| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question with standard applications. Parts (a)-(d) involve direct use of binomial formulas with p=0.2 or 0.8. Part (e) requires recognizing when a probability is negligible. Part (f) uses normal approximation with continuity correction—a standard S2 technique. All steps are routine applications of learned methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) No. of pears is \(B(10, 0.2)\) | B1 | |
| \(P(X = 5) = 0.9936 - 0.9672 = 0.0264\) | M1 A1 | |
| (b) \(P(X < 3) = P(X \leq 2) = 0.678\) | M1 A1 A1 | |
| (c) \(E(X) = 60 \times 0.2 = 12\) | B1 | |
| (d) \(\sqrt{12 \times 0.8} = \sqrt{9.6} = 3.10\) | B1 | |
| Same answer for s.d. of apples (just interchange 0.2 and 0.8) | B1 | |
| (e) \(B(60, 0.2)\), \(P(X = 5) = {60}C_5(0.2)^5(0.8)^{55} = 6.7 \times 10^{-14}\) | B1 M1 A1 A1 | |
| (f) \(B(60, 0.2) \sim N(12, 9.6)\) | B1 M1 A1 | |
| \(P(X > 15.5) = P(Z > 3.5/\sqrt{9.6}) = P(Z > 1.13) = 0.129\) | M1 A1 | Total: 18 |
(a) No. of pears is $B(10, 0.2)$ | B1 |
$P(X = 5) = 0.9936 - 0.9672 = 0.0264$ | M1 A1 |
(b) $P(X < 3) = P(X \leq 2) = 0.678$ | M1 A1 A1 |
(c) $E(X) = 60 \times 0.2 = 12$ | B1 |
(d) $\sqrt{12 \times 0.8} = \sqrt{9.6} = 3.10$ | B1 |
Same answer for s.d. of apples (just interchange 0.2 and 0.8) | B1 |
(e) $B(60, 0.2)$, $P(X = 5) = {60}C_5(0.2)^5(0.8)^{55} = 6.7 \times 10^{-14}$ | B1 M1 A1 A1 |
(f) $B(60, 0.2) \sim N(12, 9.6)$ | B1 M1 A1 |
$P(X > 15.5) = P(Z > 3.5/\sqrt{9.6}) = P(Z > 1.13) = 0.129$ | M1 A1 | **Total: 18**In an orchard, all the trees are either apple or pear trees. There are four times as many apple trees as pear trees. Find the probability that, in a random sample of 10 trees, there are
\begin{enumerate}[label=(\alph*)]
\item equal numbers of apple and pear trees, [3 marks]
\item more than 7 apple trees. [3 marks]
\end{enumerate}
In a sample of 60 trees in the orchard,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item find the expected number of pear trees. [1 mark]
\item Calculate the standard deviation of the number of pear trees and compare this result with the standard deviation of the number of apple trees. [2 marks]
\item Find the probability that exactly 35 in the sample of 60 trees are pear trees. [4 marks]
\item Find an approximate value for the probability that more than 15 of the 60 trees are pear trees. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q7 [18]}}