| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Compare approximation methods |
| Difficulty | Moderate -0.3 This is a standard S2 question testing routine application of binomial distribution and its approximations. Parts (a) and (b) require straightforward binomial probability calculations, part (c) applies normal approximation with continuity correction (standard procedure), and part (d) asks for recall of when Poisson is preferred over normal (n large, p small). All techniques are textbook exercises 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 binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a)(i) | \(B(60, 0.1)\) | B1 |
| \([P(Y = 0) + P(Y = 1)] = [(0.9)^{60} + 60(0.1)'(0.9)^{59}\) | M1 | Mark (i) and (ii) together |
| \(0.013777\ldots\) | A1 | awrt 0.0138 |
| Part (a)(ii) | ||
| Part (b) | \(\text{Po}(6)\) | B1 |
| \(P(Y \leq 1) = 0.01735\ldots\) or \(0.0174\) from tables | B1 | awrt 0.0174 |
| Part (c) | \(N(6, 5.4)\) | B1 |
| \(P(Y \leq 1) = P\left(Z < \frac{1.5 - 6}{\sqrt{5.4}}\right)\) | M1 dM1 | 1st M1 \(\pm \left(\frac{1.5 \text{ or } 1 \text{ or } 0.5 - \text{their mean}}{\text{their sd}}\right)\); 2nd M1 dependent on 1st M1 for using a continuity correction \(1 \pm 0.5\) |
| \(= P(Z < -1.936\ldots)\) | A1 | |
| \(= 1 - 0.9738 = 0.0262\) | A1 | (or from calc 0.02640...) |
| Part (d) | (Poisson approximation is more suitable since...) E.g. \(n\) is large and \(p\) is small; \(p\) is not close to 0.5; (b) is closer to the true value | B1 |
| Total (11) |
| **Part (a)(i)** | $B(60, 0.1)$ | B1 | |
| | $[P(Y = 0) + P(Y = 1)] = [(0.9)^{60} + 60(0.1)'(0.9)^{59}$ | M1 | Mark (i) and (ii) together |
| | $0.013777\ldots$ | A1 | awrt **0.0138** |
| **Part (a)(ii)** | | | |
| **Part (b)** | $\text{Po}(6)$ | B1 | 1st B1 for writing or using Po(6) (may be implied by a correct answer) |
| | $P(Y \leq 1) = 0.01735\ldots$ or $0.0174$ from tables | B1 | awrt **0.0174** |
| **Part (c)** | $N(6, 5.4)$ | B1 | mean $= 6$ and variance $= 5.4$ (may be seen in standardisation) |
| | $P(Y \leq 1) = P\left(Z < \frac{1.5 - 6}{\sqrt{5.4}}\right)$ | M1 dM1 | 1st M1 $\pm \left(\frac{1.5 \text{ or } 1 \text{ or } 0.5 - \text{their mean}}{\text{their sd}}\right)$; 2nd M1 dependent on 1st M1 for using a continuity correction $1 \pm 0.5$ |
| | $= P(Z < -1.936\ldots)$ | A1 | |
| | $= 1 - 0.9738 = 0.0262$ | A1 | (or from calc 0.02640...) |
| **Part (d)** | (Poisson approximation is more suitable since...) E.g. $n$ is large and $p$ is small; $p$ is not close to 0.5; (b) is closer to the true value | B1 | for a correct supporting reason (Condone $n$ is large together with $np < 10$) (Condone mean (6) $\approx$ variance (5.4)) |
| | | | Total (11) |
---Left-handed people make up 10\% of a population. A random sample of 60 people is taken from this population. The discrete random variable $Y$ represents the number of left-handed people in the sample.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down an expression for the exact value of $\mathrm{P}(Y \leq 1)$
\item Evaluate your expression, giving your answer to 3 significant figures. [3]
\end{enumerate}
\item Using a Poisson approximation, estimate $\mathrm{P}(Y \leq 1)$ [2]
\item Using a normal approximation, estimate $\mathrm{P}(Y \leq 1)$ [5]
\item Give a reason why the Poisson approximation is a more suitable estimate of $\mathrm{P}(Y \leq 1)$ [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2016 Q3 [11]}}