| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Poisson approximation to binomial |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial distribution (part a) followed by a standard normal approximation with continuity correction (part b). The question explicitly signals the approximation method and uses standard parameters (np=8, npq=7.6) that clearly satisfy approximation conditions. Requires only routine application of formulas with no problem-solving insight. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(X\) = number of leaf cuttings successfully taking root; \(X \sim B(10, 0.05)\) | B1 | May appear in (i) or (ii) or may be implied |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X=1) = P(X \leq 1) - P(X=0)\) or \(^{10}C_1 \times 0.05 \times 0.95^9 = 0.9139 - 0.5987 = 0.3152\) | M1 | Writing or using \(P(X \leq 1) - P(X=0)\) or \(^nC_1 \times p \times (1-p)^{n-1}\), \((0 < p < 1)\) |
| \(= 0.3152\) | A1 | awrt 0.315 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9885 = 0.0115\) | M1 | Writing or using \(1 - P(X \leq 2)\) |
| \(= 0.0115\) | A1 | awrt 0.0115 |
| Answer | Marks | Guidance |
|---|---|---|
| \(Y \sim Po(8)\) | B1 | Writing or using \(Po(8)\) or \(N(8, 7.6)\) |
| \(P(Y \geq 10) = 1 - P(Y \leq 9) = 1 - 0.7166 = 0.2834\) | M1 | Writing or using \(1 - P(Y \leq 9)\); or for M1: \(P\left(Z > \frac{9.5-8}{\sqrt{7.6}}\right)\) |
| \(= 0.2834\) | A1 | awrt 0.283; normal range (0.293, 0.295) also acceptable |
# Question 1:
## Part (a) - Setup
| Let $X$ = number of leaf cuttings successfully taking root; $X \sim B(10, 0.05)$ | B1 | May appear in (i) or (ii) or may be implied |
## Part (a)(i)
| $P(X=1) = P(X \leq 1) - P(X=0)$ or $^{10}C_1 \times 0.05 \times 0.95^9 = 0.9139 - 0.5987 = 0.3152$ | M1 | Writing or using $P(X \leq 1) - P(X=0)$ or $^nC_1 \times p \times (1-p)^{n-1}$, $(0 < p < 1)$ |
| $= 0.3152$ | A1 | awrt 0.315 |
## Part (a)(ii)
| $P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9885 = 0.0115$ | M1 | Writing or using $1 - P(X \leq 2)$ |
| $= 0.0115$ | A1 | awrt 0.0115 |
## Part (b)
| $Y \sim Po(8)$ | B1 | Writing or using $Po(8)$ or $N(8, 7.6)$ |
| $P(Y \geq 10) = 1 - P(Y \leq 9) = 1 - 0.7166 = 0.2834$ | M1 | Writing or using $1 - P(Y \leq 9)$; or for M1: $P\left(Z > \frac{9.5-8}{\sqrt{7.6}}\right)$ |
| $= 0.2834$ | A1 | awrt 0.283; normal range (0.293, 0.295) also acceptable |
---\begin{enumerate}
\item The probability of a leaf cutting successfully taking root is 0.05
\end{enumerate}
Find the probability that, in a batch of 10 randomly selected leaf cuttings, the number taking root will be\\
(a) (i) exactly 1\\
(ii) more than 2
A second random sample of 160 leaf cuttings is selected.\\
(b) Using a suitable approximation, estimate the probability of at least 10 leaf cuttings taking root.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q1 [8]}}