| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Interquartile range calculation |
| Difficulty | Standard +0.3 This is a straightforward S3 question testing standard probability density function techniques. Part (i) requires finding quartiles by integration (routine but slightly tedious due to the piecewise function), part (ii) is a standard variance calculation using E(Y²) - [E(Y)]², and part (iii) requires recognizing the symmetry to simplify E(|Y|). All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.03c Calculate mean/variance: by integration5.03d E(g(X)): general expectation formula |
| Answer | Marks | Guidance |
|---|---|---|
| \(Q_3 : \int_o^4 \frac{1}{y}dy = 1/4 \Rightarrow \left[\frac{1}{8}y^2\right]_0 = 1/4 \Rightarrow Q_3 = \sqrt{2}\) oe for LQ | \(M1\) | For 1 correct integral with limits |
| \(Q_1 = -\sqrt{2}\) | \(A1\) | For correct equation |
| IQR \(= 2\sqrt{2}\) | \(A1\) | For \(Q_1\) or \(Q_3\) allow 1.41 -1.41 |
| \(A1\) | For IQR allow 2.83 | |
| \([4]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(Y^2) = \int_{-2}^{\frac{y^4}{4}} dy + \int_0^2 \frac{y^3}{4} dy\) | \(M1\) | |
| \(= [-y^3/16] + [y^4/16]\) | \(A1\) | |
| \(= 2\) | \(A1\) | Allow final A1 even if E(Y)=0 missing, but not if 2 missing from Var(Y)=E(Y²)-[E(Y)]² |
| \(E(Y) = 0 ; \text{Var}(Y) = 2\) | \(B1 A1\) | |
| \([5]\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^0 (y^2/4)dy + \int_0^2(y^2/4)dy\) | \(M1 A1\) | Allow M1A0 for one of these integrals correct and the other incorrect or missing or correct and not added. |
| \(= 2/3 + 2/3\) | \(A1\) | If f([Y])used , B1 for y/2, M1A1 for \(\int(y/2)dy\) ,with limits, AEF |
| \(= 4/3\) | \(A1\) | SC M1A1 for 4/3 obtained from incorrect use of modulus in negative y case |
| \([4]\) |
### (i)
$Q_3 : \int_o^4 \frac{1}{y}dy = 1/4 \Rightarrow \left[\frac{1}{8}y^2\right]_0 = 1/4 \Rightarrow Q_3 = \sqrt{2}$ oe for LQ | $M1$ | For 1 correct integral with limits
$Q_1 = -\sqrt{2}$ | $A1$ | For correct equation
IQR $= 2\sqrt{2}$ | $A1$ | For $Q_1$ or $Q_3$ allow 1.41 -1.41
| $A1$ | For IQR allow 2.83
| $[4]$ |
### (ii)
$E(Y^2) = \int_{-2}^{\frac{y^4}{4}} dy + \int_0^2 \frac{y^3}{4} dy$ | $M1$ |
$= [-y^3/16] + [y^4/16]$ | $A1$ |
$= 2$ | $A1$ | Allow final A1 even if E(Y)=0 missing, but not if 2 missing from Var(Y)=E(Y²)-[E(Y)]²
$E(Y) = 0 ; \text{Var}(Y) = 2$ | $B1 A1$ |
| $[5]$ |
### (iii)
$\int_0^0 (y^2/4)dy + \int_0^2(y^2/4)dy$ | $M1 A1$ | Allow M1A0 for one of these integrals correct and the other incorrect or missing or correct and not added.
$= 2/3 + 2/3$ | $A1$ | If f([Y])used , B1 for y/2, M1A1 for $\int(y/2)dy$ ,with limits, AEF
$= 4/3$ | $A1$ | SC M1A1 for 4/3 obtained from incorrect use of modulus in negative y case
| $[4]$ |The continuous random variable $Y$ has probability density function given by
$$\text{f}(y) = \begin{cases}
-\frac{1}{4}y & -2 < y < 0, \\
\frac{1}{4}y & 0 \leqslant y \leqslant 2, \\
0 & \text{otherwise.}
\end{cases}$$
Find
\begin{enumerate}[label=(\roman*)]
\item the interquartile range of $Y$, [4]
\item Var$(Y)$, [5]
\item E$(|Y|)$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2012 Q6 [13]}}