| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a standard S1 normal distribution question requiring routine z-score calculations and table lookups. Part (a) involves straightforward standardization and reading from tables. Part (b) requires solving simultaneous equations from inverse normal values, which is slightly beyond pure recall but still a well-practiced technique at this level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | ||
| Weight, \(X \sim N(406, 4.2^2)\) | ||
| 7(a)(i) | ||
| \(\mathbb{P}(X < 400) = \mathbb{P}\left(Z < \frac{400 - 406}{4.2}\right)\) | M1 | Standardising (399.5, 400 or 400.5) with 406 and (\(\sqrt{4.2}\), 4.2 or \(4.2^2\)) and/or (406 – x) |
| \(= \mathbb{P}(Z < -1.428\) to \(-1.43)\) | m1 | \(\Phi(-z) = 1 - \Phi(z)\) |
| \(= 0.076\) to \(0.077\) | A1 | AWRT, 0.07636 |
| 3 | ||
| 7(a)(ii) | ||
| \(\mathbb{P}(402.5 < X < 407.5) = \mathbb{P}(X < 407.5) - \mathbb{P}(X < 402.5) =\) | M1 | Difference OE |
| \(\mathbb{P}(Z < 0.36) - \mathbb{P}(Z < -0.83)\) | B2, I | AWRT; ignoring signs |
| \(= 0.64058 - (1 - 0.79673) = 0.433\) to \(0.44\) | A1 | AWFW, 0.43731 |
| 4 | ||
| 7(b)(i) | ||
| \(0.975 \Rightarrow z = 1.96\) | M1 | Accept explanation in words |
| \(\mathbb{P}(Y < 310) = \mathbb{P}\left(Z < \frac{310 - \mu}{\sigma}\right)\) | M1 | Standardising 310 using \(\mu\) and \(\sigma\) |
| or \(x = \mu \pm 2\sigma\) | Accept in words | |
| Thus \(\frac{310 - \mu}{\sigma} = 1.96 \Rightarrow\) result | m1 | Equating, AG, Substitution |
| or \(310 = \mu + 1.96\sigma \Rightarrow\) result | ||
| NB: Working backwards from given equation \(\Rightarrow\) at most M1 M0 mo | 3 | |
| 7(b)(ii) | ||
| \(0.86 \Rightarrow z = 1.08\) | B1 | AWRT, 1.0803 |
| \(310 - \mu = 1.96\sigma\) | M1 | Attempt at solving 2 equations each of form \(x - \mu = z\sigma\) |
| \(307.5 - \mu = 1.08\sigma\) | ||
| \(2.5 = 0.88\sigma\) | M1 | |
| \(\sigma = 2.84\) to \(2.842\) | A1 | AWFW, 2.841 |
| \(\mu = 304.4\) to \(304.5\) | A1 | AWFW, 304.43 |
| 4 |
| **7(a)** |
|----------|
| Weight, $X \sim N(406, 4.2^2)$ | | |
| **7(a)(i)** |
|----------|
| $\mathbb{P}(X < 400) = \mathbb{P}\left(Z < \frac{400 - 406}{4.2}\right)$ | M1 | Standardising (399.5, 400 or 400.5) with 406 and ($\sqrt{4.2}$, 4.2 or $4.2^2$) and/or (406 – x) |
| $= \mathbb{P}(Z < -1.428$ to $-1.43)$ | m1 | $\Phi(-z) = 1 - \Phi(z)$ |
| $= 0.076$ to $0.077$ | A1 | AWRT, 0.07636 |
| | | 3 |
| **7(a)(ii)** |
|----------|
| $\mathbb{P}(402.5 < X < 407.5) = \mathbb{P}(X < 407.5) - \mathbb{P}(X < 402.5) =$ | M1 | Difference OE |
| $\mathbb{P}(Z < 0.36) - \mathbb{P}(Z < -0.83)$ | B2, I | AWRT; ignoring signs |
| $= 0.64058 - (1 - 0.79673) = 0.433$ to $0.44$ | A1 | AWFW, 0.43731 |
| | | 4 |
| **7(b)(i)** |
|----------|
| $0.975 \Rightarrow z = 1.96$ | M1 | Accept explanation in words |
| $\mathbb{P}(Y < 310) = \mathbb{P}\left(Z < \frac{310 - \mu}{\sigma}\right)$ | M1 | Standardising 310 using $\mu$ and $\sigma$ |
| or $x = \mu \pm 2\sigma$ | | Accept in words |
| Thus $\frac{310 - \mu}{\sigma} = 1.96 \Rightarrow$ result | m1 | Equating, AG, Substitution |
| or $310 = \mu + 1.96\sigma \Rightarrow$ result | | |
| NB: Working backwards from given equation $\Rightarrow$ at most M1 M0 mo | | 3 |
| **7(b)(ii)** |
|----------|
| $0.86 \Rightarrow z = 1.08$ | B1 | AWRT, 1.0803 |
| $310 - \mu = 1.96\sigma$ | M1 | Attempt at solving 2 equations each of form $x - \mu = z\sigma$ |
| $307.5 - \mu = 1.08\sigma$ | | |
| $2.5 = 0.88\sigma$ | M1 | |
| $\sigma = 2.84$ to $2.842$ | A1 | AWFW, 2.841 |
| $\mu = 304.4$ to $304.5$ | A1 | AWFW, 304.43 |
| | | 4 |
**Total for Q7: 14**
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## TOTAL: 757
\begin{enumerate}[label=(\alph*)]
\item The weight, $X$ grams, of soup in a carton may be modelled by a normal random variable with mean 406 and standard deviation 4.2.
Find the probability that the weight of soup in a carton:
\begin{enumerate}[label=(\roman*)]
\item is less than 400 grams;
\item is between 402.5 grams and 407.5 grams.
\end{enumerate}\item The weight, $Y$ grams, of chopped tomatoes in a tin is a normal random variable with mean $\mu$ and standard deviation $\sigma$.
\begin{enumerate}[label=(\roman*)]
\item Given that $\mathrm { P } ( Y < 310 ) = 0.975$, explain why:
$$310 - \mu = 1.96 \sigma$$
\item Given that $\mathrm { P } ( Y < 307.5 ) = 0.86$, find, to two decimal places, values for $\mu$ and $\sigma$.\\
(4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2006 Q7 [14]}}