| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This question tests standard linear combinations of normal distributions with straightforward application of formulas. Part (i) requires finding the distribution of 9M + 7W and calculating P(total < 1200), while part (ii) involves finding P(W > M) = P(W - M > 0). Both are routine applications of the key result that linear combinations of independent normals are normal, requiring only careful arithmetic with means and variances. No novel insight or complex problem-solving is needed. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(T) = 9 \times 78 + 7 \times 66 (= 1164)\) | B1 | Or \(9 \times 78 + 7 \times 66 - 1200\) |
| \(\text{Var}(T) = 9 \times 7^2 + 7 \times 5^2 (= 616)\) | B1 | |
| \(\frac{1200 - \text{`}1164\text{'}}{\sqrt{\text{`}616\text{'}}} (= 1.450)\) | M1 | \(\pm\) Allow without \(\sqrt{\phantom{0}}\) |
| \(P(z < 1.450) = \Phi(1.450) = 0.927\) (3 sf) | M1, A1 [5] | Correct tail consistent with their mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(D) = 66 - 78 (= -12)\) | B1 | Both needed |
| \(\text{Var}(D) = 7^2 + 5^2 (= 74)\) | ||
| \(\frac{0-(\text{`}-12\text{'})}{\sqrt{74}} (= 1.395)\) | M1 | \(\pm\) Allow without \(\sqrt{\phantom{0}}\) |
| \(P(D>0) = 1 - \Phi(\text{`}1.395\text{'}) = 0.0815\) (3 sf) | M1, A1 [4] | Correct tail consistent with their mean. Similar scheme for \(P(M-W) < 0\) |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(T) = 9 \times 78 + 7 \times 66 (= 1164)$ | B1 | Or $9 \times 78 + 7 \times 66 - 1200$ |
| $\text{Var}(T) = 9 \times 7^2 + 7 \times 5^2 (= 616)$ | B1 | |
| $\frac{1200 - \text{`}1164\text{'}}{\sqrt{\text{`}616\text{'}}} (= 1.450)$ | M1 | $\pm$ Allow without $\sqrt{\phantom{0}}$ |
| $P(z < 1.450) = \Phi(1.450) = 0.927$ (3 sf) | M1, A1 [5] | Correct tail consistent with their mean |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(D) = 66 - 78 (= -12)$ | B1 | Both needed |
| $\text{Var}(D) = 7^2 + 5^2 (= 74)$ | | |
| $\frac{0-(\text{`}-12\text{'})}{\sqrt{74}} (= 1.395)$ | M1 | $\pm$ Allow without $\sqrt{\phantom{0}}$ |
| $P(D>0) = 1 - \Phi(\text{`}1.395\text{'}) = 0.0815$ (3 sf) | M1, A1 [4] | Correct tail consistent with their mean. Similar scheme for $P(M-W) < 0$ |6 The weights, in kilograms, of men and women have the distributions $\mathrm { N } \left( 78,7 ^ { 2 } \right)$ and $\mathrm { N } \left( 66,5 ^ { 2 } \right)$ respectively.\\
\begin{enumerate}[label=(\roman*)]
\item The maximum load that a certain cable car can carry safely is 1200 kg . If 9 randomly chosen men and 7 randomly chosen women enter the cable car, find the probability that the cable car can operate safely.
\item Find the probability that a randomly chosen woman weighs more than a randomly chosen man.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2015 Q6 [9]}}