| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Combined probability with other distributions |
| Difficulty | Standard +0.3 This is a standard S2 binomial distribution question with straightforward applications: basic binomial probability calculations, conditional probability using the law of total probability, and normal approximation to binomial. All parts follow textbook methods with no novel insight required. Part (d) requires normal approximation which is routine at S2 level. Slightly easier than average due to clear structure and standard techniques throughout. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03d Calculate conditional probability: from first principles2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(X \sim \text{B}(12, 0.2)\) \(P(X < 3) = P(X \leq 2) = 0.5583\ldots\) awrt \(\mathbf{0.558}\) | B1 M1 A1 (3) |
| (b)(i) | \([\text{P(customer takes sugar)} ] = 0.8 \times 0.35 + 0.2 \times 0.6 [ = 0.4*]\) | B1cso (1) |
| (ii) | \([Y \sim] \text{B}(n, 0.4)\) | B1 (1) |
| (c)(i) | \(P(Y = 4) = [=^{10}C_4(0.4)^4(0.6)^6 = 0.6331 - 0.3823] = 0.2508\) awrt \(\mathbf{0.251}\) | B1 M1 A1 (1) |
| (ii) | \(P(Y \leq 6 \mid Y \geq 3) = \frac{P(3 \leq Y \leq 6)}{P(Y \geq 3)} = \frac{P(Y \leq 6) - P(Y \leq 2)}{1 - P(Y \leq 2)} = \frac{0.9452 - 0.01673 [= 0.7779]}{1 - 0.1673 [= 0.8327]} = 0.934\ldots\) awrt \(\mathbf{0.934}\) | M1 M1 A1 (3) |
| (d) | \(C \sim \text{B}(150, 0.4) \to \text{N}(60, 36)\) \(P(C \geq 75) = P\left(Z > \frac{74.5 - 60}{\sqrt{36}}\right) = [P(Z > 2.416\ldots)] = 1 - 0.9922 = 0.0078\) awrt \(\mathbf{0.0078}\) | M1 M1 M1 A1 (4) |
| Total 13 |
(a) | $X \sim \text{B}(12, 0.2)$ $P(X < 3) = P(X \leq 2) = 0.5583\ldots$ awrt $\mathbf{0.558}$ | B1 M1 A1 (3) | B1 for writing or using B(10, 0.2) / M1 for writing or using $P(X \leq 2)$
(b)(i) | $[\text{P(customer takes sugar)} ] = 0.8 \times 0.35 + 0.2 \times 0.6 [ = 0.4*]$ | B1cso (1) | B1cso for $0.8 \times 0.35 + 0.2 \times 0.6$ or equivalent. Condone use of percentages here
(ii) | $[Y \sim] \text{B}(n, 0.4)$ | B1 (1) | B1 $\text{B}(n, 0.4)$
(c)(i) | $P(Y = 4) = [=^{10}C_4(0.4)^4(0.6)^6 = 0.6331 - 0.3823] = 0.2508$ awrt $\mathbf{0.251}$ | B1 M1 A1 (1) | 1st M1 for a correct ratio expression [Do not allow $P(Y \leq 6 \cap Y \geq 3)$ on numerator] / 2nd M1 for writing or using $\frac{P(Y \leq 6) - P(Y \leq 2)}{1 - P(Y \leq 2)}$
(ii) | $P(Y \leq 6 \mid Y \geq 3) = \frac{P(3 \leq Y \leq 6)}{P(Y \geq 3)} = \frac{P(Y \leq 6) - P(Y \leq 2)}{1 - P(Y \leq 2)} = \frac{0.9452 - 0.01673 [= 0.7779]}{1 - 0.1673 [= 0.8327]} = 0.934\ldots$ awrt $\mathbf{0.934}$ | M1 M1 A1 (3) |
(d) | $C \sim \text{B}(150, 0.4) \to \text{N}(60, 36)$ $P(C \geq 75) = P\left(Z > \frac{74.5 - 60}{\sqrt{36}}\right) = [P(Z > 2.416\ldots)] = 1 - 0.9922 = 0.0078$ awrt $\mathbf{0.0078}$ | M1 M1 M1 A1 (4) | 1st M1 for using a Normal approximation to binomial with $\mu = np$ and $\sigma^2 = np(1-p)$ / 2nd M1 for standardising 74.5, 75, 75.5 with their mean and standard deviation / 3rd M1 for use of continuity correction (75 ± 0.5)
| | | Total 13 |
---\begin{enumerate}
\item At a cafe, customers ordering hot drinks order either tea or coffee.
\end{enumerate}
Of all customers ordering hot drinks, $80 \%$ order tea and $20 \%$ order coffee. Of those who order tea, $35 \%$ take sugar and of those who order coffee $60 \%$ take sugar.\\
\begin{enumerate}[label=(\alph*)]
\item A random sample of 12 customers ordering hot drinks is selected.
Find the probability that fewer than 3 of these customers order coffee.
\item
\begin{enumerate}[label=(\roman*)]
\item A randomly selected customer who orders a hot drink is chosen. Show that the probability that the customer takes sugar is 0.4
\item Write down the distribution for the number of customers who take sugar from a random sample of $n$ customers ordering hot drinks.
\end{enumerate}
\item A random sample of 10 customers ordering hot drinks is selected.\\
\begin{enumerate}[label=(\roman*)]
\item Find the probability that exactly 4 of these 10 customers take sugar.
\item Given that at least 3 of these 10 customers take sugar, find the probability that no more than 6 of these 10 customers take sugar.
\end{enumerate}
\item In a random sample of 150 customers ordering hot drinks, find, using a suitable approximation, the probability that at least half of them take sugar.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2018 Q2 [13]}}