| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Multi-part piecewise CDF |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring understanding of CDF continuity conditions, differentiation to find PDFs, probability calculations, and linearity of expectation across piecewise functions. While each individual step uses standard S2 techniques, the question requires careful coordination across five parts with multiple constants to determine and integration across different pieces. The complexity and length elevate it above average, though it remains within standard Further Maths S2 scope without requiring novel insights. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(F(3) = 0 \rightarrow \frac{1}{16}\left(3^2 - 6(3) + a\right) = 0\) | M1 | 1st M1 writing or use of \(F(3) = 0\) |
| \(a = 9\) | A1 | 1st A1 \(a = 9\) cao |
| \(F(10) = 1 \rightarrow \frac{1}{12}\left(100(10) - (5)10^5 + c\right) = 1\) | M1 | 2nd M1 writing or use of \(F(10) = 1\) |
| \(c = -488\) | A1 | 2nd A1 \(c = -488\) cao |
| (4) | ||
| (b) \(\frac{1}{16}\left(5^2 - 6(5) + "9"\right) = \frac{1}{12}\left(5 + b\right)\) \(\left | \frac{1}{12}(9 + b) = \frac{1}{12}\left(100(9) - 5(9)^2 + "-488"\right)\right | \) |
| \(b = -2\) | A1 \(b = -2\) cao | |
| (2) | ||
| (c) \(P(6 < Y \leqslant 9) = F(9) - F(6)\) | M1 M1 | 1st M1 writing or using \(F(9) - F(6)\) (may be implied by 2nd M1) |
| \(= \frac{1}{12}\left(9 + "-2"\right) - \frac{1}{12}\left(6 + "-2"\right)\) | 2nd M1 substituting 9 and 6 into \(F(y)\) with their value of \(b\). allow \(\frac{1}{12}\left(100(9) + 5(9)^2 + "-488"\right) - \frac{1}{12}\left(6 + "-2"\right)\) with their value of \(b\) and their value of \(c\) | |
| \(= \frac{1}{4}\) | A1 | A1 \(\frac{1}{4}\) oe |
| (3) | ||
| (d) \(f(y) = \frac{1}{12}\) | B1 | B1 \(\frac{1}{12}\) |
| (1) | ||
| (e) \(E(6Y - 5) = [26.5+]\int_5^9(6y - 5)"\frac{1}{12}"dy\) | M1 | 1st M1 use of \(\int (6y - 5)"\frac{1}{12}"dy\) (ignore limits) |
| \(= [26.5+]\frac{1}{12}[(3y^2 - 5y)]_5^9\) | dM1 dM1 | 2nd M1 (dep on 1st M1) attempt to integrate \((6y - 5)"\frac{1}{12}"\) with at least one \(y^n \to y^{n+1}\) |
| \(= 26.5 + \frac{1}{12}[(3(9)^2 - 5(9)) - (3(5)^2 - 5(5))]\) | 3rd M1 (dep on 1st M1) \(26.5 + \int(6y - 5)"\frac{1}{12}"dy\) | |
| \(= \frac{233}{6}\) | A1 | A1 awrt 38.8 |
| (4) | ||
| [Total 14] |
**(a)** $F(3) = 0 \rightarrow \frac{1}{16}\left(3^2 - 6(3) + a\right) = 0$ | M1 | 1st M1 writing or use of $F(3) = 0$
$a = 9$ | A1 | 1st A1 $a = 9$ cao
$F(10) = 1 \rightarrow \frac{1}{12}\left(100(10) - (5)10^5 + c\right) = 1$ | M1 | 2nd M1 writing or use of $F(10) = 1$
$c = -488$ | A1 | 2nd A1 $c = -488$ cao
| | (4)
**(b)** $\frac{1}{16}\left(5^2 - 6(5) + "9"\right) = \frac{1}{12}\left(5 + b\right)$ $\left|\frac{1}{12}(9 + b) = \frac{1}{12}\left(100(9) - 5(9)^2 + "-488"\right)\right|$ | M1 A1 | M1 use of $F(5) = F(5)$ [= $\frac{1}{12}$] or $F(9) = F(9)$ [= $\frac{12}{12}$] ft their values from (a)
$b = -2$ | | A1 $b = -2$ cao
| | (2)
**(c)** $P(6 < Y \leqslant 9) = F(9) - F(6)$ | M1 M1 | 1st M1 writing or using $F(9) - F(6)$ (may be implied by 2nd M1)
$= \frac{1}{12}\left(9 + "-2"\right) - \frac{1}{12}\left(6 + "-2"\right)$ | | 2nd M1 substituting 9 and 6 into $F(y)$ with their value of $b$. allow $\frac{1}{12}\left(100(9) + 5(9)^2 + "-488"\right) - \frac{1}{12}\left(6 + "-2"\right)$ with their value of $b$ and their value of $c$
$= \frac{1}{4}$ | A1 | A1 $\frac{1}{4}$ oe
| | (3)
**(d)** $f(y) = \frac{1}{12}$ | B1 | B1 $\frac{1}{12}$
| | (1)
**(e)** $E(6Y - 5) = [26.5+]\int_5^9(6y - 5)"\frac{1}{12}"dy$ | M1 | 1st M1 use of $\int (6y - 5)"\frac{1}{12}"dy$ (ignore limits)
$= [26.5+]\frac{1}{12}[(3y^2 - 5y)]_5^9$ | dM1 dM1 | 2nd M1 (dep on 1st M1) attempt to integrate $(6y - 5)"\frac{1}{12}"$ with at least one $y^n \to y^{n+1}$
$= 26.5 + \frac{1}{12}[(3(9)^2 - 5(9)) - (3(5)^2 - 5(5))]$ | | 3rd M1 (dep on 1st M1) $26.5 + \int(6y - 5)"\frac{1}{12}"dy$
$= \frac{233}{6}$ | A1 | A1 awrt 38.8
| | (4)
| | [Total 14]\begin{enumerate}
\item A continuous random variable $Y$ has cumulative distribution function given by
\end{enumerate}
$$\mathrm { F } ( y ) = \left\{ \begin{array} { c r }
0 & y < 3 \\
\frac { 1 } { 16 } \left( y ^ { 2 } - 6 y + a \right) & 3 \leqslant y \leqslant 5 \\
\frac { 1 } { 12 } ( y + b ) & 5 < y \leqslant 9 \\
\frac { 1 } { 12 } \left( 100 y - 5 y ^ { 2 } + c \right) & 9 < y \leqslant 10 \\
1 & y > 10
\end{array} \right.$$
where $a$, $b$ and $c$ are constants.\\
(a) Find the value of $a$ and the value of $c$\\
(b) Find the value of $b$\\
(c) Find $\mathrm { P } ( 6 < Y \leqslant 9 )$
Show your working clearly.\\
(d) Specify the probability density function, f(y), for $5 < y \leqslant 9$
Using the information
$$\int _ { 3 } ^ { 5 } ( 6 y - 5 ) f ( y ) d y + \int _ { 9 } ^ { 10 } ( 6 y - 5 ) f ( y ) d y = 26.5$$
(e) find $\mathrm { E } ( 6 Y - 5 )$
You should make your method clear.
\hfill \mbox{\textit{Edexcel S2 2023 Q5 [14]}}