| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard CDF/PDF manipulation. Parts (a)-(c) involve routine calculus operations (solving F(x)=0.5, differentiating for pdf, evaluating 1-F(1.2)). Part (d) requires recognizing independence and computing a probability to the fourth power. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| \(m^2 + 2m - 4.125 = 0\) | M1; M1 | (3) |
| \(m = \frac{-2 \pm \sqrt{4 + 16.5}}{2}\) | A1 | |
| \(m = 1.26, -3.264\) | ||
| (median) = 1.26 | ||
| (b) Differentiating \(\frac{d}{dx}\left[\frac{4}{9}(x^2 + 2x - 3)\right] = \frac{4}{9}(2x + 2)\) | M1 A1 | (3) |
| \[f(x) = \begin{cases} \frac{8}{9}(x+1) & 1 \leq x \leq 1.5 \\ 0 & \text{otherwise} \end{cases}\] | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1 - 0.3733\) | M1; A1 | (2) |
| \(= \frac{47}{75} \cdot 0.6267\) | awrt | |
| 0.627 | ||
| (d) \((0.6267)^3 = 0.154\) | M1 A1 | (2) |
**(a)** $\frac{4}{9}(m^2 + 2m - 3) = 0.5$
$m^2 + 2m - 4.125 = 0$ | M1; M1 | (3) |
$m = \frac{-2 \pm \sqrt{4 + 16.5}}{2}$ | A1 | |
$m = 1.26, -3.264$ | | |
(median) = 1.26 | | |
**(b)** Differentiating $\frac{d}{dx}\left[\frac{4}{9}(x^2 + 2x - 3)\right] = \frac{4}{9}(2x + 2)$ | M1 A1 | (3) | M1 attempt to differentiate. At least one $x^n \to x^{n-1}$ (ignore limits); A1 correct differentiation; B1 must have both parts- follow through their $F'(x)$ Condone $<$
$$f(x) = \begin{cases} \frac{8}{9}(x+1) & 1 \leq x \leq 1.5 \\ 0 & \text{otherwise} \end{cases}$$ | B1ft | |
**(c)** $P(X \geq 1.2) = 1 - F(1.2)$
$= 1 - 0.3733$ | M1; A1 | (2) | M1 finding/writing $1 - F(1.2)$ may use/write $\int_{1.2}^{1.5}(x+1)dx$ or $1 - \int_{}^{1.2} "their f(x)"dx$; Condone missing $dx$
$= \frac{47}{75} \cdot 0.6267$ | awrt | |
0.627 | | |
**(d)** $(0.6267)^3 = 0.154$ | M1 A1 | (2) | M1 (c)$^3$ If expressions are not given you need to check the calculation is correct to 2sf.; A1 awrt 0.154 or 0.155
---The lifetime, $X$, in tens of hours, of a battery has a cumulative distribution function F(x) given by
$$\text{F}(x) = \begin{cases}
0 & x < 1 \\
\frac{4}{9}(x^2 + 2x - 3) & 1 \leqslant x \leqslant 1.5 \\
1 & x > 1.5
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Find the median of $X$, giving your answer to 3 significant figures. [3]
\item Find, in full, the probability density function of the random variable $X$. [3]
\item Find P($X \geqslant 1.2$) [2]
\end{enumerate}
A camping lantern runs on 4 batteries, all of which must be working. Four new batteries are put into the lantern.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the probability that the lantern will still be working after 12 hours. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2010 Q4 [10]}}