| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | CDF of transformed variable |
| Difficulty | Standard +0.3 This is a straightforward S2 question testing standard CDF properties and basic probability calculations. Part (a) uses continuity conditions at boundaries (routine), part (b) is direct differentiation, part (c) involves equating two simple probabilities, and part (d) is a binomial probability calculation. All techniques are standard textbook exercises with no novel insight required, making it 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| If \(y=0\) then \(1 - (\alpha + \beta y^2) = 0 \therefore \alpha = 1\) | B1cso | For stating or using that when \(y=0\) then \(\alpha + \beta y^2 = 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| If \(y=5\) then \(1 - (\alpha + \beta y^2) = 1\), so \(1 + 25\beta = 0 \therefore \beta = -\frac{1}{25}\) | B1cso | For stating or using that when \(y=5\) then \(\alpha + \beta y^2 = 0\) and setting up equation leading to \(\beta = -\frac{1}{25}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(y) = \frac{1}{25}y^2\) so \(f(y) = \frac{dF(y)}{dy} = \frac{2}{25}y\) | M1 | For differentiating; implied by \(\pm\frac{2}{"25"}y\); can ft their value of \(\beta\) |
| \(f(y) = \begin{cases} \frac{2}{25}y & 0 \leqslant y \leqslant 5 \\ 0 & \text{otherwise} \end{cases}\) | A1 | For a fully correct \(f(y)\) defined for the whole range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\left(R > \frac{11}{5}\right) = P\left(Y > \frac{5}{3}\right) = 1 - \frac{1}{25} \times \left(\frac{5}{3}\right)^2 = \frac{8}{9}\) oe | B1 | For using \(F(y)\) and \(\frac{5}{3}\) to find \(P(Y > \frac{5}{3})\); allow \(\frac{8}{9}\) or any exact equivalent |
| \(\frac{3d - \frac{11}{5}}{3d - d} = \frac{8}{9}\) oe or \(\frac{\frac{11}{5} - d}{3d - d} = \frac{1}{9}\) oe | M1 | For LHS \(= p\) where \(0 < p < 1\) |
| \(d = \frac{9}{5}\) oe | A1 | For \(\frac{9}{5}\) or any exact equivalent e.g. 1.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P\left(Y < \frac{11}{5}\right) = \frac{121}{625}\) or 0.1936 | B1 | This mark could be implied by a correct answer |
| \([G = \text{number of spins with distance} < 2.2\text{ m}]\), \([P(G \geqslant 5) =]\) \(\left(\frac{1}{9}\right)^3 \times \left(\frac{121}{625}\right)^3 + 3 \times \left(\frac{1}{9}\right)^2 \times \left(\frac{8}{9}\right) \times \left(\frac{121}{625}\right)^3 + 3 \times \left(\frac{1}{9}\right)^3 \times \left(\frac{121}{625}\right)^2 \times \left(\frac{504}{625}\right)\) | M1, M1 | 1st M1: for \(p^3q^3 + np^2(1-p)q^3 + np^3q^2(1-q)\); 2nd M1: for \(p^3q^3 + 3p^2(1-p)q^3 + 3p^3q^2(1-q)\) |
| \(= 0.000\,373226\) | A1 | awrt 0.000 373 |
# Question 5:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| If $y=0$ then $1 - (\alpha + \beta y^2) = 0 \therefore \alpha = 1$ | B1cso | For stating or using that when $y=0$ then $\alpha + \beta y^2 = 1$ |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| If $y=5$ then $1 - (\alpha + \beta y^2) = 1$, so $1 + 25\beta = 0 \therefore \beta = -\frac{1}{25}$ | B1cso | For stating or using that when $y=5$ then $\alpha + \beta y^2 = 0$ and setting up equation leading to $\beta = -\frac{1}{25}$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(y) = \frac{1}{25}y^2$ so $f(y) = \frac{dF(y)}{dy} = \frac{2}{25}y$ | M1 | For differentiating; implied by $\pm\frac{2}{"25"}y$; can ft their value of $\beta$ |
| $f(y) = \begin{cases} \frac{2}{25}y & 0 \leqslant y \leqslant 5 \\ 0 & \text{otherwise} \end{cases}$ | A1 | For a fully correct $f(y)$ defined for the whole range |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P\left(R > \frac{11}{5}\right) = P\left(Y > \frac{5}{3}\right) = 1 - \frac{1}{25} \times \left(\frac{5}{3}\right)^2 = \frac{8}{9}$ oe | B1 | For using $F(y)$ and $\frac{5}{3}$ to find $P(Y > \frac{5}{3})$; allow $\frac{8}{9}$ or any exact equivalent |
| $\frac{3d - \frac{11}{5}}{3d - d} = \frac{8}{9}$ oe or $\frac{\frac{11}{5} - d}{3d - d} = \frac{1}{9}$ oe | M1 | For LHS $= p$ where $0 < p < 1$ |
| $d = \frac{9}{5}$ oe | A1 | For $\frac{9}{5}$ or any exact equivalent e.g. 1.8 |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P\left(Y < \frac{11}{5}\right) = \frac{121}{625}$ or 0.1936 | B1 | This mark could be implied by a correct answer |
| $[G = \text{number of spins with distance} < 2.2\text{ m}]$, $[P(G \geqslant 5) =]$ $\left(\frac{1}{9}\right)^3 \times \left(\frac{121}{625}\right)^3 + 3 \times \left(\frac{1}{9}\right)^2 \times \left(\frac{8}{9}\right) \times \left(\frac{121}{625}\right)^3 + 3 \times \left(\frac{1}{9}\right)^3 \times \left(\frac{121}{625}\right)^2 \times \left(\frac{504}{625}\right)$ | M1, M1 | 1st M1: for $p^3q^3 + np^2(1-p)q^3 + np^3q^2(1-q)$; 2nd M1: for $p^3q^3 + 3p^2(1-p)q^3 + 3p^3q^2(1-q)$ |
| $= 0.000\,373226$ | A1 | awrt **0.000 373** |\begin{enumerate}
\item A game uses two turntables, one red and one yellow. Each turntable has a point marked on the circumference that is lined up with an arrow at the start of the game. Jim spins both turntables and measures the distance, in metres, each point is from the arrow, around the circumference in an anticlockwise direction when the turntables stop spinning.
\end{enumerate}
The continuous random variable $Y$ represents the distance, in metres, the point is from the arrow for the yellow turntable. The cumulative distribution function of $Y$ is given by $\mathrm { F } ( y )$ where
$$F ( y ) = \left\{ \begin{array} { c r }
0 & y < 0 \\
1 - \left( \alpha + \beta y ^ { 2 } \right) & 0 \leqslant y \leqslant 5 \\
1 & y > 5
\end{array} \right.$$
(a) Explain why (i) $\alpha = 1$
$$\text { (ii) } \beta = - \frac { 1 } { 25 }$$
(b) Find the probability density function of $Y$
The continuous random variable $R$ represents the distance, in metres, the point is from the arrow for the red turntable. The distribution of $R$ is modelled by a continuous uniform distribution over the interval $[ d , 3 d ]$
Given that $\mathrm { P } \left( R > \frac { 11 } { 5 } \right) = \mathrm { P } \left( Y > \frac { 5 } { 3 } \right)$\\
(c) find the value of $d$
In the game each turntable is spun 3 times. The distance between the point and the arrow is determined for each spin. To win a prize, at least 5 of the distances the point is from the arrow when a turntable is spun must be less than $\frac { 11 } { 5 } \mathrm {~m}$\\
Jo plays the game once.\\
(d) Calculate the probability of Jo winning a prize.\\
\hfill \mbox{\textit{Edexcel S2 2021 Q5 [11]}}