| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Standard two probabilities given |
| Difficulty | Standard +0.8 Part (a) involves routine standardization and table lookup. Part (b) requires understanding inverse normal distribution, setting up simultaneous equations from two given probabilities, and solving them—this is a non-trivial problem-solving task that goes beyond standard textbook exercises, though still within S1 scope. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(V < 400) = P\left(Z < \frac{400-412}{8}\right) = P(Z < -1.5)\) | M1 | Standardising with \(\mu=412\), \(\sigma=8\) |
| \(= 1 - \Phi(1.5) = 1 - 0.9332\) | M1 | Using tables correctly |
| \(= 0.0668\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(V > 420) = P\left(Z > \frac{420-412}{8}\right) = P(Z > 1)\) | M1 | Standardising |
| \(= 1 - 0.8413 = 0.1587\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(V = 410) = 0\) | B1 | Since \(V\) is continuous |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(V < 400) = 0.05 \Rightarrow \frac{400-\mu}{\sigma} = -1.6449\) | B1 | Correct \(z\)-value \(\pm1.6449\) seen |
| So \(400 - \mu = -1.6449\sigma\) | A1 | Correct equation shown |
| \(P(V > 420) = 0.01 \Rightarrow \frac{420-\mu}{\sigma} = 2.3263\) | B1 | Correct \(z\)-value \(2.3263\) seen, leading to correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Adding equations: \(20 = 3.9712\sigma\) | M1 | Eliminating \(\mu\) by adding equations |
| \(\sigma = \frac{20}{3.9712} = 5.036 \approx 5.04\) | A1 | Accept awrt \(5.04\) |
| \(\mu = 400 + 1.6449 \times 5.036 \approx 408.3\) | A1 | Accept awrt \(408\) |
# Question 6:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(V < 400) = P\left(Z < \frac{400-412}{8}\right) = P(Z < -1.5)$ | M1 | Standardising with $\mu=412$, $\sigma=8$ |
| $= 1 - \Phi(1.5) = 1 - 0.9332$ | M1 | Using tables correctly |
| $= 0.0668$ | A1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(V > 420) = P\left(Z > \frac{420-412}{8}\right) = P(Z > 1)$ | M1 | Standardising |
| $= 1 - 0.8413 = 0.1587$ | A1 | |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(V = 410) = 0$ | B1 | Since $V$ is continuous |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(V < 400) = 0.05 \Rightarrow \frac{400-\mu}{\sigma} = -1.6449$ | B1 | Correct $z$-value $\pm1.6449$ seen |
| So $400 - \mu = -1.6449\sigma$ | A1 | Correct equation shown |
| $P(V > 420) = 0.01 \Rightarrow \frac{420-\mu}{\sigma} = 2.3263$ | B1 | Correct $z$-value $2.3263$ seen, leading to correct equation |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Adding equations: $20 = 3.9712\sigma$ | M1 | Eliminating $\mu$ by adding equations |
| $\sigma = \frac{20}{3.9712} = 5.036 \approx 5.04$ | A1 | Accept awrt $5.04$ |
| $\mu = 400 + 1.6449 \times 5.036 \approx 408.3$ | A1 | Accept awrt $408$ |6 The volume of shampoo, $V$ millilitres, delivered by a machine into bottles may be modelled by a normal random variable with mean $\mu$ and standard deviation $\sigma$.
\begin{enumerate}[label=(\alph*)]
\item Given that $\mu = 412$ and $\sigma = 8$, determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( V < 400 )$;
\item $\mathrm { P } ( V > 420 )$;
\item $\mathrm { P } ( V = 410 )$.
\end{enumerate}\item A new quality control specification requires that the values of $\mu$ and $\sigma$ are changed so that
$$\mathrm { P } ( V < 400 ) = 0.05 \quad \text { and } \quad \mathrm { P } ( V > 420 ) = 0.01$$
\begin{enumerate}[label=(\roman*)]
\item Show, with the aid of a suitable sketch, or otherwise, that
$$400 - \mu = - 1.6449 \sigma \quad \text { and } \quad 420 - \mu = 2.3263 \sigma$$
\item Hence calculate values for $\mu$ and $\sigma$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2011 Q6 [12]}}