| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Moderate -0.3 This is a straightforward Further Statistics question testing standard techniques with continuous probability distributions. Parts (a)-(d) involve routine integration and CDF manipulation that any FS2 student should handle mechanically. Part (e) requires solving a quadratic equation to find the median, which is standard. While it's Further Maths content, the question requires no novel insight—just systematic application of learned procedures across multiple parts. |
| 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/Working | Marks | Guidance |
| \(P(X < 3) = \int_1^3 \frac{1}{18}(11-2x)dx\) or area of trapezium | M1 | For integrating \(f(x)\) with correct limits or finding area of trapezium |
| \(= \left[\frac{1}{18}(11x - x^2)\right]_1^3\) | ||
| \(= \frac{7}{9}\) | A1 | Allow awrt 0.778 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Since \(P(X < 3) > 0.75\), the upper quartile is less than 3 | B1ft | For comparison of their (a) with 0.75 and concluding upper quartile is less than 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2) = \int_1^4 \frac{1}{18}x^2(11-2x)dx \left[= \frac{23}{4}\right]\) | M1 | For an attempt to find \(E(X^2)\) |
| \(\text{Var}(X) = \frac{23}{4} - \left(\frac{9}{4}\right)^2\) | M1 | For use of \(\text{Var}(X) = E(X^2) - \left(\frac{9}{4}\right)^2\) |
| \(= \frac{11}{16}\) | A1 | Allow awrt 0.688; M1 marks may be implied by correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(4) = 1 \rightarrow \frac{1}{18}(11(4) - 4^2 + c) = 1\) or \(F(1) = 0 \rightarrow \frac{1}{18}(11(1) - 1^2 + c) = 0\) | M1 | For use of \(F(4) = 1\) or \(F(1) = 0\) |
| \(c = -10\) * | A1*cso | For fully correct solution leading to given answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(m) = 0.5\) | M1 | For use of \(F(m) = 0.5\) |
| \(\frac{1}{18}(11m - m^2 - 10) = 0.5 \rightarrow m^2 - 11m + 19 = 0\) and attempt to solve | M1 | For setting up quadratic and attempt to solve |
| \(m = \frac{11 \pm \sqrt{11^2 - 4(19)}}{2}\) \([= 2.1458 \text{ or } 8.8541\ldots]\) | ||
| \(m = 2.15\) (only) | A1 | For 2.15 and rejecting the other solution |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < 3) = \int_1^3 \frac{1}{18}(11-2x)dx$ or area of trapezium | M1 | For integrating $f(x)$ with correct limits or finding area of trapezium |
| $= \left[\frac{1}{18}(11x - x^2)\right]_1^3$ | | |
| $= \frac{7}{9}$ | A1 | Allow awrt 0.778 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Since $P(X < 3) > 0.75$, the upper quartile is less than 3 | B1ft | For comparison of their (a) with 0.75 and concluding upper quartile is less than 3 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = \int_1^4 \frac{1}{18}x^2(11-2x)dx \left[= \frac{23}{4}\right]$ | M1 | For an attempt to find $E(X^2)$ |
| $\text{Var}(X) = \frac{23}{4} - \left(\frac{9}{4}\right)^2$ | M1 | For use of $\text{Var}(X) = E(X^2) - \left(\frac{9}{4}\right)^2$ |
| $= \frac{11}{16}$ | A1 | Allow awrt 0.688; M1 marks may be implied by correct answer |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(4) = 1 \rightarrow \frac{1}{18}(11(4) - 4^2 + c) = 1$ or $F(1) = 0 \rightarrow \frac{1}{18}(11(1) - 1^2 + c) = 0$ | M1 | For use of $F(4) = 1$ or $F(1) = 0$ |
| $c = -10$ * | A1*cso | For fully correct solution leading to given answer with no errors seen |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(m) = 0.5$ | M1 | For use of $F(m) = 0.5$ |
| $\frac{1}{18}(11m - m^2 - 10) = 0.5 \rightarrow m^2 - 11m + 19 = 0$ and attempt to solve | M1 | For setting up quadratic and attempt to solve |
| $m = \frac{11 \pm \sqrt{11^2 - 4(19)}}{2}$ $[= 2.1458 \text{ or } 8.8541\ldots]$ | | |
| $m = 2.15$ (only) | A1 | For 2.15 and rejecting the other solution |
---\begin{enumerate}
\item The continuous random variable $X$ has probability density function
\end{enumerate}
$$f ( x ) = \begin{cases} \frac { 1 } { 18 } ( 11 - 2 x ) & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
(a) Find $\mathrm { P } ( \mathrm { X } < 3 )$\\
(b) State, giving a reason, whether the upper quartile of $X$ is greater than 3, less than 3 or equal to 3
Given that $\mathrm { E } ( \mathrm { X } ) = \frac { 9 } { 4 }$\\
(c) use algebraic integration to find $\operatorname { Var } ( \mathrm { X } )$
The cumulative distribution function of $X$ is given by
$$F ( x ) = \left\{ \begin{array} { l r }
0 & x < 1 \\
\frac { 1 } { 18 } \left( 11 x - x ^ { 2 } + c \right) & 1 \leqslant x \leqslant 4 \\
1 & x > 4
\end{array} \right.$$
(d) Show that $\mathrm { c } = - 10$\\
(e) Find the median of $X$, giving your answer to 3 significant figures.
\section*{Q uestion 2 continued}
\hfill \mbox{\textit{Edexcel FS2 AS Q2 [11]}}