| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Identify distribution and parameters |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard binomial distribution identification, linear transformation of random variables, and normal approximation. Parts (a)-(d) are routine bookwork requiring only recall of formulas and basic calculation. Part (e) is a standard normal approximation application. While multi-part, each component is textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim B(20, 0.2)\) | M1 A1 | M1 for "binomial" or \(B(...)\); A1 for \(n=20\) and \(p=0.2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S = 4X - 1(20-X)\) | M1 | For attempt at correct expression for \(S\) using 4 and \(-1\) |
| \(S = 5X - 20\) | A1cso | For correct expression derived; no incorrect working seen. NB: 'show that' so working must be shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 4\), \(\text{Var}(X) = 3.2\) | B1, B1 | 1st B1: \(E(X)=4\); 2nd B1: \(\text{Var}(X)=3.2\) |
| \(E(S) = 5\times4 - 20 = 0\), \(\text{Var}(S) = 5^2\,\text{Var}(X) = 80\) | M1 A1 | M1: correct formula for \(E(S)\) or \(\text{Var}(S)\); A1: 0 and 80 correctly assigned |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(S \geq 20\) implies \(5X - 20 \geq 20\) | M1 | For attempt to solve inequality for \(X\) |
| So \(5X \geq 40\), i.e. \(X \geq 8\) | A1 | |
| \(P(S \geq 20) = P(X \geq 8) = 1 - P(X \leq 7)\) | M1 | For \(1 - P(X \leq 7)\) |
| \(= 1 - 0.9679 = 0.0321\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(C \sim B(100, 0.4)\), \(Y \sim N\!\left(40,\, (\sqrt{24})^2\right)\) | M1 A1 | M1: for normal approx and mean \(= 40\); A1: \(\text{Var}=24\) or sd \(=\sqrt{24}\) |
| \(P(C > 50) \simeq P(Y > 50.5)\) | M1 | For continuity correction 49.5 or 50.5 |
| \(= P\!\left(Z > \frac{50.5-40}{\sqrt{24}}\right)\) | M1 | Standardising using their mean and sd; must use 50.5, 49.5 or 50 |
| \(= P(Z > 2.14...) = 1 - 0.9838 = 0.0162\) or \(0.016044...\) (awrt \(0.016\)) | A1 | For awrt 0.016. NB: exact Bin \(= 0.01676...\); Poisson approx \(= 0.0526...\) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(20, 0.2)$ | M1 A1 | M1 for "binomial" or $B(...)$; A1 for $n=20$ and $p=0.2$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S = 4X - 1(20-X)$ | M1 | For attempt at correct expression for $S$ using 4 and $-1$ |
| $S = 5X - 20$ | A1cso | For correct expression derived; no incorrect working seen. **NB**: 'show that' so working must be shown |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 4$, $\text{Var}(X) = 3.2$ | B1, B1 | 1st B1: $E(X)=4$; 2nd B1: $\text{Var}(X)=3.2$ |
| $E(S) = 5\times4 - 20 = 0$, $\text{Var}(S) = 5^2\,\text{Var}(X) = 80$ | M1 A1 | M1: correct formula for $E(S)$ or $\text{Var}(S)$; A1: 0 and 80 correctly assigned |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $S \geq 20$ implies $5X - 20 \geq 20$ | M1 | For attempt to solve inequality for $X$ |
| So $5X \geq 40$, i.e. $X \geq 8$ | A1 | |
| $P(S \geq 20) = P(X \geq 8) = 1 - P(X \leq 7)$ | M1 | For $1 - P(X \leq 7)$ |
| $= 1 - 0.9679 = 0.0321$ | A1 | |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $C \sim B(100, 0.4)$, $Y \sim N\!\left(40,\, (\sqrt{24})^2\right)$ | M1 A1 | M1: for normal approx and mean $= 40$; A1: $\text{Var}=24$ or sd $=\sqrt{24}$ |
| $P(C > 50) \simeq P(Y > 50.5)$ | M1 | For continuity correction 49.5 or 50.5 |
| $= P\!\left(Z > \frac{50.5-40}{\sqrt{24}}\right)$ | M1 | Standardising using their mean and sd; must use 50.5, 49.5 or 50 |
| $= P(Z > 2.14...) = 1 - 0.9838 = 0.0162$ or $0.016044...$ (awrt $0.016$) | A1 | For awrt 0.016. NB: exact Bin $= 0.01676...$; Poisson approx $= 0.0526...$ |
The image you've shared appears to be only the **back cover/credits page** of an Edexcel mark scheme document. It contains only publication information (address, telephone, order code, website) and logos (Ofqual, Welsh Assembly Government, CEA) — **no actual mark scheme content** is visible on this page.
To extract mark scheme content, please share the **interior pages** of the document that contain the actual question markings, answer schemes, and mark allocations.7. As part of a selection procedure for a company, applicants have to answer all 20 questions of a multiple choice test. If an applicant chooses answers at random the probability of choosing a correct answer is 0.2 and the number of correct answers is represented by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Suggest a suitable distribution for $X$.\\
(2)
Each applicant gains 4 points for each correct answer but loses 1 point for each incorrect answer. The random variable $S$ represents the final score, in points, for an applicant who chooses answers to this test at random.
\item Show that $S = 5 X - 20$
\item Find $\mathrm { E } ( S )$ and $\operatorname { Var } ( S )$.
An applicant who achieves a score of at least 20 points is invited to take part in the final stage of the selection process.
\item Find $\mathrm { P } ( S \geqslant 20 )$\\
(4)
Cameron is taking the final stage of the selection process which is a multiple choice test consisting of 100 questions. He has been preparing for this test and believes that his chance of answering each question correctly is 0.4
\item Using a suitable approximation, estimate the probability that Cameron answers more than half of the questions correctly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2013 Q7 [17]}}