| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | PDF to CDF derivation |
| Difficulty | Standard +0.3 This is a standard S2 PDF question requiring integration to find constants and derive the CDF. While it has multiple parts and piecewise functions, each step follows routine procedures: integrate to use given probability, use total probability = 1, then integrate each piece for the CDF. The algebra is straightforward and no novel insight is required—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 |
|---|---|---|
| (a) \(\int_0^1 0.1x dx + \int_2^4 kx(8-x)dx = \frac{31}{45}\) | M1 | |
| \(\left[\frac{0.1x^2}{2}\right]_0^1 + \left[k\left(4x^2 - \frac{x^3}{3}\right)\right]_2^4 = \frac{31}{45}\) | M1 | |
| \(0.2 + k\left(64 - \frac{64}{3} - (16 - \frac{8}{3})\right) = \frac{31}{45} → k = \frac{1}{60}\) | dM1, A1 | 1st M1 sum of two integrals = 31/45 (ignore limits) It may be equated to 31/45 later in their working. Condone missing dx. 2nd M1 attempt at integration \(x → x^2\) or \(x^2 → x^3\) for at least one. 3rd dM1 dep on 1st M1 being awarded for use of correct limits. A1 \(k = \frac{1}{60}\) cao Allow 0.016 or equivalent exact value. \(k = \frac{60}{1}\) with no working gains 4/4. \(k = \frac{1}{60}\) from 0.2 = 2k(8−2) gains M0M0M0A0 |
| (b)(i) \(a = \left[\left(1 - \frac{31}{45}\right) ÷ 2\right] = \frac{7}{45}\) | B1 | \(a = \frac{7}{45}\) cao allow 0.15 or equivalent exact value |
| (ii) \(P(0_{,n} X_{,n} 5.5) = \frac{31}{45} + "a"×1.5 = \frac{83}{90}\) | M1, A1 | M1 ft "their value of a" or \(1 - 0.5×"a"\). A1 \(\frac{83}{90}\) cao Allow 0.92 or equivalent exact value |
| (c) \(\int_0^r 0.1r dt = \frac{0.1x^2}{2}\) | B1 | |
| \(\int_0^r 0.1r dt + \int_r^4 \frac{1}{60}"r(8-t)dt\), \(\frac{31}{45} + \int_r^4 \frac{7}{45}"dt\) | M1, M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{cases} 0 & x < 0 \\ 0.05x^2 & 0_{,n}x < 2 \\ \frac{1}{60}(4x^2 - \frac{x^3}{3} - \frac{4}{3}) & 2_{,n}x < 4 \\ \frac{7}{45}x + \frac{1}{15} & 4_{,n}x < 6 \\ 1 & x...6 \end{cases}\) | B1, A1, A1, (6) | 1st B1 a correct integration of 2nd line of pdf if have + C must get C = 0. 1st M1 a correct method to find 3rd line of cdf Condone incorrect integration (allow k). Allow 0.2 + \(\int_r^4 \frac{1}{60}"r(8-t)dt\) or "\(\frac{1}{60}"r(8-t)dt + C\) and F(2) = 0.2. 2nd M1 a correct method to find 4th line of cdf Condone incorrect integration (allow a). Allow \(\int_r^4 \frac{7}{45}"dt + C\) and F(6) = 1 but do not allow their F(4) + \(\int_r^6 \frac{7}{45}"dt\). |
**(a)** $\int_0^1 0.1x dx + \int_2^4 kx(8-x)dx = \frac{31}{45}$ | M1 |
$\left[\frac{0.1x^2}{2}\right]_0^1 + \left[k\left(4x^2 - \frac{x^3}{3}\right)\right]_2^4 = \frac{31}{45}$ | M1 |
$0.2 + k\left(64 - \frac{64}{3} - (16 - \frac{8}{3})\right) = \frac{31}{45} → k = \frac{1}{60}$ | dM1, A1 | 1st M1 sum of two integrals = 31/45 (ignore limits) It may be equated to 31/45 later in their working. Condone missing dx. 2nd M1 attempt at integration $x → x^2$ or $x^2 → x^3$ for at least one. 3rd dM1 dep on 1st M1 being awarded for use of correct limits. A1 $k = \frac{1}{60}$ cao Allow 0.016 or equivalent exact value. $k = \frac{60}{1}$ with no working gains 4/4. $k = \frac{1}{60}$ from 0.2 = 2k(8−2) gains M0M0M0A0
**(b)(i)** $a = \left[\left(1 - \frac{31}{45}\right) ÷ 2\right] = \frac{7}{45}$ | B1 | $a = \frac{7}{45}$ cao allow 0.15 or equivalent exact value
**(ii)** $P(0_{,n} X_{,n} 5.5) = \frac{31}{45} + "a"×1.5 = \frac{83}{90}$ | M1, A1 | M1 ft "their value of a" or $1 - 0.5×"a"$. A1 $\frac{83}{90}$ cao Allow 0.92 or equivalent exact value
**(c)** $\int_0^r 0.1r dt = \frac{0.1x^2}{2}$ | B1 |
$\int_0^r 0.1r dt + \int_r^4 \frac{1}{60}"r(8-t)dt$, $\frac{31}{45} + \int_r^4 \frac{7}{45}"dt$ | M1, M1 |
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| |
$\begin{cases} 0 & x < 0 \\ 0.05x^2 & 0_{,n}x < 2 \\ \frac{1}{60}(4x^2 - \frac{x^3}{3} - \frac{4}{3}) & 2_{,n}x < 4 \\ \frac{7}{45}x + \frac{1}{15} & 4_{,n}x < 6 \\ 1 & x...6 \end{cases}$ | B1, A1, A1, (6) | 1st B1 a correct integration of 2nd line of pdf if have + C must get C = 0. 1st M1 a correct method to find 3rd line of cdf Condone incorrect integration (allow k). Allow 0.2 + $\int_r^4 \frac{1}{60}"r(8-t)dt$ or "$\frac{1}{60}"r(8-t)dt + C$ and F(2) = 0.2. 2nd M1 a correct method to find 4th line of cdf Condone incorrect integration (allow a). Allow $\int_r^4 \frac{7}{45}"dt + C$ and F(6) = 1 but do not allow their F(4) + $\int_r^6 \frac{7}{45}"dt$.
**For the next 3 marks limits condone < for , and , for < and ...for >**
2nd B1 1st and 5th lines correct with correct limits. Allow 1 range to be otherwise for the limits, Must have consistent use of letter throughout for this mark. 1st A1 3rd line correct with correct limits Allow equivalent un-simplified expressions. 2nd A1 4th line correct with correct limits Allow equivalent un-simplified expressions
**[13 marks]**
---\begin{enumerate}
\item The continuous random variable $X$ has probability density function
\end{enumerate}
$$f ( x ) = \begin{cases} 0.1 x & 0 \leqslant x < 2 \\ k x ( 8 - x ) & 2 \leqslant x < 4 \\ a & 4 \leqslant x < 6 \\ 0 & \text { otherwise } \end{cases}$$
where $k$ and $a$ are constants.\\
It is known that $\mathrm { P } ( X < 4 ) = \frac { 31 } { 45 }$\\
(a) Find the exact value of $k$\\
(b) (i) Find the exact value of $a$\\
(ii) Find the exact value of $\mathrm { P } ( 0 \leqslant X \leqslant 5.5 )$\\
(c) Specify fully the cumulative distribution function of $X$
\hfill \mbox{\textit{Edexcel S2 2022 Q6 [13]}}