| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct comparison of probabilities |
| Difficulty | Moderate -0.3 This is a standard S1 normal distribution question covering routine calculations: finding probabilities using z-tables, inverse normal problems, and applying the Central Limit Theorem. Part (b)(i) requires conceptual understanding (times cannot be negative), but overall this is slightly easier than average due to straightforward application of well-practiced techniques with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 60) = P\!\left(Z < \frac{60-48}{20}\right)\) | M1 | Standardising (59.5, 60 or 60.5) with 48 and \((\sqrt{20}, 20\) or \(20^2)\) and/or \((48-x)\) |
| \(P(Z < 0.6) = 0.725\) to \(0.73\) | A1 | AWFW; \((0.72575)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(30 < X < 60) = P(X<60) - P(X<30) = (i) - P(X<30) = (i) - P(Z<-0.9)\) | M1 | Difference or equivalent; standardising other than 60 and 30 \(\Rightarrow\) max of M1 m1 A0 |
| \((i) - \{1 - P(Z < +0.9)\} = 0.72575 - \{1 - 0.81594\}\) | m1 | Area change |
| \(= 0.54\) to \(0.542\) | A1 | AWFW; \((0.54169)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.9 \Rightarrow z = 1.28\) to \(1.282\) | B1 | AWFW; \((1.2816)\) |
| \(z = \frac{k-48}{20} = 1.2816\) | M1 m1 | Standardising \(k\) with 48 and 20; equating \(z\)-term to \(z\)-value; not using 0.9, \(0.1\), \( |
| \(k = 73.6\) to \(74\) | A1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use of \(\mu - (2\) or \(3) \times \sigma = 37 - (50\) or \(75) < 0 \Rightarrow\) likely negative times | M1 B1 | Or equivalent justification; for (likely) negative times |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Central Limit Theorem or \(n\) large \(/> 30\) | B1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Variance of \(\bar{Y} = \frac{25^2}{35}\) | B1 | OE; stated or used |
| \(P(\bar{Y} > 40) = P\!\left(Z > \frac{40-37}{25/\sqrt{35}}\right)\) | M1 | Standardising 40 with 37 and \(25/\sqrt{35}\) and/or \((37-40)\) |
| \(P(Z > 0.71) = 1 - P(Z < 0.71)\) | m1 | Area change |
| \(= 0.238\) to \(0.24\) | A1 | AWFW; \((1 - 0.76115)\) |
# Question 7:
## Part (a)
$X \sim N(48, 20^2)$
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 60) = P\!\left(Z < \frac{60-48}{20}\right)$ | M1 | Standardising (59.5, 60 or 60.5) with 48 and $(\sqrt{20}, 20$ or $20^2)$ and/or $(48-x)$ |
| $P(Z < 0.6) = 0.725$ to $0.73$ | A1 | AWFW; $(0.72575)$ |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(30 < X < 60) = P(X<60) - P(X<30) = (i) - P(X<30) = (i) - P(Z<-0.9)$ | M1 | Difference or equivalent; standardising other than 60 and 30 $\Rightarrow$ max of M1 m1 A0 |
| $(i) - \{1 - P(Z < +0.9)\} = 0.72575 - \{1 - 0.81594\}$ | m1 | Area change |
| $= 0.54$ to $0.542$ | A1 | AWFW; $(0.54169)$ |
## Part (a)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.9 \Rightarrow z = 1.28$ to $1.282$ | B1 | AWFW; $(1.2816)$ |
| $z = \frac{k-48}{20} = 1.2816$ | M1 m1 | Standardising $k$ with 48 and 20; equating $z$-term to $z$-value; not using 0.9, $0.1$, $|1-z|$ or $\Phi(0.9) = 0.81594$ |
| $k = 73.6$ to $74$ | A1 | AWFW |
## Part (b)
$Y \sim N(37, 25^2)$
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Use of $\mu - (2$ or $3) \times \sigma = 37 - (50$ or $75) < 0 \Rightarrow$ likely negative times | M1 B1 | Or equivalent justification; for (likely) negative times |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Central Limit Theorem or $n$ large $/> 30$ | B1 | — |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Variance of $\bar{Y} = \frac{25^2}{35}$ | B1 | OE; stated or used |
| $P(\bar{Y} > 40) = P\!\left(Z > \frac{40-37}{25/\sqrt{35}}\right)$ | M1 | Standardising 40 with 37 and $25/\sqrt{35}$ and/or $(37-40)$ |
| $P(Z > 0.71) = 1 - P(Z < 0.71)$ | m1 | Area change |
| $= 0.238$ to $0.24$ | A1 | AWFW; $(1 - 0.76115)$ |7
\begin{enumerate}[label=(\alph*)]
\item Electra is employed by E \& G Ltd to install electricity meters in new houses on an estate. Her time, $X$ minutes, to install a meter may be assumed to be normally distributed with a mean of 48 and a standard deviation of 20 .
Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 60 )$;
\item $\mathrm { P } ( 30 < X < 60 )$;
\item the time, $k$ minutes, such that $\mathrm { P } ( X < k ) = 0.9$.
\end{enumerate}\item Gazali is employed by E \& G Ltd to install gas meters in the same new houses. His time, $Y$ minutes, to install a meter has a mean of 37 and a standard deviation of 25 .
\begin{enumerate}[label=(\roman*)]
\item Explain why $Y$ is unlikely to be normally distributed.
\item State why $\bar { Y }$, the mean of a random sample of 35 gas meter installations, is likely to be approximately normally distributed.
\item Determine $\mathrm { P } ( \bar { Y } > 40 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2007 Q7 [16]}}