| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparing two journey times |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal distributions with clear setup. Part (a) requires forming P-J ~ N(-1, 21) and finding P(J<P), while part (b) scales to 60 laps. The conceptual framework is standard for S3, requiring only routine manipulation of means/variances and normal table lookup, though the multi-step nature and careful arithmetic elevate it slightly above average difficulty. |
| 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 |
| \((J-P) \sim N(1, 21)\) | M1, A1 | For attempting \(J-P\) and \(E(J-P)\) or \(P-J\) and \(E(P-J)\); variance of 21 (accept \(9+12\)) |
\(P(J| dM1 |
Dependent on previous M; attempting correct probability and standardising with mean and sd |
|
| \(= P(Z < -0.2182...)\) | ||
| \(= 1 - 0.5871 = 0.4129\) | A1 | awrt \((0.413 \sim 0.414)\); calculator gives \(0.4136...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X = (J_1+J_2+...+J_{60})-(P_1+P_2+...+P_{60})\) | M1 | Clear attempt to identify correct form for \(X\) |
| \(E(X) = 60\times91 - 60\times90 = 60\) | B1 | For \(E(X)=60\); stated as \(E(X)=60\) or \(X\sim N(60,...)\) |
| \(\text{Var}(X) = 60\times9 + 60\times12 = 1260\) | A1 | For correct variance |
| \(P(X>120) = P\left(Z > \frac{120-60}{\sqrt{1260}}\right)\) | M1 | Attempting correct probability, standardising with 120, their 60 and 1260 |
| \(= P(Z > 1.69030...)\) | ||
| \(= 1 - 0.9545 = 0.0455\) | A1 | awrt \((0.0455)\) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(J-P) \sim N(1, 21)$ | M1, A1 | For attempting $J-P$ and $E(J-P)$ or $P-J$ and $E(P-J)$; variance of 21 (accept $9+12$) |
| $P(J<P) = P(J-P<0) = P\left(Z < \frac{0-1}{\sqrt{21}}\right)$ | dM1 | Dependent on previous M; attempting correct probability and standardising with mean and sd |
| $= P(Z < -0.2182...)$ | | |
| $= 1 - 0.5871 = 0.4129$ | A1 | awrt $(0.413 \sim 0.414)$; calculator gives $0.4136...$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X = (J_1+J_2+...+J_{60})-(P_1+P_2+...+P_{60})$ | M1 | Clear attempt to identify correct form for $X$ |
| $E(X) = 60\times91 - 60\times90 = 60$ | B1 | For $E(X)=60$; stated as $E(X)=60$ or $X\sim N(60,...)$ |
| $\text{Var}(X) = 60\times9 + 60\times12 = 1260$ | A1 | For correct variance |
| $P(X>120) = P\left(Z > \frac{120-60}{\sqrt{1260}}\right)$ | M1 | Attempting correct probability, standardising with 120, their 60 and 1260 |
| $= P(Z > 1.69030...)$ | | |
| $= 1 - 0.9545 = 0.0455$ | A1 | awrt $(0.0455)$ |
---2. Philip and James are racing car drivers. Philip's lap times, in seconds, are normally distributed with mean 90 and variance 9. James' lap times, in seconds, are normally distributed with mean 91 and variance 12. The lap times of Philip and James are independent. Before a race, they each take a qualifying lap.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that James' time for the qualifying lap is less than Philip's.
The race is made up of 60 laps. Assuming that they both start from the same starting line and lap times are independent,
\item find the probability that Philip beats James in the race by more than 2 minutes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q2 [9]}}