| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Waiting time applications |
| Difficulty | Moderate -0.3 This is a straightforward application of continuous uniform distribution with standard formulas for mean/SD, followed by routine use of Central Limit Theorem for sample means. All steps are textbook procedures requiring minimal problem-solving insight, though the CLT application adds slight complexity beyond pure recall. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| [Continuous] uniform on the interval \([0, 7]\) | B1 | Need uniform and correct interval; allow \(U[0,7]\); a fully correct pdf implies B1: \(f(x) = \frac{1}{7}\) for \(0 \leq x \leq 7\), 0 otherwise |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| mean \(= 3.5\) | B1 | For 3.5 |
| standard deviation \(= \sqrt{\dfrac{(7-0)^2}{12}} = \dfrac{7}{\sqrt{12}} = 2.0207\ldots\) awrt \(2.02\) | M1 A1 | M1: correct method for standard deviation; A1: awrt 2.02 (allow \(\frac{7}{\sqrt{12}}\) or \(\frac{7\sqrt{3}}{6}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| By CLT \(\bar{T} \sim N\!\left(3.5,\ \dfrac{49}{552}\right)\) | M1 | For writing or using \(N\!\left(3.5, \dfrac{49}{552}\right)\); allow \(N\!\left(3.5, \dfrac{2.02^2}{46}\right)\) or ft from (b); if Po(7) given in (a) allow \(N\!\left(7, \dfrac{7}{46}\right)\) |
| \(P(3.4 < \bar{T} < 3.6) = P\!\left(\dfrac{3.4 - 3.5}{\sqrt{49/552}} < Z < \dfrac{3.6 - 3.5}{\sqrt{49/552}}\right) = P(-0.34 < Z < 0.34)\) | M1 A1 | M1: standardising using 3.4 or 3.6 with their mean and SD; A1: fully correct expression for either 3.4 or 3.6 (may be implied by \(\pm\) awrt 0.34) |
| \(= 0.6331 - (1 - 0.6331) = 0.2662\) awrt \(0.263\) to \(0.266\) | M1 A1 | M1: for \(p-(1-p)\) or \(2(p-0.5)\); A1: awrt 0.263 to 0.266 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Large/independent/random sample allows use of CLT | B1 | Any suitable assumption |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| [Continuous] uniform on the interval $[0, 7]$ | B1 | Need uniform and correct interval; allow $U[0,7]$; a fully correct pdf implies B1: $f(x) = \frac{1}{7}$ for $0 \leq x \leq 7$, 0 otherwise |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| mean $= 3.5$ | B1 | For 3.5 |
| standard deviation $= \sqrt{\dfrac{(7-0)^2}{12}} = \dfrac{7}{\sqrt{12}} = 2.0207\ldots$ awrt $2.02$ | M1 A1 | M1: correct method for standard deviation; A1: awrt 2.02 (allow $\frac{7}{\sqrt{12}}$ or $\frac{7\sqrt{3}}{6}$) |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| By CLT $\bar{T} \sim N\!\left(3.5,\ \dfrac{49}{552}\right)$ | M1 | For writing or using $N\!\left(3.5, \dfrac{49}{552}\right)$; allow $N\!\left(3.5, \dfrac{2.02^2}{46}\right)$ or ft from (b); if Po(7) given in (a) allow $N\!\left(7, \dfrac{7}{46}\right)$ |
| $P(3.4 < \bar{T} < 3.6) = P\!\left(\dfrac{3.4 - 3.5}{\sqrt{49/552}} < Z < \dfrac{3.6 - 3.5}{\sqrt{49/552}}\right) = P(-0.34 < Z < 0.34)$ | M1 A1 | M1: standardising using 3.4 or 3.6 with their mean and SD; A1: fully correct expression for either 3.4 or 3.6 (may be implied by $\pm$ awrt 0.34) |
| $= 0.6331 - (1 - 0.6331) = 0.2662$ awrt $0.263$ to $0.266$ | M1 A1 | M1: for $p-(1-p)$ or $2(p-0.5)$; A1: awrt 0.263 to 0.266 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| Large/independent/random sample allows use of CLT | B1 | Any suitable assumption |
---\begin{enumerate}
\item Navtej travels to work by train. A train leaves the station every 7 minutes and Navtej's arrival at the station is independent of when the train is due to leave.\\
(a) Write down a suitable model for the distribution of the time, $T$ minutes, that he has to wait for a train to leave.\\
(b) Find the mean and standard deviation of $T$
\end{enumerate}
During a 10-week period, Navtej travels to work by train on 46 occasions.\\
(c) Estimate the probability that the mean length of time that he has to wait for a train to leave is between 3.4 and 3.6 minutes.\\
(d) State a necessary assumption for the calculation in part (c).
\hfill \mbox{\textit{Edexcel S3 2022 Q4 [10]}}