| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | October |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Direct variance calculation from pdf |
| Difficulty | Standard +0.3 This is a standard S2 question testing routine techniques: sketching a piecewise pdf, using variance formulas with given E(Y²), integrating to find CDFs, and finding percentiles. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| 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 |
| Correct height \(\frac{3}{14}\) on sketch | B1 | |
| Correct height \(\frac{1}{14}\) on sketch | B1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(Y) = 2\) | B1 | |
| \(\text{Var}(2Y-3) = 4\text{Var}(Y)\) | M1 | |
| \(\text{Var}(Y) = \left(\frac{131}{21} - 2^2\right)\) | M1 | |
| \(\text{Var}(2Y-3) = \frac{188}{21}\) | A1 | awrt 8.95 |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{-1}^{t} \frac{1}{14}(y+2)\,dy = \frac{1}{14}\left[\frac{y^2}{2} + 2y\right]_{-1}^{t}\) or \(\int \frac{1}{14}(y+2)\,dy = \frac{1}{14}\left[\frac{y^2}{2}+2y\right] + C\) or \(\int\frac{1}{14}(y+2)\,dy = \frac{1}{28}(y+2)^2 + C\) | M1 | |
| \(\frac{1}{14}\left[\left(\frac{t^2}{2}+2t\right) - \left(\frac{1}{2}-2\right)\right]\) or \(\frac{1}{14}\left[\frac{(-1)^2}{2}-2\right] + C = 0\ \&\ C = \frac{3}{28}\) or \(\frac{1}{28}(-1+2)^2 + C = 0\ \&\ C = -\frac{1}{28}\) leading to \(\frac{1}{14}\left(\frac{y^2}{2}+2y+\frac{3}{2}\right)\)* | A1*cso | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_1^t \frac{3}{14}\,dy + F(1) = \left[\frac{3}{14}y\right]_1^t + F(1) = \left[\left(\frac{3t}{14}\right) - \left(\frac{3}{14}\right)\right] + F(1)\) or \(\int\frac{3}{14}\,dy = \left[\frac{3}{14}y\right] + C\) and use \(F(1) =\) "their \(F(1)\)" or \(F(3) =\) "their \(F(3)\)" | M1 | |
| \(\int_3^t \frac{1}{14}(6-y)\,dy + F(3) = \frac{1}{14}\left[6y - \frac{y^2}{2}\right]_3^t + F(3) = \frac{1}{14}\left[\left(6t - \frac{t^2}{2}\right) - \left(18 - \frac{9}{2}\right)\right] + F(3)\) or \(\int\frac{1}{14}(6-y)\,dx = \frac{1}{14}\left[6y-\frac{y^2}{2}\right] + C\) or \(C - \frac{(6-y)^2}{28}\) and use \(F(5) = 1\) | M1 | |
| \(F(y) = \begin{cases} 0 & y \leq -1 \\ \frac{1}{14}\left(\frac{y^2}{2}+2y+\frac{3}{2}\right) & -1 < y \leq 1 \\ \frac{3}{14}y + \frac{1}{14} & 1 < y \leq 3 \\ \frac{3}{7}y - \frac{1}{28}y^2 - \frac{1}{4} & 3 < y \leq 5 \\ 1 & y > 5 \end{cases}\) | A1, A1, B1 | First A1 for \(1 < y \leq 3\) piece; Second A1 for \(3 < y \leq 5\) piece; B1 for \(y > 5\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \("\frac{3}{14}m + \frac{1}{14}" = 0.3\) | M1 | Setting their equation for \(1 < y \leq 3\) equal to 0.3 |
| \(m = \frac{16}{15}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(4Y \leq 5 \mid Y \leq 3) = \dfrac{\left(\frac{3}{14} \times \frac{5}{4} + \frac{1}{14}\right)}{\left(\frac{3}{14} \times 3 + \frac{1}{14}\right)} = \left[\dfrac{\frac{19}{56}}{\frac{5}{7}}\right]\) | M1 | For writing or using \(\dfrac{F\!\left(\frac{5}{4}\right)}{F(3)}\); allow use of their expression for \(3 < y \leq 5\) for the denominator |
| \(= \dfrac{19}{40}\) or \(0.475\) | A1 | cao |
# Question 6:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct height $\frac{3}{14}$ on sketch | B1 | |
| Correct height $\frac{1}{14}$ on sketch | B1 | |
| | **(2)** | |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(Y) = 2$ | B1 | |
| $\text{Var}(2Y-3) = 4\text{Var}(Y)$ | M1 | |
