| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Conditional probability with CDF |
| Difficulty | Challenging +1.2 This S2 question requires understanding of CDFs and conditional probability, but follows standard procedures: part (a) applies the conditional probability formula with CDF values, part (b) uses continuity of the CDF at y=k to solve for k, and part (c) requires finding the PDF by differentiation then computing E(Y) by integration. While multi-step and requiring careful algebra, these are well-practiced techniques for S2 students with no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
\(\left[P\!\left(Y<\tfrac{1}{4}k\mid Y| M1 A1 |
M1 for a correct probability statement or correct ratio of probabilities. A1 for \(=\frac{1}{16}\) oe or 0.0625 |
|
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{21}k^2 = -\frac{1}{15}k^2+\frac{4}{5}k-\frac{7}{5}\) | M1 | For setting the two lines of the cdf equal to each other, or \(\frac{2}{21}y\) or \(\frac{2}{15}(6-y)\) (implied by correct 3TQ) |
| \(\Rightarrow 4k^2-28k+49=0\) oe | A1 | For a correct 3TQ or \(\frac{2}{21}y\) and \(\frac{2}{15}(6-y)\) |
| \(\Rightarrow (2k-7)^2=0\) | M1 | For solving their 3TQ. If 3TQ not correct, must show correct method or set 2 lines of pdf equal |
| \(k=\frac{7}{2}\) oe | A1 | \(k=3.5\) oe. NB \(k=3.5\) with no incorrect working scores 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(y)=\begin{cases}\frac{2}{21}y & 0\leq y\leq 3.5\\ \frac{2}{15}(6-y) & 3.5< y\leq 6\\ 0 & \text{otherwise}\end{cases}\) | M1 M1 | M1 for attempting to differentiate 1 of the functions; M1 for attempting to differentiate both with one correct |
| \(E(Y)=\frac{2}{21}\int_0^{3.5}y^2\,dy+\frac{2}{15}\int_{3.5}^6(6y-y^2)\,dy\) | M1 M1 | M1 for writing/using \(E(Y)=\int_0^{3.5}y\,f(y)\,dy+\int_{3.5}^6 y\,f(y)\,dy\); M1 for attempting to integrate |
| \(\frac{2}{21}\left[\frac{y^3}{3}\right]_0^{3.5}+\frac{2}{15}\left[3y^2-\frac{y^3}{3}\right]_{3.5}^6\) | ||
| \(\frac{2}{21}\!\left(\frac{343}{24}\right)+\frac{2}{15}\!\left(\frac{325}{24}\right)=\frac{19}{6}=3.166...\) awrt 3.17 | dM1 dA1 | dM1 dependent on previous M1: substitution of limits must be 0 or 6 and ft their 3.5. A1 for \(\frac{19}{6}\) or awrt 3.17 |
# Question 6:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[P\!\left(Y<\tfrac{1}{4}k\mid Y<k\right)=\right]\frac{F\!\left(\tfrac{1}{4}k\right)}{F(k)}=\frac{\frac{1}{21}\left(\tfrac{k}{4}\right)^2}{\frac{1}{21}k^2}=\frac{1}{16}$ oe | M1 A1 | M1 for a correct probability statement or correct ratio of probabilities. A1 for $=\frac{1}{16}$ oe or 0.0625 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{21}k^2 = -\frac{1}{15}k^2+\frac{4}{5}k-\frac{7}{5}$ | M1 | For setting the two lines of the cdf equal to each other, or $\frac{2}{21}y$ or $\frac{2}{15}(6-y)$ (implied by correct 3TQ) |
| $\Rightarrow 4k^2-28k+49=0$ oe | A1 | For a correct 3TQ or $\frac{2}{21}y$ and $\frac{2}{15}(6-y)$ |
| $\Rightarrow (2k-7)^2=0$ | M1 | For solving their 3TQ. If 3TQ not correct, must show correct method or set 2 lines of pdf equal |
| $k=\frac{7}{2}$ oe | A1 | $k=3.5$ oe. NB $k=3.5$ with no incorrect working scores 4/4 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(y)=\begin{cases}\frac{2}{21}y & 0\leq y\leq 3.5\\ \frac{2}{15}(6-y) & 3.5< y\leq 6\\ 0 & \text{otherwise}\end{cases}$ | M1 M1 | M1 for attempting to differentiate 1 of the functions; M1 for attempting to differentiate both with one correct |
| $E(Y)=\frac{2}{21}\int_0^{3.5}y^2\,dy+\frac{2}{15}\int_{3.5}^6(6y-y^2)\,dy$ | M1 M1 | M1 for writing/using $E(Y)=\int_0^{3.5}y\,f(y)\,dy+\int_{3.5}^6 y\,f(y)\,dy$; M1 for attempting to integrate |
| $\frac{2}{21}\left[\frac{y^3}{3}\right]_0^{3.5}+\frac{2}{15}\left[3y^2-\frac{y^3}{3}\right]_{3.5}^6$ | | |
| $\frac{2}{21}\!\left(\frac{343}{24}\right)+\frac{2}{15}\!\left(\frac{325}{24}\right)=\frac{19}{6}=3.166...$ awrt 3.17 | dM1 dA1 | dM1 dependent on previous M1: substitution of limits must be 0 or 6 and ft their 3.5. A1 for $\frac{19}{6}$ or awrt 3.17 |\begin{enumerate}
\item The continuous random variable $Y$ has cumulative distribution function given by
\end{enumerate}
$$\mathrm { F } ( y ) = \left\{ \begin{array} { l r }
0 & y < 0 \\
\frac { 1 } { 21 } y ^ { 2 } & 0 \leqslant y \leqslant k \\
\frac { 2 } { 15 } \left( 6 y - \frac { y ^ { 2 } } { 2 } \right) - \frac { 7 } { 5 } & k < y \leqslant 6 \\
1 & y > 6
\end{array} \right.$$
(a) Find $\mathrm { P } \left( \left. Y < \frac { 1 } { 4 } k \right\rvert \, Y < k \right)$\\
(b) Find the value of $k$\\
(c) Use algebraic calculus to find $\mathrm { E } ( Y )$
\hfill \mbox{\textit{Edexcel S2 2023 Q6 [12]}}