| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find multiple parameters from system |
| Difficulty | Standard +0.3 This is a standard S2 probability density function question requiring integration and solving simultaneous equations. Part (i) is trivial recall. Part (ii) uses the two standard conditions (integral equals 1, given probability) to find constants—routine but multi-step. Part (iii) is straightforward expectation calculation. While it requires careful algebra with negative powers, it follows textbook methods 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 integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \([x \text{ represents a}]\) possible value(s) taken by \(X\) | B1 | Must refer to, or imply, both \(x\) and \(X\) or "the random variable" (1 mark) |
| Ignore extra unless definitely wrong |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^e ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^e = \frac{a}{8} + \frac{b}{24}\) | B1 | Correct indefinite integral [from any set of limits or none] |
| M1 | Integrate and substitute limits to obtain one expression | |
| \(\text{or} \int_1^e ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^e = \frac{a}{2} + \frac{b}{3}\) | M1 | Integrate and substitute limits to obtain a second expression |
| The limits must be two of \((1, \infty), (1, 2)\) or \((2, \infty)\), allow \((3, \infty)\) for "≥ 2" | ||
| \(\text{or} \int_1^2 ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^2 = \frac{3a}{8} + \frac{7b}{24}\) | M1 | Equate two expressions from definite integrals to 1 or \(\frac{3}{16}\); or \(\frac{3a}{8} + \frac{7b}{13}\) |
| \(\frac{a}{2} + \frac{b}{3} = 1\) or \(\frac{3a}{8} + \frac{7b}{16} = 13\) | A1 | Both equations correct, any equivalent simplified form, can be implied |
| ["simplified" = one \(a\) term, one \(b\) term, one number term] | ||
| Solve to get | A1 | Correctly show \(a = 1\) AG, www |
| \(a = 1\) | B1 | Correct value of \(b\) obtained from at least one correct equation: 7/7 |
| \(b = \frac{1}{2}\) | SC: One equation only: M1B1 M0M0A0 A0B1, max 3/7 | |
| Two equations, assume \(a = 1\), solve for \(b\), checked in other equation: 7/7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_1^x ax^{-2} + bx^{-1}\text{d}x = \left[-\frac{a}{x} - \frac{b}{2x^2}\right]_1^x\) | M1 | Integrate \(f(x)\), limits 1 and \(\infty\) seen somewhere |
| \(\left\{a + \frac{b}{2}\right\}\) | B1ft | Correct indefinite integral, their \(b\), can be implied by correct answer |
| A1 | Correctly obtain \(1\frac{1}{4}\) or a.r.t. 1.75 www, allow from calculator (3 marks) | |
| Expect to see: \(\int_1^x x^{-2} + \frac{3}{8}x^{-3}\text{d}x = \left[-\frac{1}{x} - \frac{3}{4x^2}\right]_1^x\) |
## (i)
$[x \text{ represents a}]$ possible value(s) taken by $X$ | B1 | Must refer to, or imply, both $x$ and $X$ or "the random variable" (1 mark)
| | Ignore extra unless definitely wrong
## (ii)
$\int_1^e ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^e = \frac{a}{8} + \frac{b}{24}$ | B1 | Correct indefinite integral [from any set of limits or none]
| M1 | Integrate and substitute limits to obtain one expression
$\text{or} \int_1^e ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^e = \frac{a}{2} + \frac{b}{3}$ | M1 | Integrate and substitute limits to obtain a second expression
| | The limits must be two of $(1, \infty), (1, 2)$ or $(2, \infty)$, allow $(3, \infty)$ for "≥ 2"
$\text{or} \int_1^2 ax^{-3} + bx^{-1}\text{d}x = \left[-\frac{a}{2x^2} - \frac{b}{3x^3}\right]_1^2 = \frac{3a}{8} + \frac{7b}{24}$ | M1 | Equate two expressions from definite integrals to 1 or $\frac{3}{16}$; or $\frac{3a}{8} + \frac{7b}{13}$
$\frac{a}{2} + \frac{b}{3} = 1$ or $\frac{3a}{8} + \frac{7b}{16} = 13$ | A1 | Both equations correct, any equivalent simplified form, can be implied
| | ["simplified" = one $a$ term, one $b$ term, one number term]
Solve to get | A1 | Correctly show $a = 1$ AG, www
$a = 1$ | B1 | Correct value of $b$ obtained from at least one correct equation: 7/7
$b = \frac{1}{2}$ | | SC: One equation only: M1B1 M0M0A0 A0B1, max 3/7
| | Two equations, assume $a = 1$, solve for $b$, checked in other equation: 7/7
## (iii)
$\int_1^x ax^{-2} + bx^{-1}\text{d}x = \left[-\frac{a}{x} - \frac{b}{2x^2}\right]_1^x$ | M1 | Integrate $f(x)$, limits 1 and $\infty$ seen somewhere
$\left\{a + \frac{b}{2}\right\}$ | B1ft | Correct indefinite integral, their $b$, can be implied by correct answer
| A1 | Correctly obtain $1\frac{1}{4}$ or a.r.t. 1.75 www, allow from calculator (3 marks)
| | Expect to see: $\int_1^x x^{-2} + \frac{3}{8}x^{-3}\text{d}x = \left[-\frac{1}{x} - \frac{3}{4x^2}\right]_1^x$
$= 1\frac{1}{4}$A continuous random variable $X$ has probability density function
$$\text{f}(x) = \begin{cases}
ax^{-3} + bx^{-4} & x \geq 1, \\
0 & \text{otherwise,}
\end{cases}$$
where $a$ and $b$ are constants.
\begin{enumerate}[label=(\roman*)]
\item Explain what the letter $x$ represents. [1]
\end{enumerate}
It is given that P$(X > 2) = \frac{3}{16}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $a = 1$, and find the value of $b$. [7]
\item Find E$(X)$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR S2 2016 Q7 [11]}}