| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Multiple independent observations |
| Difficulty | Standard +0.3 This is a standard S2 question requiring routine integration of a polynomial pdf, calculation of expectation and probability, followed by straightforward application of independence for two batteries. Part (d) involves conditional probability but is mechanical once you recognize both batteries must survive the additional time. All techniques are textbook exercises 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = \frac{1}{9}\int_1^4 (4x^2 - x^3)\, dx\) | M1 | Using \(\int xf(x)\,dx\); multiplying out; at least one of \(x^2 \to x^3\) or \(x^3 \to x^4\); ignore limits |
| \(= \frac{1}{9}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_1^4\) | A1 | Correct integration, ignore limits |
| \(= \frac{1}{9}\left[\frac{4 \times 4^3}{3} - \frac{4^4}{4}\right] - \frac{1}{9}\left[\frac{4}{3} - \frac{1}{4}\right]\) | M1d | Substituting in correct limits (allow 1 sign error) |
| \(= \frac{9}{4}\) or \(2.25\) | A1 | cao; allow equivalent fractions (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 2.5) = \frac{1}{9}\int_{2.5}^4 x(4-x)\,dx\) | M1 | Using \(\frac{1}{9}\int_{2.5}^4 x(4-x)\,dx\) or \(1 - \frac{1}{9}\int_1^{2.5} x(4-x)\,dx\); correct limits needed at some point; or \(1 - \frac{2}{9}x^2 - \frac{1}{27}x^3 - \frac{5\ddot{o}}{27\ddot{o}}\) and attempt to substitute 2.5 |
| \(= \frac{1}{9}\left[2x^2 - \frac{x^3}{3}\right]_{2.5}^4\) | A1 | Correct integration with correct limits at some point |
| \(= \frac{3}{8}\) or \(0.375\) | A1 | Allow equivalent fractions (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| P(both batteries working after 25 hours) \(= (0.375)^2\) | M1 | (their part(b))\(^2\) or writing \((P(X > 2.5))^2\) |
| \(= 0.140625\) or \(\frac{9}{64}\) | A1 | awrt 0.141 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 1.6) = \frac{1}{9}\int_{1.6}^4 x(4-x)\,dx = \frac{96}{125}\) or \(0.768\) | B1 | 0.768 or awrt 0.77 or 0.5898... or awrt 0.59; may be seen in conditional probability or implied by correct final answer |
| \(P(\text{works for 25 hours} \mid \text{worked for 16 hours}) = \frac{0.140625}{(0.768)^2}\) | M1 | \(\frac{\text{their part(c)}}{prob}\) or \(\frac{(\text{their}(b))^2}{prob}\) and numerator \(<\) denominator |
| \(= 0.2384\ldots\) | A1 | awrt 0.238 |
| NB if use one battery rather than 2 they could get B1 M0 A0 | (3 marks) |
## Question 3:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \frac{1}{9}\int_1^4 (4x^2 - x^3)\, dx$ | M1 | Using $\int xf(x)\,dx$; multiplying out; at least one of $x^2 \to x^3$ or $x^3 \to x^4$; ignore limits |
| $= \frac{1}{9}\left[\frac{4x^3}{3} - \frac{x^4}{4}\right]_1^4$ | A1 | Correct integration, ignore limits |
| $= \frac{1}{9}\left[\frac{4 \times 4^3}{3} - \frac{4^4}{4}\right] - \frac{1}{9}\left[\frac{4}{3} - \frac{1}{4}\right]$ | M1d | Substituting in correct limits (allow 1 sign error) |
| $= \frac{9}{4}$ or $2.25$ | A1 | cao; allow equivalent fractions **(4 marks)** |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 2.5) = \frac{1}{9}\int_{2.5}^4 x(4-x)\,dx$ | M1 | Using $\frac{1}{9}\int_{2.5}^4 x(4-x)\,dx$ or $1 - \frac{1}{9}\int_1^{2.5} x(4-x)\,dx$; correct limits needed at some point; or $1 - \frac{2}{9}x^2 - \frac{1}{27}x^3 - \frac{5\ddot{o}}{27\ddot{o}}$ and attempt to substitute 2.5 |
| $= \frac{1}{9}\left[2x^2 - \frac{x^3}{3}\right]_{2.5}^4$ | A1 | Correct integration with correct limits at some point |
| $= \frac{3}{8}$ or $0.375$ | A1 | Allow equivalent fractions **(3 marks)** |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| P(both batteries working after 25 hours) $= (0.375)^2$ | M1 | (their part(b))$^2$ or writing $(P(X > 2.5))^2$ |
| $= 0.140625$ or $\frac{9}{64}$ | A1 | awrt 0.141 **(2 marks)** |
### Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 1.6) = \frac{1}{9}\int_{1.6}^4 x(4-x)\,dx = \frac{96}{125}$ or $0.768$ | B1 | 0.768 or awrt 0.77 or 0.5898... or awrt 0.59; may be seen in conditional probability or implied by correct final answer |
| $P(\text{works for 25 hours} \mid \text{worked for 16 hours}) = \frac{0.140625}{(0.768)^2}$ | M1 | $\frac{\text{their part(c)}}{prob}$ or $\frac{(\text{their}(b))^2}{prob}$ and numerator $<$ denominator |
| $= 0.2384\ldots$ | A1 | awrt 0.238 |
| **NB** if use one battery rather than 2 they could get B1 M0 A0 | | **(3 marks)** |
---3. The lifetime, $X$, in tens of hours, of a battery is modelled by the probability density function
$$f ( x ) = \left\{ \begin{array} { c c }
\frac { 1 } { 9 } x ( 4 - x ) & 1 \leqslant x \leqslant 4 \\
0 & \text { otherwise }
\end{array} \right.$$
Use algebraic integration to find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { E } ( X )$
\item $\mathrm { P } ( X > 2.5 )$
A radio runs using 2 of these batteries, both of which must be working. Two fully-charged batteries are put into the radio.
\item Find the probability that the radio will be working after 25 hours of use.
Given that the radio is working after 16 hours of use,
\item find the probability that the radio will be working after being used for another 9 hours.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2017 Q3 [12]}}