| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with standard formulas. Parts (a)-(c) require only direct application of the rule that linear combinations of independent normals are normal (calculating means and variances, then using tables), part (d) is basic binomial probability, and part (e) tests conceptual understanding of independence. While multi-part, each step is routine for S3 level with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(T\) = total time taken; \(T \sim N(41+81+57, \ 5.2^2+4.2^2+6.6^2)\) so \(T \sim N(179, 88.24)\) | M1 A1 | M1 for setting up normal distribution with mean \(41+81+57 (=179)\); A1 for correct variance 88.24 or s.d. awrt 9.39 |
| \(P(T > 180) = P\left(Z > \frac{180-179}{\sqrt{88.24}}\right)\) | M1 | For standardising with 180, their mean and their standard deviation |
| \(= 1 - 0.5438 = 0.4562\) awrt 0.456 to 0.458 | M1 A1 | Use of \(1-p\) with \(0.5 < p < 1\); awrt 0.456 to 0.458 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(Y\) = difference between run and swim, \(Y \sim N(16, 70.6)\) or \(D = R - S - 20\), \(D \sim N(-4, 70.6)\) | B1 | For \(N(\pm16, 70.6)\) or \(N(\pm4, 70.6)\). May be seen in a calculation |
| \(P(Y > 20) = P\left(Z > \frac{20-16}{\sqrt{70.6}}\right)\) or \(P(D > 0) = P\left(Z > \frac{0-(-4)}{\sqrt{70.6}}\right)\) | M1 | For standardisation with \(\pm20\) or 0, their mean and s.d. (variance must be \(> 0\)). Must be compatible |
| \(= 1 - 0.6844 = 0.3156\) awrt 0.316/0.317 | M1 A1 | Use of \(1-p\); awrt 0.316/0.317 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(T > t) = 0.95 \Rightarrow P\left(Z > \frac{t-179}{\sqrt{88.24}}\right) = 0.95 \Rightarrow \frac{t-179}{\sqrt{88.24}} = -1.6449\) | M1 B1 | M1 for standardising using their mean and s.d. \(= z\) value \(1 < |
| \(t = 163.548\ldots\) awrt 164 | A1 | awrt 164 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(X\) = number of times greater than 3 hours in 6 attempts; \(X \sim B(6, \text{"0.456"})\) | B1ft | ft their answer to part (a) to 3sf |
| \(P(X \geq 1) = 1 - P(X=0) = 1 - \text{"0.5438"}^6\) | M1 | Use of \(P(X \geq 1) = 1 - P(X=0)\) \([= 1-(1-\text{their}(a))^6]\). Allow \(P(X\geq1) = P(X=1)+P(X=2)+\ldots+P(X=6)\) |
| \(= 0.9741\ldots\) awrt 0.974/0.975 | A1 | awrt 0.974/0.975 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. The times for each event are not now likely to be independent | M1 | Reference to events no longer being independent (he might get tired after each event or events now follow consecutively)/calculation does not include time between events |
| Jane is correct/calculation is not valid | A1 | Correct conclusion with corresponding reason |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $T$ = total time taken; $T \sim N(41+81+57, \ 5.2^2+4.2^2+6.6^2)$ so $T \sim N(179, 88.24)$ | M1 A1 | M1 for setting up normal distribution with mean $41+81+57 (=179)$; A1 for correct variance 88.24 or s.d. awrt 9.39 |
| $P(T > 180) = P\left(Z > \frac{180-179}{\sqrt{88.24}}\right)$ | M1 | For standardising with 180, their mean and their standard deviation |
| $= 1 - 0.5438 = 0.4562$ awrt 0.456 to 0.458 | M1 A1 | Use of $1-p$ with $0.5 < p < 1$; awrt 0.456 to 0.458 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $Y$ = difference between run and swim, $Y \sim N(16, 70.6)$ or $D = R - S - 20$, $D \sim N(-4, 70.6)$ | B1 | For $N(\pm16, 70.6)$ or $N(\pm4, 70.6)$. May be seen in a calculation |
| $P(Y > 20) = P\left(Z > \frac{20-16}{\sqrt{70.6}}\right)$ or $P(D > 0) = P\left(Z > \frac{0-(-4)}{\sqrt{70.6}}\right)$ | M1 | For standardisation with $\pm20$ or 0, their mean and s.d. (variance must be $> 0$). Must be compatible |
| $= 1 - 0.6844 = 0.3156$ awrt 0.316/0.317 | M1 A1 | Use of $1-p$; awrt 0.316/0.317 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(T > t) = 0.95 \Rightarrow P\left(Z > \frac{t-179}{\sqrt{88.24}}\right) = 0.95 \Rightarrow \frac{t-179}{\sqrt{88.24}} = -1.6449$ | M1 B1 | M1 for standardising using their mean and s.d. $= z$ value $1 < |z| < 2$; B1 for correct $z$ value $\pm1.6449$ or better, must have compatible sign |
| $t = 163.548\ldots$ awrt 164 | A1 | awrt 164 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $X$ = number of times greater than 3 hours in 6 attempts; $X \sim B(6, \text{"0.456"})$ | B1ft | ft their answer to part (a) to 3sf |
| $P(X \geq 1) = 1 - P(X=0) = 1 - \text{"0.5438"}^6$ | M1 | Use of $P(X \geq 1) = 1 - P(X=0)$ $[= 1-(1-\text{their}(a))^6]$. Allow $P(X\geq1) = P(X=1)+P(X=2)+\ldots+P(X=6)$ |
| $= 0.9741\ldots$ awrt 0.974/0.975 | A1 | awrt 0.974/0.975 |
## Part (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. The times for each event are not now likely to be independent | M1 | Reference to events no longer being independent (he might get tired after each event or events now follow consecutively)/calculation does not include time between events |
| Jane is correct/calculation is not valid | A1 | Correct conclusion with corresponding reason |
---\begin{enumerate}
\item Charlie is training for three events: a 1500 m swim, a 40 km bike ride and a 10 km run.
\end{enumerate}
From past experience his times, in minutes, for each of the three events independently have the following distributions.
$$\begin{aligned}
& S \sim \mathrm {~N} \left( 41,5.2 ^ { 2 } \right) \text { represents the time for the swim } \\
& B \sim \mathrm {~N} \left( 81,4.2 ^ { 2 } \right) \text { represents the time for the bike ride } \\
& R \sim \mathrm {~N} \left( 57,6.6 ^ { 2 } \right) \text { represents the time for the run }
\end{aligned}$$
(a) Find the probability that Charlie's total time for a randomly selected swim, bike ride and run exceeds 3 hours.\\
(b) Find the probability that the time for a randomly selected swim will be at least 20 minutes quicker than the time for a randomly selected run.
Given that $\mathrm { P } ( S + B + R > t ) = 0.95$\\
(c) find the value of $t$
A triathlon consists of a 1500 m swim, immediately followed by a 40 km bike ride, immediately followed by a 10 km run.
Charlie uses the answer to part (a) to find the probability that, in 6 successive independent triathlons, his time will exceed 3 hours on at least one occasion.\\
(d) Find the answer Charlie should obtain.
Jane says that Charlie should not have used the answer to part (a) for the calculation in part (d).\\
(e) Explain whether or not Jane is correct.
\hfill \mbox{\textit{Edexcel S3 2022 Q5 [17]}}