| Exam Board | OCR |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating the Poisson to the Normal distribution |
| Type | Combined Poisson approximation and exact calculation |
| Difficulty | Standard +0.8 This S2 question requires multiple approximations (Poisson to Normal) with continuity corrections, reverse table lookup, and chaining results across three parts. The conceptual demand of recognizing when and how to apply normal approximations, plus the multi-stage reasoning linking parts (i)-(iii), places this above average difficulty but within reach of well-prepared S2 students. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Po}(11)\) | M1 | Po(11) stated or clearly implied |
| \(1 - P(\leq r) = 0.854\) gives \(r = 14\), so \(n = 15\) | M1, A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Po}(44) \approx N(44, 44)\) | M1, A1 | Normal, mean attempted \(2.2 \times 20\); both parameters 44, allow var \(= \sqrt{44}\) or \(44^2\) |
| \(\Phi\left(\frac{37.5-44}{\sqrt{44}}\right) = \Phi(-0.980)\) | M1, A1 | Standardise their 44, allow cc, \(\sqrt{}\) errors |
| \(= \mathbf{0.1635}\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(B(40, 0.146) \approx N(5.84, 4.98736)\) | M1, M1 | B(40, 0.146) stated or implied; Normal, attempt at mean \(= np\) |
| \(1 - \Phi\left(\frac{7.5-5.84}{\sqrt{4.98736}}\right) = 1 - \Phi(0.7433)\) | A1, M1, A1 | Both parameters correct; standardise with \(np\) and \(npq\), allow cc, \(\sqrt{}\) errors |
| \(= \mathbf{0.2286}\) | A1 | 6 |
# Question 7:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Po}(11)$ | M1 | Po(11) stated or clearly implied |
| $1 - P(\leq r) = 0.854$ gives $r = 14$, so $n = 15$ | M1, A1 | **3** | Find $1 - 0.146$ in tables, e.g. answer 14; $n = 15$ only, allow "$\geq 15$" |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Po}(44) \approx N(44, 44)$ | M1, A1 | Normal, mean attempted $2.2 \times 20$; both parameters 44, allow var $= \sqrt{44}$ or $44^2$ |
| $\Phi\left(\frac{37.5-44}{\sqrt{44}}\right) = \Phi(-0.980)$ | M1, A1 | Standardise their 44, allow cc, $\sqrt{}$ errors |
| $= \mathbf{0.1635}$ | A1 | **5** | Answer in range [0.163, 0.164] |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $B(40, 0.146) \approx N(5.84, 4.98736)$ | M1, M1 | B(40, 0.146) stated or implied; Normal, attempt at mean $= np$ |
| $1 - \Phi\left(\frac{7.5-5.84}{\sqrt{4.98736}}\right) = 1 - \Phi(0.7433)$ | A1, M1, A1 | Both parameters correct; standardise with $np$ and $npq$, allow cc, $\sqrt{}$ errors |
| $= \mathbf{0.2286}$ | A1 | **6** | Answer in range [0.228, 0.229] |
---7 The number of customer complaints received by a company per day is denoted by $X$. Assume that $X$ has the distribution $\operatorname { Po } ( 2.2 )$.\\
(i) In a week of 5 working days, the probability there are at least $n$ customer complaints is 0.146 correct to 3 significant figures. Use tables to find the value of $n$.\\
(ii) Use a suitable approximation to find the probability that in a period of 20 working days there are fewer than 38 customer complaints.
A week of 5 working days in which at least $n$ customer complaints are received, where $n$ is the value found in part (i), is called a 'bad' week.\\
(iii) Use a suitable approximation to find the probability that, in 40 randomly chosen weeks, more than 7 are bad.
\hfill \mbox{\textit{OCR S2 2011 Q7 [14]}}