| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Verify algebraic PDF formula |
| Difficulty | Moderate -0.3 This is a routine S3 question testing standard PDF/CDF verification techniques. Part (i) requires integrating ln(y) using integration by parts (a standard A-level technique), parts (ii)-(iii) are straightforward applications of definitions, and part (iv) uses the transformation formula for CDFs. While it requires multiple steps and integration by parts, all techniques are standard bookwork with no novel problem-solving 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 |
| \(f\) is non-negative over \([1, e]\) | B1 | |
| Attempt to show area, between 1 and \(e\), \(= 1\) | M1 | |
| \(\int_1^e \ln y\,dy = y\ln y - \int dy\) | M1 | Integrate by parts. Allow 1 error. |
| \(= \left[y\ln y - y\right]_1^e\) | ||
| \(= (e\ln e - e) - (1\ln 1 - 1) = 1\) | A1 | cwo |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(y) = \int_1^y \ln t\,dt = \left[t\ln t - t\right]_1^y\) | M1 | Limits must be correct. If indef integral, must have \(+c\) and attempt to evaluate |
| \(= (y\ln y - y) - (1\ln 1 - 1)\) | A1 | \(1\ln 1 - 1 + c = 0\) OR \(e\ln e - e + c = 1\) |
| \(= y\ln y - y + 1\) AG, over \([1, e]\) | A1 | Needs proper justification. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F(2.45) = 0.745,\ F(2.46) = 0.754\) | M1 | or \(1 - F = 0.255, 0.246\) or \(y\ln y - y + 1 = 0.75\) oe |
| and \(0.745 < 0.75 < 0.754\) and result follows | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(G(x) = P(X < x)\) | ||
| \(= P(\ln Y < x)\) | ||
| \(= P(Y < e^x)\) | M1 | |
| \(= F(e^x)\) | M1 | Allow M0M1A1B1 |
| \(= (e^x \ln e^x - e^x + 1)\ \Rightarrow\ xe^x - e^x + 1\); over \([0,1]\) | A1;B1 | |
| [4] |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f$ is non-negative over $[1, e]$ | B1 | |
| Attempt to show area, between 1 and $e$, $= 1$ | M1 | |
| $\int_1^e \ln y\,dy = y\ln y - \int dy$ | M1 | Integrate by parts. Allow 1 error. |
| $= \left[y\ln y - y\right]_1^e$ | | |
| $= (e\ln e - e) - (1\ln 1 - 1) = 1$ | A1 | cwo |
| | **[4]** | |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(y) = \int_1^y \ln t\,dt = \left[t\ln t - t\right]_1^y$ | M1 | Limits must be correct. If indef integral, must have $+c$ and attempt to evaluate |
| $= (y\ln y - y) - (1\ln 1 - 1)$ | A1 | $1\ln 1 - 1 + c = 0$ OR $e\ln e - e + c = 1$ |
| $= y\ln y - y + 1$ AG, over $[1, e]$ | A1 | Needs proper justification. |
| | **[3]** | |
---
# Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F(2.45) = 0.745,\ F(2.46) = 0.754$ | M1 | or $1 - F = 0.255, 0.246$ or $y\ln y - y + 1 = 0.75$ oe |
| and $0.745 < 0.75 < 0.754$ and result follows | A1 | |
| | **[2]** | |
---
# Question 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $G(x) = P(X < x)$ | | |
| $= P(\ln Y < x)$ | | |
| $= P(Y < e^x)$ | M1 | |
| $= F(e^x)$ | M1 | Allow M0M1A1B1 |
| $= (e^x \ln e^x - e^x + 1)\ \Rightarrow\ xe^x - e^x + 1$; over $[0,1]$ | A1;B1 | |
| | **[4]** | |
---5 The continuous random variable $Y$ has probability density function given by
$$\mathrm { f } ( y ) = \begin{cases} \ln ( y ) & 1 \leqslant y \leqslant \mathrm { e } \\ 0 & \text { otherwise } \end{cases}$$
(i) Verify that this is a valid probability density function.\\
(ii) Show that the (cumulative) distribution function of $Y$ is given by
$$\mathrm { F } ( y ) = \begin{cases} 0 & y < 1 \\ y \ln y - y + 1 & 1 \leqslant y \leqslant \mathrm { e } \\ 1 & \text { otherwise } \end{cases}$$
(iii) Verify that the upper quartile of $Y$ lies in the interval [2.45, 2.46].\\
(iv) Find the (cumulative) distribution function of $X$ where $X = \ln Y$.
\hfill \mbox{\textit{OCR S3 2013 Q5 [13]}}