| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Parameter interpretation in context |
| Difficulty | Standard +0.3 This is a straightforward S2 question requiring interpretation of a parameter, verification that the pdf integrates to 1 (routine integration of x^{-2}), and calculation of E(X) using a given value. All steps are standard textbook exercises with no novel problem-solving 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Longest lifetime | B1 [1] | Must be in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_1^a \frac{k}{x^2}\,dx = 1\) | M1 | Integrate \(f(x)\) and equate to 1; ignore limits |
| \(k\left[-\frac{1}{x}\right]_1^a = 1\) | A1 | Correct integral and limits |
| \(k\left[-\frac{1}{a}+1\right]=1\) | ||
| \(k\left[\frac{-1+a}{a}\right]=1\) or \(k(-1+a)=a\) | ||
| \(k = \frac{a}{a-1}\) AG | A1 [3] | Must be convinced (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{5}{3}\int_1^{2.5}\frac{1}{x}\,dx\) or \(k\int_1^{2.5}\frac{1}{x}\,dx\) | M1 | Integrate \(xf(x)\); ignore limits |
| \(= \frac{5}{3}\left[\ln x\right]_1^{2.5}\) or \(k[\ln x]_1^{2.5}\) | A1 | Correct integral and limits (accept "\(k\)" or "their \(k\)") |
| \(= \frac{5}{3}\ln 2.5\) or 1.53 (3 s.f.) | A1 [3] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Longest lifetime | B1 [1] | Must be in context |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_1^a \frac{k}{x^2}\,dx = 1$ | M1 | Integrate $f(x)$ and equate to 1; ignore limits |
| $k\left[-\frac{1}{x}\right]_1^a = 1$ | A1 | Correct integral and limits |
| $k\left[-\frac{1}{a}+1\right]=1$ | | |
| $k\left[\frac{-1+a}{a}\right]=1$ or $k(-1+a)=a$ | | |
| $k = \frac{a}{a-1}$ **AG** | A1 [3] | Must be convinced (AG) |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5}{3}\int_1^{2.5}\frac{1}{x}\,dx$ or $k\int_1^{2.5}\frac{1}{x}\,dx$ | M1 | Integrate $xf(x)$; ignore limits |
| $= \frac{5}{3}\left[\ln x\right]_1^{2.5}$ or $k[\ln x]_1^{2.5}$ | A1 | Correct integral and limits (accept "$k$" or "their $k$") |
| $= \frac{5}{3}\ln 2.5$ or 1.53 (3 s.f.) | A1 [3] | |
---5 The lifetime, $X$ years, of a certain type of battery has probability density function given by
$$f ( x ) = \begin{cases} \frac { k } { x ^ { 2 } } & 1 \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$
where $k$ and $a$ are positive constants.\\
(i) State what the value of $a$ represents in this context.\\
(ii) Show that $k = \frac { a } { a - 1 }$.\\
(iii) Experience has shown that the longest that any battery of this type lasts is 2.5 years. Find the mean lifetime of batteries of this type.
\hfill \mbox{\textit{CAIE S2 2014 Q5 [7]}}