| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2023 |
| 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 standard Further Maths statistics question on piecewise CDFs requiring routine techniques: continuity conditions, direct CDF evaluation, quartile finding by solving F(x)=0.25 and F(x)=0.75, and equation solving. While it has multiple parts (12 marks total), each part follows textbook methods with no novel insight required. The algebra is straightforward, making it slightly easier than average even for Further Maths. |
| 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 |
| (a) \(\frac{1}{480}b^4 + \frac{7}{15} = 1\) | M1 | M1 for setting \(F(b) = 1\) |
| \(b^4 = 256\) | ||
| \(b = 4\) | A1 | At least one step between M1 and \(b = 4\); Convincing. |
| (b) \(P(X < 2.5) = F(2.5) = \frac{1}{480} \times 2.5^4 + \frac{7}{15}\) | M1 | Attempt to substitute 2.5 into \(F(x)\). |
| \(= 0.548\) or \(\frac{1403}{2560}\) | A1 | 3sf or better |
| (c) Lower quartile \(= 1\) | B1 | |
| (d) \(\frac{1}{480}u^4 + \frac{7}{15} = 0.75\) | M1 | |
| \(u^4 = 136\) | A1 | |
| \(u = 3.41\) | A1 | |
| (e) \(P(X > 3.5) = 1 - F(3.5) = 1 - \left(\frac{1}{480} \times 3.5^4 + \frac{7}{15}\right)\) | M1 | si |
| \(= 1 - 0.779... = 0.2207...\) | A1 | |
| \(P(X < k) = 0.2207...\) | M1 | si FT their 0.2207... provided > 0.125 and < 0.5 |
| \(\frac{1}{4}k = 0.2207...\) | ||
| \(k = 0.883\) | A1 | cao. Do not allow 0.884. \((k = \frac{113}{128})\) |
| Total [12] |
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(a)** $\frac{1}{480}b^4 + \frac{7}{15} = 1$ | M1 | M1 for setting $F(b) = 1$ |
| $b^4 = 256$ | | |
| $b = 4$ | A1 | At least one step between M1 and $b = 4$; Convincing. |
| **(b)** $P(X < 2.5) = F(2.5) = \frac{1}{480} \times 2.5^4 + \frac{7}{15}$ | M1 | Attempt to substitute 2.5 into $F(x)$. |
| $= 0.548$ or $\frac{1403}{2560}$ | A1 | 3sf or better |
| **(c)** Lower quartile $= 1$ | B1 | |
| **(d)** $\frac{1}{480}u^4 + \frac{7}{15} = 0.75$ | M1 | |
| $u^4 = 136$ | A1 | |
| $u = 3.41$ | A1 | |
| **(e)** $P(X > 3.5) = 1 - F(3.5) = 1 - \left(\frac{1}{480} \times 3.5^4 + \frac{7}{15}\right)$ | M1 | si |
| $= 1 - 0.779... = 0.2207...$ | A1 | |
| $P(X < k) = 0.2207...$ | M1 | si FT their 0.2207... provided > 0.125 and < 0.5 |
| $\frac{1}{4}k = 0.2207...$ | | |
| $k = 0.883$ | A1 | cao. Do not allow 0.884. $(k = \frac{113}{128})$ |
| **Total [12]** | | |A continuous random variable $X$ has cumulative distribution function $F$ given by
$$F(x) = \begin{cases}
0 & \text{for } x < 0, \\
\frac{1}{4}x & \text{for } 0 \leqslant x \leqslant 2, \\
\frac{1}{480}x^4 + \frac{7}{15} & \text{for } 2 < x \leqslant b, \\
1 & \text{for } x > b.
\end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $b = 4$. [2]
\item Find P$(X \leqslant 2 \cdot 5)$. [2]
\item Write down the value of the lower quartile of $X$. [1]
\item Find the value of the upper quartile of $X$. [3]
\item Find, correct to three significant figures, the value of $k$ that satisfies the equation P$(X > 3 \cdot 5) = \text{P}(X < k)$. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2023 Q4 [12]}}