| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Find quantiles from CDF |
| Difficulty | Standard +0.3 This is a straightforward FS2 question testing standard CDF techniques: solving F(x)=0.5 for the median, applying conditional probability formula, transforming variables using the relationship F_Y(y)=P(1/X≤y)=P(X≥1/y), and finding the mode from the pdf. All steps are routine applications of learned 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 pdf5.03g Cdf of transformed variables |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(1.5m - 0.25m^2 - 1.25 = 0.5 \;\Rightarrow\; 0.25m^2 - 1.5m + 1.75 = 0\) | M1 | 1.1b — Correct equation for \(m\) |
| \(m = 3 - \sqrt{2}\) (reject \(m = 3+\sqrt{2}\)) | A1 | 2.2a — Only isw if exact answer given then rounded |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(X < 1.6 \mid X > 1.2) = \dfrac{P(1.2 < X < 1.6)}{P(X > 1.2)}\) | M1 | 3.1a — Determining the two probabilities required |
| \(\dfrac{F(1.6) - F(1.2)}{1 - F(1.2)}\) | M1 | 1.1b — Correct ratio of probabilities |
| \(= \dfrac{32}{81}\) | A1 | 1.1b — Allow awrt \(0.395\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(Y,\, y) = P\!\left(\frac{1}{X},\, y\right)\) | M1: Realising \(P\!\left(X \ldots \frac{1}{y}\right)\) is necessary | |
| \(= P\!\left(X \leq \tfrac{1}{y}\right) = 1 - F\!\left(\tfrac{1}{y}\right)\) | M1 | 3.1a |
| \(= 1 - \left(\frac{1.5}{y} - 0.25\!\left(\tfrac{1}{y}\right)^2 - 1.25\right)\) | M1 | 1.1b — Correct use of \(F(x)\) |
| \(F(y) = \begin{cases} 0 & y < \tfrac{1}{3} \\ 2.25 - \dfrac{1.5}{y} + 0.25\!\left(\tfrac{1}{y}\right)^2 & \tfrac{1}{3} \leq y \leq 1 \\ 1 & y > 1 \end{cases}\) | A1 A1 | 2.1 / 1.1b — A1 correct limits (allow \(\leq\) for \(<\)); A1 all lines of cdf correct (ignoring limits), must be in terms of \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f(y) = \dfrac{\mathrm{d}}{\mathrm{d}y}(F(y)) = 1.5y^{-2} - 0.5y^{-3} \;\Rightarrow\; f'(y) = -3y^{-3} + 1.5y^{-4}\) | M1 | 3.1a — Realising cdf must be differentiated twice |
| \(f'(y) = -3y^{-3} + 1.5y^{-4} = 0\) | depM1 | 1.1b — Equating \(f'(y)=0\) with attempt to solve |
| \(1.5y^{-4} = 3y^{-3} \;\Rightarrow\; \dfrac{1.5}{y^4} = \dfrac{3}{y^3}\) | ||
| Mode \(= 0.5\) (since \(f''(0.5) = -48 < 0\)) | A1 | 1.1b — cao |
# Question 8(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $1.5m - 0.25m^2 - 1.25 = 0.5 \;\Rightarrow\; 0.25m^2 - 1.5m + 1.75 = 0$ | M1 | 1.1b — Correct equation for $m$ |
| $m = 3 - \sqrt{2}$ (reject $m = 3+\sqrt{2}$) | A1 | 2.2a — Only isw if exact answer given then rounded |
# Question 8(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(X < 1.6 \mid X > 1.2) = \dfrac{P(1.2 < X < 1.6)}{P(X > 1.2)}$ | M1 | 3.1a — Determining the two probabilities required |
| $\dfrac{F(1.6) - F(1.2)}{1 - F(1.2)}$ | M1 | 1.1b — Correct ratio of probabilities |
| $= \dfrac{32}{81}$ | A1 | 1.1b — Allow awrt $0.395$ |
# Question 8(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(Y,\, y) = P\!\left(\frac{1}{X},\, y\right)$ | | M1: Realising $P\!\left(X \ldots \frac{1}{y}\right)$ is necessary |
| $= P\!\left(X \leq \tfrac{1}{y}\right) = 1 - F\!\left(\tfrac{1}{y}\right)$ | M1 | 3.1a |
| $= 1 - \left(\frac{1.5}{y} - 0.25\!\left(\tfrac{1}{y}\right)^2 - 1.25\right)$ | M1 | 1.1b — Correct use of $F(x)$ |
| $F(y) = \begin{cases} 0 & y < \tfrac{1}{3} \\ 2.25 - \dfrac{1.5}{y} + 0.25\!\left(\tfrac{1}{y}\right)^2 & \tfrac{1}{3} \leq y \leq 1 \\ 1 & y > 1 \end{cases}$ | A1 A1 | 2.1 / 1.1b — A1 correct limits (allow $\leq$ for $<$); A1 all lines of cdf correct (ignoring limits), must be in terms of $y$ |
# Question 8(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(y) = \dfrac{\mathrm{d}}{\mathrm{d}y}(F(y)) = 1.5y^{-2} - 0.5y^{-3} \;\Rightarrow\; f'(y) = -3y^{-3} + 1.5y^{-4}$ | M1 | 3.1a — Realising cdf must be differentiated twice |
| $f'(y) = -3y^{-3} + 1.5y^{-4} = 0$ | depM1 | 1.1b — Equating $f'(y)=0$ with attempt to solve |
| $1.5y^{-4} = 3y^{-3} \;\Rightarrow\; \dfrac{1.5}{y^4} = \dfrac{3}{y^3}$ | | |
| Mode $= 0.5$ (since $f''(0.5) = -48 < 0$) | A1 | 1.1b — cao |\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function given by
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r }
0 & x < 1 \\
1.5 x - 0.25 x ^ { 2 } - 1.25 & 1 \leqslant x \leqslant 3 \\
1 & x > 3
\end{array} \right.$$
(a) Find the exact value of the median of $X$\\
(b) Find $\mathrm { P } ( X < 1.6 \mid X > 1.2 )$
The random variable $Y = \frac { 1 } { X }$\\
(c) Specify fully the cumulative distribution function of $Y$\\
(d) Hence or otherwise find the mode of $Y$
\hfill \mbox{\textit{Edexcel FS2 2022 Q8 [12]}}