| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a standard S3 question requiring routine application of PDF/CDF techniques: finding a constant by integration, deriving a CDF by integrating the piecewise PDF, and using transformation of variables. Part (iii) involves the standard Y = g(T) transformation method. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total area from \(t=0\) to \(4\) is \(1\): \(\frac{1}{2}a + 3a = 1\) | M1 | Any method |
| Solve to give \(a = 2/7\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 \quad t<0\) | ||
| \(t^2/7 \quad 0 \leq t \leq 1\) | B1 | Ft \(a\) for B1 and A1 |
| \("1/7" + \int_1^t 2dt/7 = 2t/7 - 1/7\) | M1 A1 | |
| \(1 \quad t>4\) | B1 | For \(t<0\) and \(t>4\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(Y) = P(Y \leq y) = P(T \leq y^2)= F(y^2)\) | M1 A1 | Allow \(<\). Seen or implied. Possible to score M0A0 then M1A1M1A1B1. \(\int_0^1 2tdt/7 + \int_1^4 2dt/7=1\) M1, \(2ydy=dt\) oe M1 |
| \(G(y) = \begin{cases} 0 & y<0 \\ y^4/7 & 0 \leq y \leq 1 \\ (2y^2-1)/7 & 1 < y \leq 2 \\ 1 & y \geq 2 \end{cases}\) | M1 A1ft | For using F correctly. For correct expressions ft \(a\) and \(F(t)\). \((\int 4y^3dy/7 + \int 4ydy/7=1)\), \(g(y)=4y^3/7, 4y/7\) A1, \(0\leq y\leq1,\ 1 |
| \(g(y) = G'(y) = \begin{cases} 4y^3/7 & 0 \leq y \leq 1 \\ 4y/7 & 1 < y \leq 2 \\ 0 & \text{otherwise} \end{cases}\) | M1 A1 B1 | For differentiating. Correctly, allow from eg \(2y^2/7\). Correct ranges for \(y\) seen |
| [7] |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total area from $t=0$ to $4$ is $1$: $\frac{1}{2}a + 3a = 1$ | M1 | Any method |
| Solve to give $a = 2/7$ | A1 | |
| | **[2]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 \quad t<0$ | | |
| $t^2/7 \quad 0 \leq t \leq 1$ | B1 | Ft $a$ for B1 and A1 |
| $"1/7" + \int_1^t 2dt/7 = 2t/7 - 1/7$ | M1 A1 | |
| $1 \quad t>4$ | B1 | For $t<0$ and $t>4$ |
| | **[4]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(Y) = P(Y \leq y) = P(T \leq y^2)= F(y^2)$ | M1 A1 | Allow $<$. Seen or implied. Possible to score M0A0 then M1A1M1A1B1. $\int_0^1 2tdt/7 + \int_1^4 2dt/7=1$ M1, $2ydy=dt$ oe M1 |
| $G(y) = \begin{cases} 0 & y<0 \\ y^4/7 & 0 \leq y \leq 1 \\ (2y^2-1)/7 & 1 < y \leq 2 \\ 1 & y \geq 2 \end{cases}$ | M1 A1ft | For using F correctly. For correct expressions ft $a$ and $F(t)$. $(\int 4y^3dy/7 + \int 4ydy/7=1)$, $g(y)=4y^3/7, 4y/7$ A1, $0\leq y\leq1,\ 1<y\leq2$ B1, $G(y)=y^4/7$ B1, $2y^2/7+c$ and $G(1)=1/7$ M1, $(2y^2-1)/7$ A1 |
| $g(y) = G'(y) = \begin{cases} 4y^3/7 & 0 \leq y \leq 1 \\ 4y/7 & 1 < y \leq 2 \\ 0 & \text{otherwise} \end{cases}$ | M1 A1 B1 | For differentiating. Correctly, allow from eg $2y^2/7$. Correct ranges for $y$ seen |
| | **[7]** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{054e0081-afce-4a87-93f5-650dad40b313-3_508_611_262_719}
The diagram shows the probability density function f of the continuous random variable $T$, given by
$$f ( t ) = \begin{cases} a t & 0 \leqslant t \leqslant 1 \\ a & 1 < t \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
where $a$ is a constant.\\
(i) Find the value of $a$.\\
(ii) Obtain the cumulative distribution function of $T$.\\
(iii) Find the cumulative distribution of $Y$, where $Y = T ^ { \frac { 1 } { 2 } }$, and hence find the probability density function of $Y$.
\hfill \mbox{\textit{OCR S3 2012 Q6 [13]}}