| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Sampling distribution of mean or linear combination |
| Difficulty | Standard +0.3 This is a straightforward sampling distribution problem requiring enumeration of outcomes (4 cases), probability calculations using independence, and a simple expectation calculation. While it involves multiple steps, each is routine application of basic probability rules with no conceptual challenges beyond standard S2 material. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2(6)+6}{3}=6\), \(\frac{2(9)+9}{3}=9\), \(\frac{2(6)+9}{3}=7\), \(\frac{2(9)+6}{3}=8\) | B1 | At least three correct values for \(y\) of either 6, 7, 8 or 9 |
| Correct values for \(y\) of 6, 7, 8 and 9 only | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{(6,6)\Rightarrow P(Y=6)\}=(0.35)^2\); \(\{(6,9)\Rightarrow P(Y=7)\}=(0.65)(0.35)\); \(\{(9,6)\Rightarrow P(Y=8)\}=(0.35)(0.65)\); \(\{(9,9)\Rightarrow P(Y=9)\}=(0.65)^2\) | M1 | At least one of either \((0.35)^2\), \((0.65)(0.35)\), \((0.35)(0.65)\) or \((0.65)^2\) |
| M1 | At least two of either \((0.35)^2\), \((0.65)(0.35)\), \((0.35)(0.65)\) or \((0.65)^2\) | |
| \(P(Y=6)=0.1225\), \(P(Y=7)=0.2275\), \(P(Y=8)=0.2275\), \(P(Y=9)=0.4225\) | A1 | At least two correct probabilities linked to correct sample |
| or \(P(Y=y)\): \(\frac{49}{400}\), \(\frac{91}{400}\), \(\frac{91}{400}\), \(\frac{169}{400}\) | A1 | At least 3 correct |
| B1ft | See notes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{E(Y)\}=6(0.1225)+7(0.2275)+8(0.2275)+9(0.4225)=7.95\) or \(\frac{159}{20}\) | M1; A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(Y)=\frac{2}{3}E(X_1)+\frac{1}{3}E(X_2)=\frac{2}{3}E(X)+\frac{1}{3}E(X)=E(X)\) | ||
| \(=6(0.35)+9(0.65)=7.95\) or \(\frac{159}{20}\) | M1; A1 cao |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2(6)+6}{3}=6$, $\frac{2(9)+9}{3}=9$, $\frac{2(6)+9}{3}=7$, $\frac{2(9)+6}{3}=8$ | B1 | At least three correct values for $y$ of either 6, 7, 8 or 9 |
| Correct values for $y$ of 6, 7, 8 and 9 only | B1 | |
**[2]**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{(6,6)\Rightarrow P(Y=6)\}=(0.35)^2$; $\{(6,9)\Rightarrow P(Y=7)\}=(0.65)(0.35)$; $\{(9,6)\Rightarrow P(Y=8)\}=(0.35)(0.65)$; $\{(9,9)\Rightarrow P(Y=9)\}=(0.65)^2$ | M1 | At least one of either $(0.35)^2$, $(0.65)(0.35)$, $(0.35)(0.65)$ or $(0.65)^2$ |
| | M1 | At least two of either $(0.35)^2$, $(0.65)(0.35)$, $(0.35)(0.65)$ or $(0.65)^2$ |
| $P(Y=6)=0.1225$, $P(Y=7)=0.2275$, $P(Y=8)=0.2275$, $P(Y=9)=0.4225$ | A1 | At least two correct probabilities linked to correct sample |
| or $P(Y=y)$: $\frac{49}{400}$, $\frac{91}{400}$, $\frac{91}{400}$, $\frac{169}{400}$ | A1 | At least 3 correct |
| | B1ft | See notes |
**[5]**
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{E(Y)\}=6(0.1225)+7(0.2275)+8(0.2275)+9(0.4225)=7.95$ or $\frac{159}{20}$ | M1; A1 cao | |
**[2]**
## Part (c) – Alternative Method
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(Y)=\frac{2}{3}E(X_1)+\frac{1}{3}E(X_2)=\frac{2}{3}E(X)+\frac{1}{3}E(X)=E(X)$ | | |
| $=6(0.35)+9(0.65)=7.95$ or $\frac{159}{20}$ | M1; A1 cao | |
**[2]**
### Question 5 Notes:
- **(a) Note**: You can mark parts (a) and (b) together
- **(b) 1st M1**: At least one of $(0.35)^2$, $(0.65)(0.35)$, $(0.35)(0.65)$ or $(0.65)^2$. Can be implied.
- **(b) 2nd M1**: At least two of $(0.35)^2$, $(0.65)(0.35)$, $(0.35)(0.65)$ or $(0.65)^2$. Can be implied.
- **(b) 1st A1**: At least two correct probabilities given which either must be linked to a correct sample $(x_1,x_2)$ or their followed through $y$-value
- **(b) 2nd A1**: At least 3 correct probabilities corresponding to the correct value of $y$
- **(b) B1ft**: Either all 4 correct probabilities corresponding to the correct value of $y$, or 6, 7, 8 and 9 with two correct probabilities, two other probabilities and $\sum p(y)=1$. B1ft is dependent on 1st M1 2nd M1 1st A1.
- **(b) Note**: A table is not required but $y$-values must be linked with their probabilities for 2nd A1 B1
- **(b) Note**: Eg: (6,6) by itself does not count as an acceptable value of $y$
- **(c) M1**: A correct follow through expression for $E(Y)$ using their distribution. Also allow M1 for a correct expression for $E(X)$
- **(c) A1**: 7.95 cao. Allow $\frac{159}{20}$
---5. A bag contains a large number of counters with $35 \%$ of the counters having a value of 6 and $65 \%$ of the counters having a value of 9
A random sample of size 2 is taken from the bag and the value of each counter is recorded as $X _ { 1 }$ and $X _ { 2 }$ respectively.
The statistic $Y$ is calculated using the formula
$$Y = \frac { 2 X _ { 1 } + X _ { 2 } } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item List all the possible values of $Y$.
\item Find the sampling distribution of $Y$.
\item Find $\mathrm { E } ( Y )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2015 Q5 [9]}}