| $\text{Var}(Y) = \left(\frac{131}{21} - 2^2\right)$ | M1 | |
| $\text{Var}(2Y-3) = \frac{188}{21}$ | A1 | awrt 8.95 |
| | **(4)** | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{-1}^{t} \frac{1}{14}(y+2)\,dy = \frac{1}{14}\left[\frac{y^2}{2} + 2y\right]_{-1}^{t}$ or $\int \frac{1}{14}(y+2)\,dy = \frac{1}{14}\left[\frac{y^2}{2}+2y\right] + C$ or $\int\frac{1}{14}(y+2)\,dy = \frac{1}{28}(y+2)^2 + C$ | M1 | |
| $\frac{1}{14}\left[\left(\frac{t^2}{2}+2t\right) - \left(\frac{1}{2}-2\right)\right]$ or $\frac{1}{14}\left[\frac{(-1)^2}{2}-2\right] + C = 0\ \&\ C = \frac{3}{28}$ or $\frac{1}{28}(-1+2)^2 + C = 0\ \&\ C = -\frac{1}{28}$ leading to $\frac{1}{14}\left(\frac{y^2}{2}+2y+\frac{3}{2}\right)$* | A1*cso | |
| | **(2)** | |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_1^t \frac{3}{14}\,dy + F(1) = \left[\frac{3}{14}y\right]_1^t + F(1) = \left[\left(\frac{3t}{14}\right) - \left(\frac{3}{14}\right)\right] + F(1)$ or $\int\frac{3}{14}\,dy = \left[\frac{3}{14}y\right] + C$ and use $F(1) =$ "their $F(1)$" or $F(3) =$ "their $F(3)$" | M1 | |
| $\int_3^t \frac{1}{14}(6-y)\,dy + F(3) = \frac{1}{14}\left[6y - \frac{y^2}{2}\right]_3^t + F(3) = \frac{1}{14}\left[\left(6t - \frac{t^2}{2}\right) - \left(18 - \frac{9}{2}\right)\right] + F(3)$ or $\int\frac{1}{14}(6-y)\,dx = \frac{1}{14}\left[6y-\frac{y^2}{2}\right] + C$ or $C - \frac{(6-y)^2}{28}$ and use $F(5) = 1$ | M1 | |
| $F(y) = \begin{cases} 0 & y \leq -1 \\ \frac{1}{14}\left(\frac{y^2}{2}+2y+\frac{3}{2}\right) & -1 < y \leq 1 \\ \frac{3}{14}y + \frac{1}{14} & 1 < y \leq 3 \\ \frac{3}{7}y - \frac{1}{28}y^2 - \frac{1}{4} & 3 < y \leq 5 \\ 1 & y > 5 \end{cases}$ | A1, A1, B1 | First A1 for $1 < y \leq 3$ piece; Second A1 for $3 < y \leq 5$ piece; B1 for $y > 5$ |
| | **(5)** | |
## Question (e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $"\frac{3}{14}m + \frac{1}{14}" = 0.3$ | M1 | Setting their equation for $1 < y \leq 3$ equal to 0.3 |
| $m = \frac{16}{15}$ | A1 | cao |
**(2 marks)**
---
## Question (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(4Y \leq 5 \mid Y \leq 3) = \dfrac{\left(\frac{3}{14} \times \frac{5}{4} + \frac{1}{14}\right)}{\left(\frac{3}{14} \times 3 + \frac{1}{14}\right)} = \left[\dfrac{\frac{19}{56}}{\frac{5}{7}}\right]$ | M1 | For writing or using $\dfrac{F\!\left(\frac{5}{4}\right)}{F(3)}$; allow use of their expression for $3 < y \leq 5$ for the denominator |
| $= \dfrac{19}{40}$ or $0.475$ | A1 | cao |
**(2 marks)**
**Total: 17 marks**6. The continuous random variable $Y$ has probability density function $\mathrm { f } ( y )$ given by
$$f ( y ) = \begin{cases} \frac { 1 } { 14 } ( y + 2 ) & - 1 < y \leqslant 1 \\ \frac { 3 } { 14 } & 1 < y \leqslant 3 \\ \frac { 1 } { 14 } ( 6 - y ) & 3 < y \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the probability density function $\mathrm { f } ( \mathrm { y } )$
Given that $\mathrm { E } \left( Y ^ { 2 } \right) = \frac { 131 } { 21 }$
\item find $\operatorname { Var } ( 2 Y - 3 )$
The cumulative distribution function of $Y$ is $\mathrm { F } ( y )$
\item Show that $\mathrm { F } ( y ) = \frac { 1 } { 14 } \left( \frac { y ^ { 2 } } { 2 } + 2 y + \frac { 3 } { 2 } \right)$ for $- 1 < y \leqslant 1$
\item Find $\mathrm { F } ( y )$ for all values of $y$
\item Find the exact value of the 30th percentile of $Y$
\item Find $\mathrm { P } ( 4 Y \leqslant 5 \mid Y \leqslant 3 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2021 Q6 [17]}}