| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | At least one success |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial distribution with standard probability calculations. Parts (a) and (b) are direct formula substitutions, (c) requires solving np ≥ 5, and (d) uses the complement rule 1 - (0.85)^n > 0.95. All parts are routine textbook exercises requiring only basic binomial knowledge and calculator work, making this easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X=0) = 0.85^{10}\) or from tables | M1 | \((p)^{10}\) with \(0 < p < 1\) |
| \(= 0.1969\) | A1 | awrt 0.197 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X > 3) = 1 - P(X \leq 3)\) | M1 | Writing or using \(1 - P(X \leq 3)\) |
| \(= 1 - 0.6477 = 0.3523\) | A1 | awrt 0.352 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n \times 0.15 = 5\) | M1 | \(np = 5\), \(0 < p < 1\) |
| \(n = 33\) or \(34\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 - P(X=0) > 0.95\) | M1 | Writing or using \(1 - P(X=0) > 0.95\) or \(P(X=0) < 0.05\). Also accepted: \(=\) or \(\geq\) instead of \(>\), and \(=\) or \(\leq\) instead of \(<\) |
| \(1 - (0.85)^n > 0.95\), i.e. \(0.85^n < 0.05\) | A1 | Writing or using \(1-(0.85)^n > 0.95\) or \((0.85)^n < 0.05\) |
| \(n > 18.4\) | NB answer of 18.4 gets M1A1A0 | |
| \(n = 19\) | A1 | cao. Answer of 19 gets M1A1A1 unless from clearly incorrect working |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=0) = 0.85^{10}$ or from tables | M1 | $(p)^{10}$ with $0 < p < 1$ |
| $= 0.1969$ | A1 | awrt 0.197 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X > 3) = 1 - P(X \leq 3)$ | M1 | Writing or using $1 - P(X \leq 3)$ |
| $= 1 - 0.6477 = 0.3523$ | A1 | awrt 0.352 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n \times 0.15 = 5$ | M1 | $np = 5$, $0 < p < 1$ |
| $n = 33$ or $34$ | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - P(X=0) > 0.95$ | M1 | Writing or using $1 - P(X=0) > 0.95$ or $P(X=0) < 0.05$. Also accepted: $=$ or $\geq$ instead of $>$, and $=$ or $\leq$ instead of $<$ |
| $1 - (0.85)^n > 0.95$, i.e. $0.85^n < 0.05$ | A1 | Writing or using $1-(0.85)^n > 0.95$ or $(0.85)^n < 0.05$ |
| $n > 18.4$ | | NB answer of 18.4 gets M1A1A0 |
| $n = 19$ | A1 | cao. Answer of 19 gets M1A1A1 unless from clearly incorrect working |
---3. The probability of a telesales representative making a sale on a customer call is 0.15
Find the probability that
\begin{enumerate}[label=(\alph*)]
\item no sales are made in 10 calls,
\item more than 3 sales are made in 20 calls.
Representatives are required to achieve a mean of at least 5 sales each day.
\item Find the least number of calls each day a representative should make to achieve this requirement.
\item Calculate the least number of calls that need to be made by a representative for the probability of at least 1 sale to exceed 0.95
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2012 Q3 [9]}}