| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Expectation and variance from distribution |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard formulas for expectation and variance. Part (a) requires routine calculation of E(X) and Var(X) from a probability distribution. Part (b) applies standard results for linear combinations of random variables (Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)). Part (c) involves combining independent normal distributions and finding a probability using tables. All techniques are direct applications of bookwork with no problem-solving insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| \(\boldsymbol { x }\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.16 | 0.15 | 0.25 | 0.25 | 0.15 | 0.04 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X) = 2.2\) | B1 | CAO |
| \(\text{Var}(X) = E(X^2) - 2.2^2 = 6.8 - 4.84 = 1.96\) | M1, A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(S) = E(X) + 2.0 = 4.2\) | B1F | F on (a) |
| \(\text{Var}(S) = \text{Var}(X) + 1.5 + 2\times(-0.43) = 2.6\) | M1, A1F | Used for \(S\) or \(D\); F on (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(D) = E(X) - 2.0 = 0.2\) | B1F | F on (a) |
| \(\text{Var}(D) = \text{Var}(X) + 1.5 - 2\times(-0.43) = 4.32\) | M1, A1F | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(L \sim N(2.31,\ 0.89^2)\), \(M \sim N(2.04,\ 0.43^2)\) | ||
| \(T = L + M \sim N(4.35,\ 0.977)\) | B1 B1 | Both CAO; \(\text{SD}=0.98843\) |
| \(P(T>5) = P\!\left(Z > \dfrac{5-4.35}{\sqrt{0.977}}\right)\) | M1 | Standardising 5 or 5.01 using C's mean & SD |
| \(= P(Z>0.66) = 1 - P(Z<0.66)\) | m1 | Area change |
| \(= 0.25\) to \(0.26\) | A1 | 5 |
# Question 6:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = 2.2$ | B1 | CAO |
| $\text{Var}(X) = E(X^2) - 2.2^2 = 6.8 - 4.84 = 1.96$ | M1, A1 | 3 | Used or equivalent; CAO |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(S) = E(X) + 2.0 = 4.2$ | B1F | F on (a) |
| $\text{Var}(S) = \text{Var}(X) + 1.5 + 2\times(-0.43) = 2.6$ | M1, A1F | Used for $S$ or $D$; F on (a) |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(D) = E(X) - 2.0 = 0.2$ | B1F | F on (a) |
| $\text{Var}(D) = \text{Var}(X) + 1.5 - 2\times(-0.43) = 4.32$ | M1, A1F | 5 | F on (a) |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $L \sim N(2.31,\ 0.89^2)$, $M \sim N(2.04,\ 0.43^2)$ | | |
| $T = L + M \sim N(4.35,\ 0.977)$ | B1 B1 | Both CAO; $\text{SD}=0.98843$ |
| $P(T>5) = P\!\left(Z > \dfrac{5-4.35}{\sqrt{0.977}}\right)$ | M1 | Standardising 5 or 5.01 using C's mean & SD |
| $= P(Z>0.66) = 1 - P(Z<0.66)$ | m1 | Area change |
| $= 0.25$ to $0.26$ | A1 | 5 | AWFW (0.25540) |
---6 The table shows the probability distribution for the number of weekday (Monday to Friday) morning newspapers, $X$, purchased by the Reed household per week.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.16 & 0.15 & 0.25 & 0.25 & 0.15 & 0.04 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find values for $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\item The number of weekday (Monday to Friday) evening newspapers, $Y$, purchased by the same household per week is such that
$$\mathrm { E } ( Y ) = 2.0 , \quad \operatorname { Var } ( Y ) = 1.5 \quad \text { and } \quad \operatorname { Cov } ( X , Y ) = - 0.43$$
Find values for the mean and variance of:
\begin{enumerate}[label=(\roman*)]
\item $S = X + Y$;
\item $\quad D = X - Y$.
\end{enumerate}\item The total cost per week, $L$, of the Reed household's weekday morning and evening newspapers may be assumed to be normally distributed with a mean of $\pounds 2.31$ and a standard deviation of $\pounds 0.89$.
The total cost per week, $M$, of the household's weekend (Saturday and Sunday) newspapers may be assumed to be independent of $L$ and normally distributed with a mean of $\pounds 2.04$ and a standard deviation of $\pounds 0.43$.
Determine the probability that the total cost per week of the Reed household's newspapers is more than $\pounds 5$.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2009 Q6 [13]}}