| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Challenging +1.2 This question requires understanding of linear combinations of normal distributions and independence, but follows standard S3 procedures: (a) form difference of sums T₁+T₂-(T₃+T₄), find combined variance using independence, standardize; (b) find P(first team wins once) then raise to power 4. The calculations are straightforward once the setup is recognized, making this moderately above average for S3 but not requiring novel insight. |
| Spec | 5.04b Linear combinations: of normal distributions |
| mean | standard deviation | |
| \(1 ^ { \text {st } }\) leg \(- A\) | 63.1 | 1.2 |
| \(2 ^ { \text {nd } }\) leg \(- B\) | 65.7 | 1.5 |
| \(3 ^ { \text {rd } } \operatorname { leg } - C\) | 65.4 | 1.8 |
| \(4 ^ { \text {th } }\) leg - \(D\) | 62.5 | 0.9 |
| Answer | Marks |
|---|---|
| \(= N(0.9, 7.74)\) | M1 A2 M1 |
| Answer | Marks |
|---|---|
| \(= P(Z < -0.32) = 1 - 0.6255 = 0.3745\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(G \sim N(259.0 - 256.7, 3.4^2 + 7.74) = N(2.3, 19.3)\) | M1 M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= P(Z > \neg 0.52) = 0.6985\) | M1 A1 | |
| \(P(\text{first team wins all four}) = (0.6985)^4 = 0.238\) | M1 A1 | (14) |
(a) let $E =$ how much longer for first two legs than next two
$\therefore E \sim N(63.1 + 65.7 - 65.4 - 62.5, 1.2^2 + 1.5^2 + 1.8^2 + 0.9^2)$
$= N(0.9, 7.74)$ | M1 A2 M1 |
$P(E < 0) = P(Z < \frac{0-0.9}{\sqrt{7.74}})$
$= P(Z < -0.32) = 1 - 0.6255 = 0.3745$ | M1 A1 |
(b) let $F =$ total time for first team
$\therefore F \sim N(63.1 + 65.7 + 65.4 + 62.5, 7.74) = N(256.7, 7.74)$
let $G =$ how much longer second team take in total
$G \sim N(259.0 - 256.7, 3.4^2 + 7.74) = N(2.3, 19.3)$ | M1 M1 A1 |
$P(\text{first team wins one race}) = P(G > 0) = P(Z > \frac{0-2.3}{\sqrt{19.3}})$
$= P(Z > \neg 0.52) = 0.6985$ | M1 A1 |
$P(\text{first team wins all four}) = (0.6985)^4 = 0.238$ | M1 A1 | (14)
---6. Four swimmers, $A , B , C$ and $D$, are to be used in a $4 \times 100$ metres freestyle relay. The time for each swimmer to complete a leg follows a normal distribution. The mean and standard deviation, in seconds, of the time for each swimmer to complete a leg and the order in which they are to swim are shown in the table below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
& mean & standard deviation \\
\hline
$1 ^ { \text {st } }$ leg $- A$ & 63.1 & 1.2 \\
\hline
$2 ^ { \text {nd } }$ leg $- B$ & 65.7 & 1.5 \\
\hline
$3 ^ { \text {rd } } \operatorname { leg } - C$ & 65.4 & 1.8 \\
\hline
$4 ^ { \text {th } }$ leg - $D$ & 62.5 & 0.9 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the total time for first two legs is less than the total time for the last two.\\
(6 marks)\\
The total time for another team to complete this relay is normally distributed with a mean of 259.0 seconds and a standard deviation of 3.4 seconds. The two teams are to compete over four races.
\item Find the probability that the first team wins all four races, assuming that the team's performances are not affected by previous results.\\
(8 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q6 [14]}}