| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Expected profit or cost problem |
| Difficulty | Standard +0.3 This is a straightforward S2 question involving a given discrete probability distribution. Part (a) requires simple probability summation, (b) uses standard E(X) and E(X²) formulas, (c) applies expectation to a linear profit function, and (d) compares two expected values. All steps are routine applications of A-level statistics formulas with no novel problem-solving required. The Poisson context is given but not deeply explored. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(\boldsymbol { x }\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.135 | 0.271 | 0.271 | \(a\) | \(b\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| All probabilities sum to 1: \(0.135 + 0.271 + 0.271 + a + b = 1\) | M1 | Setting up sum = 1 |
| Regular customer always buys 1 copy, so additional demand is Poisson(2), \(X = 1 +\) additional demand | M1 | Using Poisson distribution correctly |
| \(P(X=4) = a = e^{-2}\frac{2^3}{3!} = 0.180\) | A1 | |
| \(P(X=5) = b = 1 - 0.135 - 0.271 - 0.271 - 0.180 = 0.143\) | A1 | Including \(P(X \geq 5)\) correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 1(0.135) + 2(0.271) + 3(0.271) + 4(0.180) + 5(0.143)\) | M1 | Correct method for \(E(X)\) |
| \(E(X) = 0.135 + 0.542 + 0.813 + 0.720 + 0.715 = 2.925\) | A1 | |
| \(E(X^2) = 1(0.135) + 4(0.271) + 9(0.271) + 16(0.180) + 25(0.143)\) | M1 | Correct method for \(E(X^2)\) |
| \(E(X^2) = 0.135 + 1.084 + 2.439 + 2.880 + 3.575 = 10.113\) | A1 | |
| \(\text{Var}(X) = 10.113 - (2.925)^2 = 1.557\) | M1 | Using \(\text{Var}(X) = E(X^2) - [E(X)]^2\) |
| \(\text{SD}(X) = \sqrt{1.557} = 1.248\) | A1 | awrt 1.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Profit \(= 100X - 50(5-X)\) pence when \(X < 5\); \(= 500\) when \(X = 5\) | M1 | Correct profit expression |
| Mean profit \(= 100E(X) - 50(5 - E(X))\) pence \(= 150E(X) - 250\) | M1 | Using linearity of expectation |
| \(= 150(2.925) - 250 = 188.75\) p \(= £1.89\) | A1 | awrt £1.89 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| With 4 copies stocked, profit \(= 100x - 50(4-x)\) for \(x \leq 4\); \(= 400\) for \(x \geq 4\) | M1 | Setting up new profit structure |
| \(E(\text{profit}) = 150(1)(0.135) + 150(2)(0.271) + 150(3)(0.271) + 400(0.180 + 0.143)\) | M1 | Correct calculation |
| \(= 20.25 + 81.30 + 121.95 + 129.20 = 352.70\) p | A1 | |
| \(352.70 > 188.75 \times \frac{4}{5}\)... comparing correctly | M1 | Valid comparison shown |
| Mean profit with 4 copies \(≈ £3.53 > £1.89\), so stocking 4 copies gives greater profit | A1 | Correct conclusion with supporting figures |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| All probabilities sum to 1: $0.135 + 0.271 + 0.271 + a + b = 1$ | M1 | Setting up sum = 1 |
| Regular customer always buys 1 copy, so additional demand is Poisson(2), $X = 1 +$ additional demand | M1 | Using Poisson distribution correctly |
| $P(X=4) = a = e^{-2}\frac{2^3}{3!} = 0.180$ | A1 | |
| $P(X=5) = b = 1 - 0.135 - 0.271 - 0.271 - 0.180 = 0.143$ | A1 | Including $P(X \geq 5)$ correctly |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 1(0.135) + 2(0.271) + 3(0.271) + 4(0.180) + 5(0.143)$ | M1 | Correct method for $E(X)$ |
| $E(X) = 0.135 + 0.542 + 0.813 + 0.720 + 0.715 = 2.925$ | A1 | |
| $E(X^2) = 1(0.135) + 4(0.271) + 9(0.271) + 16(0.180) + 25(0.143)$ | M1 | Correct method for $E(X^2)$ |
| $E(X^2) = 0.135 + 1.084 + 2.439 + 2.880 + 3.575 = 10.113$ | A1 | |
| $\text{Var}(X) = 10.113 - (2.925)^2 = 1.557$ | M1 | Using $\text{Var}(X) = E(X^2) - [E(X)]^2$ |
| $\text{SD}(X) = \sqrt{1.557} = 1.248$ | A1 | awrt 1.25 |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Profit $= 100X - 50(5-X)$ pence when $X < 5$; $= 500$ when $X = 5$ | M1 | Correct profit expression |
| Mean profit $= 100E(X) - 50(5 - E(X))$ pence $= 150E(X) - 250$ | M1 | Using linearity of expectation |
| $= 150(2.925) - 250 = 188.75$ p $= £1.89$ | A1 | awrt £1.89 |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| With 4 copies stocked, profit $= 100x - 50(4-x)$ for $x \leq 4$; $= 400$ for $x \geq 4$ | M1 | Setting up new profit structure |
| $E(\text{profit}) = 150(1)(0.135) + 150(2)(0.271) + 150(3)(0.271) + 400(0.180 + 0.143)$ | M1 | Correct calculation |
| $= 20.25 + 81.30 + 121.95 + 129.20 = 352.70$ p | A1 | |
| $352.70 > 188.75 \times \frac{4}{5}$... comparing correctly | M1 | Valid comparison shown |
| Mean profit with 4 copies $≈ £3.53 > £1.89$, so stocking 4 copies gives greater profit | A1 | Correct conclusion with supporting figures |
These pages (22, 23, and 24) are **answer space pages** from an AQA exam paper (P/Jun15/MS2B). They contain no mark scheme content — they are blank lined answer spaces provided for students to write their responses to Question 7, followed by an "END OF QUESTIONS" notice and a blank page.
There is **no mark scheme content to extract** from these images. To obtain the mark scheme for this paper, you would need the separate AQA mark scheme document for P/Jun15/MS2B.7 Each week, a newsagent stocks 5 copies of the magazine Statistics Weekly. A regular customer always buys one copy. The demand for additional copies may be modelled by a Poisson distribution with mean 2.
The number of copies sold in a week, $X$, has the probability distribution shown in the table, where probabilities are stated correct to three decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.135 & 0.271 & 0.271 & $a$ & $b$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that, correct to three decimal places, the values of $a$ and $b$ are 0.180 and 0.143 respectively.
\item Find the values of $\mathrm { E } ( X )$ and $\mathrm { E } \left( X ^ { 2 } \right)$, showing the calculations needed to obtain these values, and hence calculate the standard deviation of $X$.
\item The newsagent makes a profit of $\pounds 1$ on each copy of Statistics Weekly that is sold and loses 50 p on each copy that is not sold. Find the mean weekly profit for the newsagent from sales of this magazine.
\item Assuming that the weekly demand remains the same, show that the mean weekly profit from sales of Statistics Weekly will be greater if the newsagent stocks only 4 copies.\\[0pt]
[5 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{6cdf244b-168a-4be5-8ef8-8125daae8608-24_2488_1728_219_141}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2015 Q7 [15]}}