| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| 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 This question tests standard transformations of normal distributions and properties of linear combinations. Parts (i)-(ii) involve routine application of E(aX+b) and Var(aX+b) formulas to find constants. Part (iii) requires recognizing that Y₂ - Y₁ ~ N(0, 200) and calculating a probability, which is a standard technique in S3. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(Y) = aE(X) + b \Rightarrow 40 = 48.5a + b\) | \(M1\) | If Var(Y)=...+b used allow final M1 for a=0.794 b=1.49 from GC |
| \(\text{Var}(Y) = a^2\text{Var}(X)\) OR SDY \(= a \cdot \text{SDX}, (10=12.5a)\) | \(M1\) | Solve simultaneously |
| \(\therefore a=0.8 \text{ } b=1.2\) | \(A1 A1\) | |
| \([4]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(N(40, 100)\) | \(B1\) | Allow \(10^2\) |
| \([1]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Use \(Y_2 - Y_1 \sim N(0, \sigma^2)\) | \(M1 A1\) | Allow use of X instead of Y. |
| \(\sigma^2=200\) | \(A1\) | \(\sigma^2=312.5\) |
| \(P(Y_2 - Y_1 \geq 10) =\) | \(M1\) | \(10/a (=12.5)\) |
| \(=1 - \phi([10/\sqrt{"200"}])\) | \(["12.5"/ \sqrt{"312.5"}]\) | |
| \(= 0.2399\) | \(A1\) | ART 0.240 |
| \([5]\) |
### (i)
$E(Y) = aE(X) + b \Rightarrow 40 = 48.5a + b$ | $M1$ | If Var(Y)=...+b used allow final M1 for a=0.794 b=1.49 from GC
$\text{Var}(Y) = a^2\text{Var}(X)$ OR SDY $= a \cdot \text{SDX}, (10=12.5a)$ | $M1$ | Solve simultaneously
$\therefore a=0.8 \text{ } b=1.2$ | $A1 A1$ |
| $[4]$ |
### (ii)
$N(40, 100)$ | $B1$ | Allow $10^2$
| $[1]$ |
### (iii)
Use $Y_2 - Y_1 \sim N(0, \sigma^2)$ | $M1 A1$ | Allow use of X instead of Y.
$\sigma^2=200$ | $A1$ | $\sigma^2=312.5$
$P(Y_2 - Y_1 \geq 10) =$ | $M1$ | $10/a (=12.5)$
$=1 - \phi([10/\sqrt{"200"}])$ | | $["12.5"/ \sqrt{"312.5"}]$
$= 0.2399$ | $A1$ | ART 0.240
| $[5]$ |$X$ is a continuous random variable with the distribution N$(48.5, 12.5^2)$. The values of $X$ are transformed to standardised values of $Y$, using the equation $Y = aX + b$, where $a$ and $b$ are constants with $a > 0$.
\begin{enumerate}[label=(\roman*)]
\item Find values of $a$ and $b$ for which the mean and standard deviation of $Y$ are 40 and 10 respectively. [4]
\item State the distribution of $Y$. [1]
\end{enumerate}
Two randomly chosen standardised values are denoted by $Y_1$ and $Y_2$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Calculate the probability that $Y_2$ is at least 10 greater than $Y_1$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2012 Q4 [10]}}