| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Pure expectation and variance calculation |
| Difficulty | Moderate -0.8 This is a straightforward application of standard results for linear combinations of independent normal variables and sampling distributions. Part (a) requires simple addition of means and variances, part (b) is a routine normal probability calculation, and part (c) applies the sampling distribution of the mean formula. All steps are direct recall with no problem-solving or insight required. |
| Spec | 5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean \(= 15.0 + 32.0 + 8.6\ [= 55.6]\) | B1 | Allow unsimplified |
| \(\text{Var} = 1.1^2 + 3.5^2 + 1.2^2\ [= 14.9]\) | B1 | Allow unsimplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{60 - \text{"55.6"}}{\sqrt{\text{"14.9"}}}\ [= 1.140]\) | M1 | FT *their* 55.6 and 14.9. Ignore continuity correction |
| \(1 - \phi(\text{"1.140"})\) | M1 | For correct probability area consistent with *their* working |
| \(0.127\) (3 sf) | A1 | CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{54.5 - \text{"55.6"}}{\sqrt{\frac{\text{"14.9"}}{15}}}\) or \(\frac{817.5 - 834}{\sqrt{223.5}}\ [= -1.104]\) | M1 | FT *their* 55.6 and 14.9. No mixed methods |
| \(1 - \phi(\text{"1.104"})\) | M1 | For correct probability area consistent with their working |
| \(0.135\) (3 sf) | A1 | As final answer |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean $= 15.0 + 32.0 + 8.6\ [= 55.6]$ | B1 | Allow unsimplified |
| $\text{Var} = 1.1^2 + 3.5^2 + 1.2^2\ [= 14.9]$ | B1 | Allow unsimplified |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{60 - \text{"55.6"}}{\sqrt{\text{"14.9"}}}\ [= 1.140]$ | M1 | FT *their* 55.6 and 14.9. Ignore continuity correction |
| $1 - \phi(\text{"1.140"})$ | M1 | For correct probability area consistent with *their* working |
| $0.127$ (3 sf) | A1 | CWO |
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{54.5 - \text{"55.6"}}{\sqrt{\frac{\text{"14.9"}}{15}}}$ or $\frac{817.5 - 834}{\sqrt{223.5}}\ [= -1.104]$ | M1 | FT *their* 55.6 and 14.9. No mixed methods |
| $1 - \phi(\text{"1.104"})$ | M1 | For correct probability area consistent with their working |
| $0.135$ (3 sf) | A1 | As final answer |4 Wendy's journey to work consists of three parts: walking to the train station, riding on the train and then walking to the office. The times, in minutes, for the three parts of her journey are independent and have the distributions $\mathrm { N } \left( 15.0,1.1 ^ { 2 } \right) , \mathrm { N } \left( 32.0,3.5 ^ { 2 } \right)$ and $\mathrm { N } \left( 8.6,1.2 ^ { 2 } \right)$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the mean and variance of the total time for Wendy's journey.\\
If Wendy's journey takes more than 60 minutes, she is late for work.
\item Find the probability that, on a randomly chosen day, Wendy will be late for work.
\item Find the probability that the mean of Wendy's journey times over 15 randomly chosen days will be less than 54.5 minutes.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q4 [8]}}