| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine z-score calculations and use of tables. Part (a) is straightforward standardization, part (b) is inverse normal (finding a value given probability), and part (c) requires solving simultaneous equations from two z-score relationships—all textbook techniques with no novel insight required. Slightly above average difficulty due to the algebraic manipulation in part (c), but still a typical exam question. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X>168) = P\!\left(Z > \frac{168-160}{5}\right)\) | M1 | For attempt to standardise 168 with 160 and 5, i.e. \(\pm\!\left(\frac{168-160}{5}\right)\) or implied by 1.6 |
| \(= P(Z > 1.6)\) | A1 | For \(P(Z>1.6)\) or \(P(Z<-1.6)\), i.e. \(z=1.6\) and correct inequality or 1.6 on shaded diagram |
| \(= 0.0548\) | A1 | awrt 0.0548. Correct answer implies all 3 marks |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
\(P(X| M1 |
For attempting \(\pm\!\left(\frac{w-160}{5}\right)\) = recognisable \(z\) value \(( |
|
| \(\frac{w-160}{5} = -2.3263\) | B1 | For \(z = \pm2.3263\) or better. Note: \(1-2.3263=\frac{w-160}{5}\) is M0B0 |
| \(w = 148.37\) | A1 | awrt 148. For awrt 148 only with no working award M1B0A1 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{160-\mu}{\sigma} = 2.3263\) | M1 | For attempting to standardise 160 or 152 with \(\mu\) and \(\sigma\), allow \(\pm\), and equate to \(z\) value \(( |
| \(\frac{152-\mu}{\sigma} = -1.2816\) | B1 | For awrt \(\pm2.33\) or \(\pm2.32\) seen |
| B1 | For awrt \(\pm1.28\) seen | |
| \(160-\mu = 2.3263\sigma\) | ||
| \(152-\mu = -1.2816\sigma\) | ||
| \(8 = 3.6079\sigma\) | M1 | For attempt to solve two linear equations in \(\mu\) and \(\sigma\) leading to equation in just one variable |
| \(\sigma = 2.21\ldots\) | A1 | awrt 2.22. Award when first seen |
| \(\mu = 154.84\ldots\) | A1 | awrt 155. Correct answer only for part (c) scores all 6 marks |
| (6) | NB \(\sigma=2.21\) commonly from \(z=2.34\), scores M1B0B1M1A0A1. Both M marks required for A marks |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X>168) = P\!\left(Z > \frac{168-160}{5}\right)$ | M1 | For attempt to standardise 168 with 160 and 5, i.e. $\pm\!\left(\frac{168-160}{5}\right)$ or implied by 1.6 |
| $= P(Z > 1.6)$ | A1 | For $P(Z>1.6)$ or $P(Z<-1.6)$, i.e. $z=1.6$ and correct inequality or 1.6 on shaded diagram |
| $= 0.0548$ | A1 | awrt 0.0548. Correct answer implies all 3 marks |
| | **(3)** | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X<w) = P\!\left(Z < \frac{w-160}{5}\right)$ | M1 | For attempting $\pm\!\left(\frac{w-160}{5}\right)$ = recognisable $z$ value $(|z|>1)$ |
| $\frac{w-160}{5} = -2.3263$ | B1 | For $z = \pm2.3263$ or better. Note: $1-2.3263=\frac{w-160}{5}$ is M0B0 |
| $w = 148.37$ | A1 | awrt 148. For awrt 148 only with no working award M1B0A1 |
| | **(3)** | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{160-\mu}{\sigma} = 2.3263$ | M1 | For attempting to standardise 160 or 152 with $\mu$ and $\sigma$, allow $\pm$, and equate to $z$ value $(|z|>1)$ |
| $\frac{152-\mu}{\sigma} = -1.2816$ | B1 | For awrt $\pm2.33$ or $\pm2.32$ seen |
| | B1 | For awrt $\pm1.28$ seen |
| $160-\mu = 2.3263\sigma$ | | |
| $152-\mu = -1.2816\sigma$ | | |
| $8 = 3.6079\sigma$ | M1 | For attempt to solve two linear equations in $\mu$ and $\sigma$ leading to equation in just one variable |
| $\sigma = 2.21\ldots$ | A1 | awrt 2.22. Award when first seen |
| $\mu = 154.84\ldots$ | A1 | awrt 155. Correct answer only for part (c) scores all 6 marks |
| | **(6)** | NB $\sigma=2.21$ commonly from $z=2.34$, scores M1B0B1M1A0A1. **Both M marks required for A marks** |\begin{enumerate}
\item The weight, $X$ grams, of soup put in a tin by machine $A$ is normally distributed with a mean of 160 g and a standard deviation of 5 g .\\
A tin is selected at random.\\
(a) Find the probability that this tin contains more than 168 g .
\end{enumerate}
The weight stated on the tin is $w$ grams.\\
(b) Find $w$ such that $\mathrm { P } ( X < w ) = 0.01$
The weight, $Y$ grams, of soup put into a carton by machine $B$ is normally distributed with mean $\mu$ grams and standard deviation $\sigma$ grams.\\
(c) Given that $\mathrm { P } ( Y < 160 ) = 0.99$ and $\mathrm { P } ( Y > 152 ) = 0.90$ find the value of $\mu$ and the value of $\sigma$.\\
\hfill \mbox{\textit{Edexcel S1 2011 Q8 [12]}}