6)\) | M1 | |
| \(= 1 - 0.9452\) | ||
| \(= 0.0548\) | A1 | awrt 0.0548 |
| (3 marks) | ||
| (b) \(0.5 - 0.2 = 0.3\) | M1 | 0.3 or 0.7 seen |
| \(\frac{w - 232}{5} = 0.5244\) | B1; M1 | 0.5244 seen; Second M1 standardise with 232 and 5 and equate to z value of (0.52 to 0.53) or (0.84 to 0.85) |
| \(w = 234.622\) | A1 | awrt 235. Al dependent upon second M mark for awrt 235 but see note below. Common errors involving probabilities and not z values: \(P(Z<0.2)=0.5793\) used instead of z value gives awrt 235 but award M0B0M0A0. \(P(Z<0.8)=0.7881\) used instead of z value award M0B0M0A0. M1B0M0A0 for 0.6179, M1B0M0A0 for 0.7580 |
| (4 marks) | ||
| (c) \(0.2 \times (1 - 0.2)\) | M1 | |
| \(2 \times 0.8 \times (1 - 0.8) = 0.32\) | M1 A1 | M1 for '2× \(p(1-p)\)'; A1 0.32 correct answer only |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| or \(S_H > 0\) with some calculation | B1 B1 | "Increase" |
| (2 marks) |
| Answer | Marks |
|---|---|
| or awrt -86.4 or \(S_{WW} < 0\) | B2 |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. \(\sigma_w = 0.126\) and \(\bar{w} = 11.175\) so fake coin is over 69 sds away from the mean | B1 B1 | '(very) unlikely' or 'impossible' |
| (2 marks) |
**(a)** $P(W < 224) = P\left(z < \frac{224 - 232}{5}\right)$ | M1 | M1 for standardising with 232 and 5. (i.e. not $5^2$ or $\sqrt{5}$). Accept $\pm \frac{w-232}{5}$.
| | |
| $= P(z < -1.6)$ | M1 | |
| | |
| $= 1 - 0.9452$ | | |
| | |
| $= 0.0548$ | A1 | awrt 0.0548
| | (3 marks) |
**(b)** $0.5 - 0.2 = 0.3$ | M1 | 0.3 or 0.7 seen
| | |
| $\frac{w - 232}{5} = 0.5244$ | B1; M1 | 0.5244 seen; Second M1 standardise with 232 and 5 and equate to z value of (0.52 to 0.53) or (0.84 to 0.85)
| | |
| $w = 234.622$ | A1 | awrt 235. Al dependent upon second M mark for awrt 235 but see note below. Common errors involving probabilities and not z values: $P(Z<0.2)=0.5793$ used instead of z value gives awrt 235 but award M0B0M0A0. $P(Z<0.8)=0.7881$ used instead of z value award M0B0M0A0. M1B0M0A0 for 0.6179, M1B0M0A0 for 0.7580
| | (4 marks) |
**(c)** $0.2 \times (1 - 0.2)$ | M1 | |
| | |
| $2 \times 0.8 \times (1 - 0.8) = 0.32$ | M1 A1 | M1 for '2× $p(1-p)$'; A1 0.32 correct answer only
| | (3 marks) |
**Total: 10 marks**
---
## Special Cases (Question 5):
**Special Case 1:** Attempt to calculate $S_H$
$$\sum hw = 1669.62, \sum t = 153, \sum w = 91.75 \text{ or } S_H = 1660.62 - \frac{153 \times 91.75}{9} \text{ or awrt } 101$$
or $S_H > 0$ with some calculation | B1 B1 | "Increase"
| | (2 marks) |
**Special Case 2:** Attempt to calculate $S_{WW}$
$$\sum w^2 = 1248.96625 - 400 = 848.96625 \text{ or awrt } 849 \text{ or } S_{WW} = 848.96625 - \frac{91.75^2}{9}$$
or awrt -86.4 or $S_{WW} < 0$ | B2 | |
| | (2 marks) |
**Special Case 3:** Argument based on standard deviation.
e.g. $\sigma_w = 0.126$ and $\bar{w} = 11.175$ so fake coin is over 69 sds away from the mean | B1 B1 | '(very) unlikely' or 'impossible'
| | (2 marks) |\begin{enumerate}
\item A manufacturer fills jars with coffee. The weight of coffee, $W$ grams, in a jar can be modelled by a normal distribution with mean 232 grams and standard deviation 5 grams.\\
(a) Find $\mathrm { P } ( W < 224 )$.\\
(b) Find the value of $w$ such that $\mathrm { P } ( 232 < W < w ) = 0.20$
\end{enumerate}
Two jars of coffee are selected at random.\\
(c) Find the probability that only one of the jars contains between 232 grams and $w$ grams of coffee.
\hfill \mbox{\textit{Edexcel S1 2012 Q7 [10]}}