| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Standard +0.3 This is a straightforward CDF question requiring application of continuity conditions (F(3)=0, F(9)=1) to find two unknowns, followed by a routine quartile calculation. The algebra is simple and the method is standard textbook material for Further Statistics, making it slightly easier than average despite being a Further Maths topic. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F(3) = 0\) or \(F(9) = 1\) | M1 | 3.1a |
| \(c - 4.5(3^n) = 0\) and \(c - 4.5(9^n) = 1\) | A1 | 1.1b |
| Eliminating \(c\): \(1 + 4.5(9^n) = 4.5(3^n)\) or Eliminating \(n\): \(\dfrac{\log(\frac{c-1}{4.5})}{\log(9)} = \dfrac{\log(\frac{c}{4.5})}{\log(3)}\) | M1 | 1.1b |
| Forming 3-term quadratic: \(4.5(3^{2n}) - 4.5(3^n) + 1 = 0\) or \(4.5c^2 - 20.25c + 20.25 = 0\) | M1 | 3.1a |
| Solving TQE: \(3^n = \frac{1}{3} \to n = \ldots\) or \(3^n = \frac{2}{3} \to n = \ldots\) or \(c = 1.5\) or \(c = 3\) | M1 | 1.1b |
| \(n = -1\) only (reject other solution as \(n\) is an integer) | A1 | 2.3 |
| \(c = 1.5\) only | A1ft | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{Let } q = \text{lower quartile}]\ \ 1.5 - 4.5(q^{-1}) = 0.25\) | M1 | 1.1b |
| \(q = 3.6\) | A1ft | 1.1b |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F(3) = 0$ or $F(9) = 1$ | M1 | 3.1a |
| $c - 4.5(3^n) = 0$ and $c - 4.5(9^n) = 1$ | A1 | 1.1b |
| Eliminating $c$: $1 + 4.5(9^n) = 4.5(3^n)$ **or** Eliminating $n$: $\dfrac{\log(\frac{c-1}{4.5})}{\log(9)} = \dfrac{\log(\frac{c}{4.5})}{\log(3)}$ | M1 | 1.1b |
| Forming 3-term quadratic: $4.5(3^{2n}) - 4.5(3^n) + 1 = 0$ **or** $4.5c^2 - 20.25c + 20.25 = 0$ | M1 | 3.1a |
| Solving TQE: $3^n = \frac{1}{3} \to n = \ldots$ or $3^n = \frac{2}{3} \to n = \ldots$ **or** $c = 1.5$ or $c = 3$ | M1 | 1.1b |
| $n = -1$ only (reject other solution as $n$ is an integer) | A1 | 2.3 |
| $c = 1.5$ only | A1ft | 1.1b |
**(7 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{Let } q = \text{lower quartile}]\ \ 1.5 - 4.5(q^{-1}) = 0.25$ | M1 | 1.1b |
| $q = 3.6$ | A1ft | 1.1b |
**(2 marks)**\begin{enumerate}
\item The continuous random variable $X$ has cumulative distribution function
\end{enumerate}
$$\mathrm { F } ( x ) = \left\{ \begin{array} { l r }
0 & x < 3 \\
c - 4.5 x ^ { n } & 3 \leqslant x \leqslant 9 \\
1 & x > 9
\end{array} \right.$$
where $c$ is a positive constant and $n$ is an integer.\\
(a) Showing all stages of your working, find the value of $c$ and the value of $n$\\
(b) Find the lower quartile of $X$
\hfill \mbox{\textit{Edexcel FS2 AS 2018 Q4 [9]}}