| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Linear transformation to achieve target parameters |
| Difficulty | Standard +0.3 Part (a) is straightforward algebra using standard deviation and mean relationships for linear transformations. Part (b) requires knowledge that sums of independent normals are normal, standardization, and using tables—all standard S2 techniques with no novel insight required. The two-part structure and routine application of formulas places this slightly above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{Var}(Y) = b^2\text{Var}(X) = 4\text{Var}(X)\) \(b = 2 (> 0)\) \(E(Y) = a + 2E(X), E(X) = 0.8\) \(a + 14.24 = 7.92\) \(a = -6.32\) | B1, M1, A1 [3] | Correct answer. Using E(Y) = a + 2E(X), their "2". Correct equation in \(a\), \(b\) subst in, using 0.8 correctly. Correct answer |
| (b) (i) \(R + S \sim N(3\mu, 13)\) \(z = -1.282\) or \(-1.281\) \(1.282 = \frac{3\mu - 1}{\sqrt{13}}\) \(\mu = 1.87\) | B1, B1, M1, A1 [4] | Correct mean and variance. Correct \(z\), must be \(-\)ve (or \(-\)ve implied). Standardising and solving, their mean, z, 13 (but combined), allow "13" or "√13". Correct answer. |
| (ii) \(S - R \sim N(\mu, 13)\) \(P[(S-R) > 0] = P\left(z > \frac{0-\mu}{\sqrt{13}}\right) = \Phi(0.5197) = 0.699\) | B1 ft, M1, A1 [3] | Correct mean and variance, ft their mean. Standardising and correct area (i.e. \(> 0.5\) for correct), but consistent with their values. Accept 0.698 |
(a) $\text{Var}(Y) = b^2\text{Var}(X) = 4\text{Var}(X)$ $b = 2 (> 0)$ $E(Y) = a + 2E(X), E(X) = 0.8$ $a + 14.24 = 7.92$ $a = -6.32$ | B1, M1, A1 [3] | Correct answer. Using E(Y) = a + 2E(X), their "2". Correct equation in $a$, $b$ subst in, using 0.8 correctly. Correct answer
(b) (i) $R + S \sim N(3\mu, 13)$ $z = -1.282$ or $-1.281$ $1.282 = \frac{3\mu - 1}{\sqrt{13}}$ $\mu = 1.87$ | B1, B1, M1, A1 [4] | Correct mean and variance. Correct $z$, must be $-$ve (or $-$ve implied). Standardising and solving, their mean, z, 13 (but combined), allow "13" or "√13". Correct answer.
(ii) $S - R \sim N(\mu, 13)$ $P[(S-R) > 0] = P\left(z > \frac{0-\mu}{\sqrt{13}}\right) = \Phi(0.5197) = 0.699$ | B1 ft, M1, A1 [3] | Correct mean and variance, ft their mean. Standardising and correct area (i.e. $> 0.5$ for correct), but consistent with their values. Accept 0.6987
\begin{enumerate}[label=(\alph*)]
\item Random variables $Y$ and $X$ are related by $Y = a + b X$, where $a$ and $b$ are constants and $b > 0$. The standard deviation of $Y$ is twice the standard deviation of $X$. The mean of $Y$ is 7.92 and is 0.8 more than the mean of $X$. Find the values of $a$ and $b$.
\item Random variables $R$ and $S$ are such that $R \sim \mathrm {~N} \left( \mu , 2 ^ { 2 } \right)$ and $S \sim \mathrm {~N} \left( 2 \mu , 3 ^ { 2 } \right)$. It is given that $\mathrm { P } ( R + S > 1 ) = 0.9$.
\begin{enumerate}[label=(\roman*)]
\item Find $\mu$.
\item Hence find $\mathrm { P } ( S > R )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2009 Q7 [10]}}