| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Multi-part piecewise CDF |
| Difficulty | Standard +0.3 This is a standard S2 piecewise CDF question requiring continuity conditions to find constants, direct CDF evaluation, and conditional probability. All techniques are routine for this module with no novel insight needed, making it slightly easier than average A-level difficulty. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(d = 7\) | B1 | Realising \(d = 7\) |
| \(\frac{c}{3} - \frac{7}{6} = 1 - \frac{1}{6}("7" - c)^2\) | M1 | Forming an equation in \(c\) with their \(d\) or \(d\) |
| \(\frac{c}{3} - \frac{7}{6} = 1 - \frac{1}{6}("49" - 2\times"7"c + c^2)\) oe or \(c^2 - 12c + 36 = 0\) oe | dM1 | Dependent on previous M1. Multiplying out brackets, reducing to 3 term quadratic correct for their \(d\) or \(d\) |
| \((c-6)^2 = 0 \therefore c = 6\) | A1* | All previous marks must be awarded. For solving correct 3TQ and statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(X > 3.5) = 1 - \frac{1}{6}(3.5 - 3)^2\) | M1 | Substitution of 3.5 into correct expression |
| \(= \frac{23}{24}\) oe | A1 | Allow equivalent fractions or awrt 0.958 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P(3.5 < X < 5.5) = \left(\frac{5.5}{3} - \frac{7}{6}\right) - \left(\frac{1}{6}(3.5-3)^2\right) \left[= \frac{5}{8}\right]\) oe | M1 | Correct method to calculate \(P(3.5 < X < 5.5)\); useful figures are \(\frac{2}{3}\) and \(\frac{1}{24}\) |
| \(P(X > 4.5 \mid 3.5 < X < 5.5) = \dfrac{\left(\frac{5.5}{3} - \frac{7}{6}\right) - \left(\frac{4.5}{3} - \frac{7}{6}\right)}{"\frac{5}{8}"}\) | M1 | Correct method using their \(\frac{5}{8}\); useful figure is \(\frac{1}{3}\) |
| \(= \frac{8}{15}\) oe | A1 | Allow equivalent fractions or awrt 0.533 |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $d = 7$ | B1 | Realising $d = 7$ |
| $\frac{c}{3} - \frac{7}{6} = 1 - \frac{1}{6}("7" - c)^2$ | M1 | Forming an equation in $c$ with their $d$ or $d$ |
| $\frac{c}{3} - \frac{7}{6} = 1 - \frac{1}{6}("49" - 2\times"7"c + c^2)$ oe or $c^2 - 12c + 36 = 0$ oe | dM1 | Dependent on previous M1. Multiplying out brackets, reducing to 3 term quadratic correct for their $d$ or $d$ |
| $(c-6)^2 = 0 \therefore c = 6$ | A1* | All previous marks must be awarded. For solving correct 3TQ and statement |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(X > 3.5) = 1 - \frac{1}{6}(3.5 - 3)^2$ | M1 | Substitution of 3.5 into correct expression |
| $= \frac{23}{24}$ oe | A1 | Allow equivalent fractions or awrt 0.958 |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(3.5 < X < 5.5) = \left(\frac{5.5}{3} - \frac{7}{6}\right) - \left(\frac{1}{6}(3.5-3)^2\right) \left[= \frac{5}{8}\right]$ oe | M1 | Correct method to calculate $P(3.5 < X < 5.5)$; useful figures are $\frac{2}{3}$ and $\frac{1}{24}$ |
| $P(X > 4.5 \mid 3.5 < X < 5.5) = \dfrac{\left(\frac{5.5}{3} - \frac{7}{6}\right) - \left(\frac{4.5}{3} - \frac{7}{6}\right)}{"\frac{5}{8}"}$ | M1 | Correct method using their $\frac{5}{8}$; useful figure is $\frac{1}{3}$ |
| $= \frac{8}{15}$ oe | A1 | Allow equivalent fractions or awrt 0.533 |
---\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function given by
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r }
0 & x < 3 \\
\frac { 1 } { 6 } ( x - 3 ) ^ { 2 } & 3 \leqslant x < 4 \\
\frac { x } { 3 } - \frac { 7 } { 6 } & 4 \leqslant x < c \\
1 - \frac { 1 } { 6 } ( d - x ) ^ { 2 } & c \leqslant x < 7 \\
1 & x \geqslant 7
\end{array} \right.$$
where $c$ and $d$ are constants.\\
(a) Show that $c = 6$\\
(b) Find $\mathrm { P } ( X > 3.5 )$\\
(c) Find $\mathrm { P } ( X > 4.5 \mid 3.5 < X < 5.5 )$
\hfill \mbox{\textit{Edexcel S2 2022 Q5 [9]}}