| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Power transformation (Y = X^n, n≥2) |
| Difficulty | Standard +0.8 This S3 question requires understanding of transformations of random variables (particularly the CDF method for Y=X²), working with piecewise PDFs, and connecting E(Y) via both direct and transformed approaches. The symmetry insight for part (i) and the differentiation in part (ii) require careful technique, placing this above average difficulty but within reach of well-prepared S3 students. |
| Spec | 5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P( | X | < a) = P(-a < X < a)\) |
| \(= \int_{-a}^0(1+x)dx + \int_0^a(1-x)dx\) | A1 | For integrals or equivalent trapezi |
| \(= \left[x + \frac{1}{2}x^2\right]_0^a + \left[x - \frac{1}{2}x^2\right]_0^a = 2a - a^2\) | A1 | 3 |
| (ii) \(P(Y \leq y) = P(X^2 \leq y) = P\left( | X | \leq \sqrt{y}\right) = 2\sqrt{y} - y\) |
| A1 | For correct expression \(2\sqrt{y} - y\) | |
| Hence the pdf of Y is \(\frac{d}{dy}\left(2\sqrt{y} - y\right) = \frac{1}{\sqrt{y}} - 1\) | M1 | For differentiation of previous expression |
| A1 | 4 | For showing the given answer correctly |
| (iii) \(E(Y) = \int_0^1 y^2 - y\,dy = \left[\frac{1}{3}y^3 - \frac{1}{2}y^2\right]_0^1 = \frac{1}{6}\) | M1 | For the correct integral in terms of \(y\) |
| A1 | For correct answer \(\frac{1}{6}\) | |
| \(E(X^2) = \int_{-1}^0(x^2 + x^3)dx + \int_0^1(x^2 - x^3)dx\) | M1 | For the correct integrals in terms of \(x\) |
| \(= \left[\frac{1}{3}x^3 + \frac{1}{4}x^4\right]_{-1}^0 + \left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]_0^1 = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}\) | A1 | 4 |
| (iv) \(E(\sqrt{Y}) = \int_0^1 y^{\frac{1}{2}}g(y)\,dy = \int_0^1(1-y^{\frac{1}{2}})dy\) | M1 | For forming the correct integral |
| \(= \left[y - \frac{2}{3}y^{\frac{3}{2}}\right]_0^1 = \frac{1}{3}\) | A1 | 2 |
**(i)** $P(|X| < a) = P(-a < X < a)$ | M1 | For consideration of two areas, or equiv
$= \int_{-a}^0(1+x)dx + \int_0^a(1-x)dx$ | A1 | For integrals or equivalent trapezi
$= \left[x + \frac{1}{2}x^2\right]_0^a + \left[x - \frac{1}{2}x^2\right]_0^a = 2a - a^2$ | A1 | 3 | For showing the given answer correctly
**(ii)** $P(Y \leq y) = P(X^2 \leq y) = P\left(|X| \leq \sqrt{y}\right) = 2\sqrt{y} - y$ | M1 | For expression of $P\left(X^2 \leq y\right)$ in terms of $y$
| A1 | For correct expression $2\sqrt{y} - y$
Hence the pdf of Y is $\frac{d}{dy}\left(2\sqrt{y} - y\right) = \frac{1}{\sqrt{y}} - 1$ | M1 | For differentiation of previous expression
| A1 | 4 | For showing the given answer correctly
**(iii)** $E(Y) = \int_0^1 y^2 - y\,dy = \left[\frac{1}{3}y^3 - \frac{1}{2}y^2\right]_0^1 = \frac{1}{6}$ | M1 | For the correct integral in terms of $y$
| A1 | For correct answer $\frac{1}{6}$
$E(X^2) = \int_{-1}^0(x^2 + x^3)dx + \int_0^1(x^2 - x^3)dx$ | M1 | For the correct integrals in terms of $x$
$= \left[\frac{1}{3}x^3 + \frac{1}{4}x^4\right]_{-1}^0 + \left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]_0^1 = \frac{1}{12} + \frac{1}{12} = \frac{1}{6}$ | A1 | 4 | For the correct answer correctly obtained
**(iv)** $E(\sqrt{Y}) = \int_0^1 y^{\frac{1}{2}}g(y)\,dy = \int_0^1(1-y^{\frac{1}{2}})dy$ | M1 | For forming the correct integral
$= \left[y - \frac{2}{3}y^{\frac{3}{2}}\right]_0^1 = \frac{1}{3}$ | A1 | 2 | For the correct answer $\frac{1}{3}$5 The continuous random variable $X$ has a triangular distribution with probability density function given by
$$f ( x ) = \left\{ \begin{array} { l r }
1 + x & - 1 \leqslant x \leqslant 0 \\
1 - x & 0 \leqslant x \leqslant 1 \\
0 & \text { otherwise }
\end{array} \right.$$
(i) Show that, for $0 \leqslant a \leqslant 1$,
$$\mathrm { P } ( | X | \leqslant a ) = 2 a - a ^ { 2 } .$$
The random variable $Y$ is given by $Y = X ^ { 2 }$.\\
(ii) Express $\mathrm { P } ( Y \leqslant y )$ in terms of $y$, for $0 \leqslant y \leqslant 1$, and hence show that the probability density function of $Y$ is given by
$$g ( y ) = \frac { 1 } { \sqrt { } y } - 1 , \quad \text { for } 0 < y \leqslant 1 .$$
(iii) Use the probability density function of $Y$ to find $\mathrm { E } ( Y )$, and show how the value of $\mathrm { E } ( Y )$ may also be obtained directly using the probability density function of $X$.\\
(iv) Find $\mathrm { E } ( \sqrt { } Y )$.
\hfill \mbox{\textit{OCR S3 Q5 [13]}}