| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find or specify CDF |
| Difficulty | Standard +0.3 This is a standard Further Maths probability question requiring routine integration and application of PDF/CDF properties. Part (a) requires basic reasoning about non-negativity, part (b)(i) uses the standard normalization condition, and parts (ii)-(iii) involve straightforward polynomial integration with no novel insights required. Slightly above average difficulty due to being Further Maths content, but mechanically routine. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | The pdf must be positive (or zero) \(f(r) \geq 0\) | B1 |
| Therefore \((b - 4) \geq 0\) \(b \geq 4\) | B1 | B1 for Correct statement leading to correct conclusion. ALTERNATIVE B1 for "If \(b < 4\), \(f(r)\) is negative." B1 for stating that is not possible. |
| 4(b)(i) | \(\int_1^4 kr(4-r)dr = 1\) | M1 |
| \(\int_1^4 (4kr - kr^2)dr = 1\) | A1 | A1 Correct integration. |
| \(k\left[\frac{4r^2}{2} - \frac{r^3}{3}\right]_1^4 = 1\) | m1 | m1 substitution of correct limits and = 1. |
| \(k\left[\left(\frac{64}{2} - \frac{64}{3}\right) - \left(\frac{4}{2} - \frac{1}{3}\right)\right] = 1\) | A1 | Convincing |
| \(k = \frac{1}{9}\) *ag | ||
| 4(b)(ii) | \(F(r) = \frac{1}{9}\int_1^r t(4-t)dt\) | M1 |
| \(= \frac{1}{9}\left[\frac{4t^2}{2} - \frac{t^3}{3}\right]\) | A1 | A1 Correct integration. |
| \(= \frac{1}{9}\left[2r^2 - \frac{r^3}{3} - \left(2 - \frac{1}{3}\right)\right]\) | m1 | m1 substituting correct limits Condone upper limit = \(x\) for m1 only |
| \(= \frac{1}{9}\left(2r^2 - \frac{r^3}{3} - \frac{5}{3}\right)\) | A1 | oe Mark final expression for \(1 \leq r \leq 4\) |
| \(P(2 \leq R \leq 3) = F(3) - F(2) = \frac{22}{27} - \frac{11}{27} = \frac{11}{27}\) | M1 oe | |
| A1 | FT their \(F(r)\) for equivalent difficulty and provided probability is valid. | |
| Total [12] |
4(a) | The pdf must be positive (or zero) $f(r) \geq 0$ | B1 | B1 for implying that the pdf must be positive or zero (or cannot be negative)
| Therefore $(b - 4) \geq 0$ $b \geq 4$ | B1 | B1 for Correct statement leading to correct conclusion. ALTERNATIVE B1 for "If $b < 4$, $f(r)$ is negative." B1 for stating that is not possible.
4(b)(i) | $\int_1^4 kr(4-r)dr = 1$ | M1 | M1 Attempt at integration at least one power of $r$ increasing by 1. Limits and = 1 not required here.
| $\int_1^4 (4kr - kr^2)dr = 1$ | A1 | A1 Correct integration.
| $k\left[\frac{4r^2}{2} - \frac{r^3}{3}\right]_1^4 = 1$ | m1 | m1 substitution of correct limits and = 1.
| $k\left[\left(\frac{64}{2} - \frac{64}{3}\right) - \left(\frac{4}{2} - \frac{1}{3}\right)\right] = 1$ | A1 | Convincing
| $k = \frac{1}{9}$ *ag |
4(b)(ii) | $F(r) = \frac{1}{9}\int_1^r t(4-t)dt$ | M1 | M1 Attempt at integrating $f(t)$ at least one power of $t$ increasing by 1. Limits not required here.
| $= \frac{1}{9}\left[\frac{4t^2}{2} - \frac{t^3}{3}\right]$ | A1 | A1 Correct integration.
| $= \frac{1}{9}\left[2r^2 - \frac{r^3}{3} - \left(2 - \frac{1}{3}\right)\right]$ | m1 | m1 substituting correct limits Condone upper limit = $x$ for m1 only
| $= \frac{1}{9}\left(2r^2 - \frac{r^3}{3} - \frac{5}{3}\right)$ | A1 | oe Mark final expression for $1 \leq r \leq 4$
| $P(2 \leq R \leq 3) = F(3) - F(2) = \frac{22}{27} - \frac{11}{27} = \frac{11}{27}$ | M1 oe |
| | A1 | FT their $F(r)$ for equivalent difficulty and provided probability is valid.
| **Total [12]** |
---4. The continuous random variable $R$ has probability density function $f ( r )$ given by
$$f ( r ) = \begin{cases} k r ( b - r ) & \text { for } 1 \leqslant r \leqslant 4 , \\ 0 & \text { otherwise } , \end{cases}$$
where $k$ and $b$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item Explain why $b \geqslant 4$.
\item Given that $b = 4$,
\begin{enumerate}[label=(\roman*)]
\item show that $k = \frac { 1 } { 9 }$,
\item find an expression for $F ( r )$, valid for $1 \leqslant r \leqslant 4$, where $F$ denotes the cumulative distribution function of $R$,
\item find the probability that $R$ lies between 2 and 3 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2022 Q4 [12]}}