| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Single probability inequality |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard binomial distribution knowledge and normal approximation. Parts (a)-(c) are routine recall and calculation, while part (d) requires applying the normal approximation with continuity correction—a standard technique at this level. The calculations are mechanical with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial2.04f Find normal probabilities: Z transformation5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Occurrences of the disease are independent; The probability of catching the disease remains constant | B1, B1 | (2 marks) |
| (b) \(X \sim \text{Bin}(10,0.03)\); \(P(X = 2) = \frac{10 \times 9}{2}(0.03)^2(0.97)^8 = 0.0317\) | B1, M1A1 | A1 awrt 0.0317 |
| (c) \(E(X) = 100 \times 0.03 = 3\); \(\text{Var}(X) = 100 \times 0.03 \times 0.97 = 2.91\) | B1cao, B1cao | (2 marks) |
| (d) \(\lambda = 100 \times 0.03 = 3\); \(Y \sim \text{Po}(3)\); \(P(Y > 5) = 1 - P(Y \leq 5) = 1 - 0.9161 = 0.0839\) | B1 (use of), dM1, A1 | (3 marks) |
**(a)** Occurrences of the disease are independent; The probability of catching the disease remains constant | B1, B1 | (2 marks)
**(b)** $X \sim \text{Bin}(10,0.03)$; $P(X = 2) = \frac{10 \times 9}{2}(0.03)^2(0.97)^8 = 0.0317$ | B1, M1A1 | A1 awrt 0.0317 | (3 marks)
**(c)** $E(X) = 100 \times 0.03 = 3$; $\text{Var}(X) = 100 \times 0.03 \times 0.97 = 2.91$ | B1cao, B1cao | (2 marks)
**(d)** $\lambda = 100 \times 0.03 = 3$; $Y \sim \text{Po}(3)$; $P(Y > 5) = 1 - P(Y \leq 5) = 1 - 0.9161 = 0.0839$ | B1 (use of), dM1, A1 | (3 marks)
**Notes:**
- (a) B1 independent; B1 probability remains constant. One of these must have the context of disease. No context only one correct B0B0. If only one mark awarded give the first B1. SC if they are both correct without context award B1B0.
- (b) B1 for writing or using $B(10,0.03)$; M1 for writing or using $(p)^2(1-p)^8\frac{10!}{2!8!}$, allow $^{10}C_2$, etc.; A1 awrt 0.0317.
- (d) B1 for using Poisson. Any mean. Common values which imply Poisson used are 0.9665 and 0.8153. dM1 for writing or using $1 - P(X < 5)$ - use of binomial gets M0. This is dependent on them being awarded the previous B mark. A1 awrt 0.0839. **SC: Use of Normal in (d)** Can get B0 M1 A0 - for M1 we must see $1 - P(X < 5)$ or $1 - P(X \leq 5.5)$ or or get awrt 0.071.
**[10 marks total]**
---A disease occurs in 3\% of a population.
\begin{enumerate}[label=(\alph*)]
\item State any assumptions that are required to model the number of people with the disease in a random sample of size $n$ as a binomial distribution.
[2]
\item Using this model, find the probability of exactly 2 people having the disease in a random sample of 10 people.
[3]
\item Find the mean and variance of the number of people with the disease in a random sample of 100 people.
[2]
\end{enumerate}
A doctor tests a random sample of 100 patients for the disease. He decides to offer all patients a vaccination to protect them from the disease if more than 5 of the sample have the disease.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Using a suitable approximation, find the probability that the doctor will offer all patients a vaccination.
[3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2011 Q1 [10]}}