| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Independent probability calculations |
| Difficulty | Moderate -0.3 This is a straightforward S1 normal distribution question covering standard techniques: z-score calculations, inverse normal problems, and basic Central Limit Theorem application. All parts are routine textbook exercises requiring only direct application of formulas with no problem-solving insight needed. Slightly easier than average due to the step-by-step structure and standard context. |
| 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 |
| \(P(X < 45) = P\left(Z < \frac{45-37}{8}\right)\) | M1 | Standardising (44.5, 45 or 45.5) with 37 and \((\sqrt{8}\), 8 or \(8^2)\) and/or \((37-x)\) |
| \(= P(Z < 1)\) | A1 | CAO; ignore sign |
| \(= 0.841\) | A1 | AWRT (0.84134) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(30 < X < 45) = \text{(i)} - P(X < 30)\) | M1 | Used; OE |
| \(= \text{(i)} - P(Z < -0.875)\) | ||
| \(= \text{(i)} - [1 - (0.80785 \text{ to } 0.81057)]\) | m1 | Area change |
| \(= 0.648\) to \(0.652\) | A1 | AWFW (0.65056) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.12 \Rightarrow z = 1.17\) to \(1.18\) | B1 | AWFW; ignore sign (1.1750) |
| \(z = \frac{45 - 40}{\sigma}\) | M1 | Standardising 45 with 40 and \(\sigma\) |
| \(= 1.175\) | m1 | Equating \(z\)-term to \(z\)-value but not using 0.12, 0.88 or \( |
| \(\sigma = 4.23\) to \(4.28\) | A1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Route A: \(P(X > 45) = 1 - \text{(a)(i)}\); Route B: \(P(Y > 45) = 0.12\) | B1 | OE; must use 45 |
| Monica should use Route B (smaller prob) | B1\(\checkmark\) | \(\checkmark\) on (a)(i); allow Route \(Y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean of \(\bar{W} = 18\) | B1 | CAO; can be implied by use in standardising |
| Variance of \(\bar{W} = \frac{12^2}{36} = 4\) | B1 | CAO; OE |
| \(P(\bar{W} > 20) = P\left(Z > \frac{20-18}{2}\right)\) | M1 | Standardising 20 with 18 and 2 and/or \((18-20)\) |
| \(= P(Z > 1) = 0.159\) | A1\(\checkmark\) | AWRT (0.15866); \(\checkmark\) on (a)(i) if used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| In part (d) | B1 | CAO; OE |
# Question 6:
## Part 6(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 45) = P\left(Z < \frac{45-37}{8}\right)$ | M1 | Standardising (44.5, 45 or 45.5) with 37 and $(\sqrt{8}$, 8 or $8^2)$ and/or $(37-x)$ |
| $= P(Z < 1)$ | A1 | CAO; ignore sign |
| $= 0.841$ | A1 | AWRT (0.84134) |
## Part 6(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(30 < X < 45) = \text{(i)} - P(X < 30)$ | M1 | Used; OE |
| $= \text{(i)} - P(Z < -0.875)$ | | |
| $= \text{(i)} - [1 - (0.80785 \text{ to } 0.81057)]$ | m1 | Area change |
| $= 0.648$ to $0.652$ | A1 | AWFW (0.65056) |
## Part 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.12 \Rightarrow z = 1.17$ to $1.18$ | B1 | AWFW; ignore sign (1.1750) |
| $z = \frac{45 - 40}{\sigma}$ | M1 | Standardising 45 with 40 and $\sigma$ |
| $= 1.175$ | m1 | Equating $z$-term to $z$-value but not using 0.12, 0.88 or $|1-z|$ |
| $\sigma = 4.23$ to $4.28$ | A1 | AWFW |
## Part 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| **Route A:** $P(X > 45) = 1 - \text{(a)(i)}$; **Route B:** $P(Y > 45) = 0.12$ | B1 | OE; must use 45 |
| Monica should use **Route B** (smaller prob) | B1$\checkmark$ | $\checkmark$ on (a)(i); allow Route $Y$ |
## Part 6(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean of $\bar{W} = 18$ | B1 | CAO; can be implied by use in standardising |
| Variance of $\bar{W} = \frac{12^2}{36} = 4$ | B1 | CAO; OE |
| $P(\bar{W} > 20) = P\left(Z > \frac{20-18}{2}\right)$ | M1 | Standardising 20 with 18 and 2 and/or $(18-20)$ |
| $= P(Z > 1) = 0.159$ | A1$\checkmark$ | AWRT (0.15866); $\checkmark$ on (a)(i) if used |
## Part 6(e):
| Answer | Mark | Guidance |
|--------|------|----------|
| In part (d) | B1 | CAO; OE |
---6 When Monica walks to work from home, she uses either route A or route B.
\begin{enumerate}[label=(\alph*)]
\item Her journey time, $X$ minutes, by route A may be assumed to be normally distributed with a mean of 37 and a standard deviation of 8 .
Determine:
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { P } ( X < 45 )$;
\item $\mathrm { P } ( 30 < X < 45 )$.
\end{enumerate}\item Her journey time, $Y$ minutes, by route B may be assumed to be normally distributed with a mean of 40 and a standard deviation of $\sigma$.
Given that $\mathrm { P } ( Y > 45 ) = 0.12$, calculate the value of $\sigma$.
\item If Monica leaves home at 8.15 am to walk to work hoping to arrive by 9.00 am , state, with a reason, which route she should take.
\item When Monica travels to work from home by car, her journey time, $W$ minutes, has a mean of 18 and a standard deviation of 12 .
Estimate the probability that, for a random sample of 36 journeys to work from home by car, Monica's mean time is more than 20 minutes.
\item Indicate where, if anywhere, in this question you needed to make use of the Central Limit Theorem.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2007 Q6 [17]}}