| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Standard +0.3 This is a straightforward multi-part binomial distribution question requiring standard techniques: (a) direct probability calculations using tables/calculator, (b) solving (1-p)^12 = 0.05 for p, and (c) using variance formula np(1-p) = 1.92 to form a quadratic. All methods are routine S2 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X < 5) = 0.8424\) | B1 | awrt 0.842 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X \geq 7) = 1 - P(X \leq 6)\) | M1 | Writing or using \(1 - P(X \leq 6)\). Do not accept \(1 - P(X < 7)\) unless \(1 - P(X \leq 6)\) has been used |
| \(= 1 - 0.9857\) | ||
| \(= 0.0143\) | A1 | awrt 0.0143 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=0) = (1-p)^{12}\) | ||
| \((1-p)^{12} = 0.05\) | M1 | 1st M1: \((1-p)^n = 0.05\) |
| \((1-p) = \sqrt[12]{0.05}\) | M1 | 2nd M1: taking \(n\)th root. If logs used, must get correct expression for \(1-p\) |
| \(p = 0.221\) | A1 | awrt 0.221 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Variance} = 12p(1-p)\) | ||
| \(12p(1-p) = 1.92\) | M1 | 1st M1: \(12p(1-p) = 1.92\) o.e. |
| \(12p^2 - 12p + 1.92 = 0\) or \(p^2 - p + 0.16 = 0\) or \(25p^2 - 25p + 4 = 0\) | ||
| \(p = \frac{12 \pm \sqrt{12^2 - 4 \times 12 \times 1.92}}{24}\) or \((5p-1)(5p-4) = 0\) | M1 | 2nd M1: solving a quadratic by factorising/completing the square/formula. Working must be correct for their quadratic, or formula written correctly with at most 1 error |
| \(p = 0.2\) or \(0.8\) | A1, A1 | 1st A1 for 0.2; 2nd A1 for 0.8 |
## Question 3:
### Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X < 5) = 0.8424$ | B1 | awrt 0.842 |
### Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X \geq 7) = 1 - P(X \leq 6)$ | M1 | Writing or using $1 - P(X \leq 6)$. Do not accept $1 - P(X < 7)$ unless $1 - P(X \leq 6)$ has been used |
| $= 1 - 0.9857$ | | |
| $= 0.0143$ | A1 | awrt 0.0143 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=0) = (1-p)^{12}$ | | |
| $(1-p)^{12} = 0.05$ | M1 | 1st M1: $(1-p)^n = 0.05$ |
| $(1-p) = \sqrt[12]{0.05}$ | M1 | 2nd M1: taking $n$th root. If logs used, must get correct expression for $1-p$ |
| $p = 0.221$ | A1 | awrt 0.221 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Variance} = 12p(1-p)$ | | |
| $12p(1-p) = 1.92$ | M1 | 1st M1: $12p(1-p) = 1.92$ o.e. |
| $12p^2 - 12p + 1.92 = 0$ or $p^2 - p + 0.16 = 0$ or $25p^2 - 25p + 4 = 0$ | | |
| $p = \frac{12 \pm \sqrt{12^2 - 4 \times 12 \times 1.92}}{24}$ or $(5p-1)(5p-4) = 0$ | M1 | 2nd M1: solving a quadratic by factorising/completing the square/formula. Working must be correct for **their** quadratic, or formula written correctly with at most 1 error |
| $p = 0.2$ or $0.8$ | A1, A1 | 1st A1 for 0.2; 2nd A1 for 0.8 |
---\begin{enumerate}
\item A random variable $X$ has the distribution $\mathrm { B } ( 12 , p )$.\\
(a) Given that $p = 0.25$ find\\
(i) $\mathrm { P } ( X < 5 )$\\
(ii) $\mathrm { P } ( X \geqslant 7 )$\\
(b) Given that $\mathrm { P } ( X = 0 ) = 0.05$, find the value of $p$ to 3 decimal places.\\
(c) Given that the variance of $X$ is 1.92 , find the possible values of $p$.\\
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q3 [10]}}