| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.8 This is a straightforward S2 probability distribution question requiring only standard calculations: summing probabilities, computing mean/variance from a discrete distribution, and applying linear transformations. Part (d) tests conceptual understanding but is accessible. All techniques are routine textbook exercises with no problem-solving insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | 0 | 0.19 | 0.26 | 0.20 | 0.13 | 0.07 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | 0.35 | B1 |
| Total: 1 | ||
| (b) | \(P(< 3) = 0.45\) | B1 |
| '0.35' × '0.45' (= 0.1575) × 2 = 0.315 | M1 A1 | Their 0.35 and 0.45; CAO or equivalent fraction or % |
| Total: 3 | ||
| (c) | Mean = \(1 \times 0.19 + 2 \times 0.26 + 3 \times 0.20 + 4 \times 0.13 + 5 \times 0.07 + 6 \times 0.15\) | M1 |
| \(= 0.19 + 0.52 + 0.60 + 0.52 + 0.35 + 0.90 = 3.08\) | A1 | CAO. AG. |
| Variance = \(1^2 \times 0.19 + 2^2 \times 0.26 + 3^2 \times 0.20 + 4^2 \times 0.13 + 5^2 \times 0.07 + 6^2 \times 0.15 - 3.08^2\) | M1 | PI; AWFW 2.77 to 2.78 |
| \(= 2.77(36)\) | A1 | |
| Total: 4 | ||
| (d) | No probability of 0 books borrowed. | |
| ..and no probability of more than 6 books. | Cannot borrow more than 6 books | |
| Wrong shape or probability increases at top end. | E2,1,0 | E1 for each of these 3 distinct points up to a maximum of 2 marks |
| Total: 2 | ||
| (e)(i) | \(10 \times 3.08 = 30.8\) (pence) | B1 |
| Total: 1 | ||
| (ii) | '2.7736' × 100 = '277.36' then \(\sqrt{277.36}\) | M1 |
| \(= 16.7\) (pence) or | A1 | AWFW 16 to 17 |
| \(\sqrt{'2.7736'} = '1.66...'\), '1.67' × 10 = 16.7 (pence) | (M1) (A1) | For √ their variance and × 10; AWFW 16 to 17 |
| Where working is in £ in (i) or (ii) or both: | ||
| (i) £0.308 | (B1) | Must show £ sign |
| (ii) '2.7736' × 0.01 = 0.027736; \(\sqrt{0.027736}\) = £0.167 | (M1) (A1) | For their variance × 0.01 and √; AWFW 0.16 to 0.17. Must show £ sign |
| Or \(\sqrt{'2.7736'} = '1.66...'\), '1.67' × 0.1 = £0.167 | (M1) (A1) | For √ their variance and × 0.1; AWFW 0.16 to 0.17. Must show £ sign |
| Total: 3 | ||
| TOTAL FOR Q3: 13 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | 0.35 | B1 | CAO or equivalent fraction or % |
| | **Total: 1** | | |
| (b) | $P(< 3) = 0.45$ | B1 | |
| | '0.35' × '0.45' (= 0.1575) × 2 = 0.315 | M1 A1 | Their 0.35 and 0.45; CAO or equivalent fraction or % |
| | **Total: 3** | | |
| (c) | Mean = $1 \times 0.19 + 2 \times 0.26 + 3 \times 0.20 + 4 \times 0.13 + 5 \times 0.07 + 6 \times 0.15$ | M1 | This working, or the next line, must be seen (at least 3 products) |
| | $= 0.19 + 0.52 + 0.60 + 0.52 + 0.35 + 0.90 = 3.08$ | A1 | CAO. AG. |
| | Variance = $1^2 \times 0.19 + 2^2 \times 0.26 + 3^2 \times 0.20 + 4^2 \times 0.13 + 5^2 \times 0.07 + 6^2 \times 0.15 - 3.08^2$ | M1 | PI; AWFW 2.77 to 2.78 |
| | $= 2.77(36)$ | A1 | |
| | **Total: 4** | | |
| (d) | No probability of 0 books borrowed. | | Cannot borrow no books |
| | ..and no probability of more than 6 books. | | Cannot borrow more than 6 books |
| | Wrong shape or probability increases at top end. | E2,1,0 | E1 for each of these 3 distinct points up to a maximum of 2 marks |
| | **Total: 2** | | |
| (e)(i) | $10 \times 3.08 = 30.8$ (pence) | B1 | CAO; 31, without 30.8 seen, scores B0 |
| | **Total: 1** | | |
| (ii) | '2.7736' × 100 = '277.36' then $\sqrt{277.36}$ | M1 | For their variance × 100 and √ |
| | $= 16.7$ (pence) or | A1 | AWFW 16 to 17 |
| | $\sqrt{'2.7736'} = '1.66...'$, '1.67' × 10 = 16.7 (pence) | (M1) (A1) | For √ their variance and × 10; AWFW 16 to 17 |
| | Where working is in £ in (i) or (ii) or both: | | |
| | (i) £0.308 | (B1) | Must show £ sign |
| | (ii) '2.7736' × 0.01 = 0.027736; $\sqrt{0.027736}$ = £0.167 | (M1) (A1) | For their variance × 0.01 and √; AWFW 0.16 to 0.17. Must show £ sign |
| | Or $\sqrt{'2.7736'} = '1.66...'$, '1.67' × 0.1 = £0.167 | (M1) (A1) | For √ their variance and × 0.1; AWFW 0.16 to 0.17. Must show £ sign |
| | **Total: 3** | | |
| | **TOTAL FOR Q3: 13** | | |
---Members of a library may borrow up to 6 books. Past experience has shown that the number of books borrowed, $X$, follows the distribution shown in the table.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
P(X = x) & 0 & 0.19 & 0.26 & 0.20 & 0.13 & 0.07 & 0.15 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a member borrows more than 3 books. [1 mark]
\item Assume that the numbers of books borrowed by two particular members are independent.
Find the probability that one of these members borrows more than 3 books and the other borrows fewer than 3 books. [3 marks]
\item Show that the mean of $X$ is 3.08, and calculate the variance of $X$. [4 marks]
\item One of the library staff notices that the values of the mean and the variance of $X$ are similar and suggests that a Poisson distribution could be used to model $X$.
Without further calculations, give two reasons why a Poisson distribution would not be suitable to model $X$. [2 marks]
\item The library introduces a fee of 10 pence for each book borrowed.
Assuming that the probabilities do not change, calculate:
\begin{enumerate}[label=(\roman*)]
\item the mean amount that will be paid by a member;
\item the standard deviation of the amount that will be paid by a member.
\end{enumerate}
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2016 Q3 [13]}}