| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | All components random including container |
| Difficulty | Standard +0.3 This is a standard application of linear combinations of normal random variables requiring students to form difference distributions (part a) and sum distributions (part b). While it involves multiple steps and careful handling of variances, the techniques are routine for S3 level with no novel insight required—slightly easier than average due to straightforward setup. |
| 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 |
| Attempt to rank depths against distances | M1 | For ranking attempt |
| \(d\) values calculated from ranks | M1 | Must be using ranks |
| \(\sum d^2 = 8\) | M1A1 | For sum of 8 (or 104 for reverse ranking) |
| \(r_s = 1 - \frac{6 \times 8}{7 \times 48}\) | M1 | Correct formula with their \(\sum d^2\); if answer not correct an expression is required |
| \(= \frac{6}{7} = 0.857142...\) awrt 0.857 | A1 | Sign should correspond to ranking (use of 104 should get \(-0.857\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \rho = 0,\ H_1: \rho > 0\) | B1 | Both hypotheses in terms of \(\rho\); \(H_1\) must be one tail and compatible with ranking |
| Critical value at 1% level is 0.8929 | B1 | Accept \(\pm\) |
| \(r_s < 0.8929\) so not significant evidence to reject \(H_0\) | M1 | Correct statement relating \(r_s\) to cv; cv must satisfy \( |
| The researcher's claim is not correct (at 1% level) / insufficient evidence for researcher's claim / insufficient evidence that water gets deeper further from inner bank | A1ft | Must mention "researcher" and "claim" or "distance (from bank)" and "depth (of water)"; use of "association" is A0 |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to rank depths against distances | M1 | For ranking attempt |
| $d$ values calculated from ranks | M1 | Must be using ranks |
| $\sum d^2 = 8$ | M1A1 | For sum of 8 (or 104 for reverse ranking) |
| $r_s = 1 - \frac{6 \times 8}{7 \times 48}$ | M1 | Correct formula with their $\sum d^2$; if answer not correct an expression is required |
| $= \frac{6}{7} = 0.857142...$ awrt **0.857** | A1 | Sign should correspond to ranking (use of 104 should get $-0.857$) |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \rho = 0,\ H_1: \rho > 0$ | B1 | Both hypotheses in terms of $\rho$; $H_1$ must be one tail and compatible with ranking |
| Critical value at 1% level is 0.8929 | B1 | Accept $\pm$ |
| $r_s < 0.8929$ so not significant evidence to reject $H_0$ | M1 | Correct statement relating $r_s$ to cv; cv must satisfy $|cv|<1$ |
| The researcher's claim is not correct (at 1% level) / insufficient evidence for researcher's claim / insufficient evidence that water gets deeper further from inner bank | A1ft | Must mention "researcher" and "claim" or "distance (from bank)" and "depth (of water)"; use of "association" is A0 |
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4. A farm produces potatoes. The potatoes are packed into sacks.
The weight of a sack of potatoes is modelled by a normal distribution with mean 25.6 kg and standard deviation 0.24 kg
\begin{enumerate}[label=(\alph*)]
\item Find the probability that two randomly chosen sacks of potatoes differ in weight by more than 0.5 kg
Sacks of potatoes are randomly selected and packed onto pallets.
The weight of an empty pallet is modelled by a normal distribution with mean 20.0 kg and standard deviation 0.32 kg
Each full pallet of potatoes holds 30 sacks of potatoes.
\item Find the probability that the total weight of a randomly chosen full pallet of potatoes is greater than 785 kg\\
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{0434a6c1-686a-449d-ba16-dbb8e60288e8-15_2258_51_313_36}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2018 Q4 [11]}}