| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | January |
| 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 S2 question on CDFs requiring routine techniques: continuity condition for part (a), direct substitution for (b)/(c), solving F(x)=0.5 for median, and equation solving for part (e). All steps are standard with no novel insight required, making it slightly easier than average. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | \(F(2) = 1\) or verify | M1 |
| \(\frac{1}{20}(d^4) + \frac{1}{5} = 1\) or \(\frac{1}{20}(2^4) + \frac{1}{5} = 1\); \(\therefore d = 2\) | A1 cso | A1 also for correct conclusion with no incorrect working seen (must reject \(d = -2\) if given) |
| Part (b) | \(F(1.5) = \frac{1}{20}(1.5)^4 + \frac{1}{5} = \frac{29}{64}\) | M1 A1 |
| Part (c) | \(1\) | B1 |
| Part (d) | \(\frac{1}{20}x^4 + \frac{1}{5} = 0.5\) | M1 |
| \(x^4 = 6 \Rightarrow x = 1.56508\ldots\) | M1 A1 | 2nd M1 solving equation as far as \(x^4 = k\); awrt 1.57 |
| Part (e) | \(P(X > 1.9) = 1 - F(1.9) = 1 - 0.851605 = 0.148395\) | M1 A1 |
| \(P(X < k) = 0.148395\) | ||
| \(\frac{1}{4}k = 0.148395\); \(k = 0.59358\) | M1 A1 | 2nd M1 for writing or using \(\frac{1}{4}k = 1 - F(1.9)\); 2nd A1 awrt 0.594 |
| Total (12) |
| **Part (a)** | $F(2) = 1$ or verify | M1 | |
| | $\frac{1}{20}(d^4) + \frac{1}{5} = 1$ or $\frac{1}{20}(2^4) + \frac{1}{5} = 1$; $\therefore d = 2$ | A1 cso | A1 also for correct conclusion with no incorrect working seen (must reject $d = -2$ if given) |
| **Part (b)** | $F(1.5) = \frac{1}{20}(1.5)^4 + \frac{1}{5} = \frac{29}{64}$ | M1 A1 | M1 for substituting 1.5 into third line of F(x) on its own; awrt **0.453** |
| **Part (c)** | $1$ | B1 | |
| **Part (d)** | $\frac{1}{20}x^4 + \frac{1}{5} = 0.5$ | M1 | 1st M1 setting third line of F(x) = 0.5 |
| | $x^4 = 6 \Rightarrow x = 1.56508\ldots$ | M1 A1 | 2nd M1 solving equation as far as $x^4 = k$; awrt **1.57** |
| **Part (e)** | $P(X > 1.9) = 1 - F(1.9) = 1 - 0.851605 = 0.148395$ | M1 A1 | 1st M1 for writing or using $1 - F(1.9)$; 1st A1 awrt 0.148 (may be implied by $k =$ awrt 0.59) |
| | $P(X < k) = 0.148395$ | | |
| | $\frac{1}{4}k = 0.148395$; $k = 0.59358$ | M1 A1 | 2nd M1 for writing or using $\frac{1}{4}k = 1 - F(1.9)$; 2nd A1 awrt **0.594** |
| | | | Total (12) |
---A continuous random variable $X$ has cumulative distribution function
$$\mathrm{F}(x) = \begin{cases}
0 & x < 0 \\
\frac{1}{4}x & 0 \leq x \leq 1 \\
\frac{1}{20}x^4 + \frac{1}{5} & 1 < x \leq d \\
1 & x > d
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $d = 2$ [2]
\item Find $\mathrm{P}(X < 1.5)$ [2]
\item Write down the value of the lower quartile of $X$ [1]
\item Find the median of $X$ [3]
\item Find, to 3 significant figures, the value of $k$ such that $\mathrm{P}(X > 1.9) = \mathrm{P}(X < k)$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q4 [12]}}