| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Compare mean and median using probability |
| Difficulty | Standard +0.3 This is a straightforward continuous probability distribution question requiring standard techniques: finding k by integration, calculating E(X), finding the median from cumulative distribution, and computing a probability. All steps are routine S2 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k\int_2^6 x^{-1}dx = 1\) | M1 | Attempt integrate \(f(x)\) & = 1. Ignore limits |
| \(k[\ln x]_2^6 = 1\) | A1 | Correct sub of correct limits in correct integral leading to correct ans. No errors seen. |
| \(k(\ln 6 - \ln 2) = 1\) or \(k\ln 3 = 1\) | ||
| \(k = \frac{1}{\ln 3}\) AG | ||
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{\ln 3}\int_2^6 1\, dx\) | M1 | Attempt integ \(xf(x)\). Ignore limits |
| \(= \frac{1}{\ln 3}[x]_2^6 \quad (= \frac{1}{\ln 3}(6-2))\) | A1 | Correct integral and limits |
| \(= \frac{4}{\ln 3} = 3.64\) AG | A1 | No errors seen |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < E(X)) = \frac{1}{\ln 3}\int_2^{3.64} x^{-1}dx\) | M1 | Attempt integ \(f(x)\) from 2 to \(\frac{4}{\ln 3}\) or 3.64 |
| \(= \frac{1}{\ln 3}[\ln x]_2^{3.64}\) | A1 | Correct sub of correct limits into correct integral |
| \(= \frac{1}{\ln 3}(\ln 3.64 - \ln 2) \quad (= 0.545)\) | ||
| \(P(m < X < E(X)) = \text{"0.545"} - 0.5\) | M1 | Subt 0.5 from their \(P(X < E(X))\); art 0.045, ft their \(P(X < E(X))\) \((> 0.5)\) |
| \(= 0.045\) (2 sfs) | A1 | Equivalent method: M1 method for median — need 0.5 and limits 2 to \(m\) or \(m\) to 6; A1 \(\sqrt{12}\) or 3.464; M1 calc area from "3.464" to 3.64; A1 0.045 or better, not 0.046 |
| Total: 4 |
# Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k\int_2^6 x^{-1}dx = 1$ | **M1** | Attempt integrate $f(x)$ & = 1. Ignore limits |
| $k[\ln x]_2^6 = 1$ | **A1** | Correct sub of correct limits in correct integral leading to correct ans. No errors seen. |
| $k(\ln 6 - \ln 2) = 1$ or $k\ln 3 = 1$ | | |
| $k = \frac{1}{\ln 3}$ **AG** | | |
| **Total: 2** | | |
# Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{\ln 3}\int_2^6 1\, dx$ | **M1** | Attempt integ $xf(x)$. Ignore limits |
| $= \frac{1}{\ln 3}[x]_2^6 \quad (= \frac{1}{\ln 3}(6-2))$ | **A1** | Correct integral and limits |
| $= \frac{4}{\ln 3} = 3.64$ **AG** | **A1** | No errors seen |
| **Total: 3** | | |
# Question 6(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < E(X)) = \frac{1}{\ln 3}\int_2^{3.64} x^{-1}dx$ | **M1** | Attempt integ $f(x)$ from 2 to $\frac{4}{\ln 3}$ or 3.64 |
| $= \frac{1}{\ln 3}[\ln x]_2^{3.64}$ | **A1** | Correct sub of correct limits into correct integral |
| $= \frac{1}{\ln 3}(\ln 3.64 - \ln 2) \quad (= 0.545)$ | | |
| $P(m < X < E(X)) = \text{"0.545"} - 0.5$ | **M1** | Subt 0.5 from their $P(X < E(X))$; art 0.045, ft their $P(X < E(X))$ $(> 0.5)$ |
| $= 0.045$ (2 sfs) | **A1** | Equivalent method: M1 method for median — need 0.5 and limits 2 to $m$ or $m$ to 6; A1 $\sqrt{12}$ or 3.464; M1 calc area from "3.464" to 3.64; A1 0.045 or better, not 0.046 |
| **Total: 4** | | |6 The random variable $X$ has probability density function given by
$$f ( x ) = \begin{cases} k x ^ { - 1 } & 2 \leqslant x \leqslant 6 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { \ln 3 }$.\\
(ii) Show that $\mathrm { E } ( X ) = 3.64$, correct to 3 significant figures.\\
(iii) Given that the median of $X$ is $m$, find $\mathrm { P } ( m < X < \mathrm { E } ( X ) )$.\\
\hfill \mbox{\textit{CAIE S2 2018 Q6 [9]}}