| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.3 This is a straightforward S1 question requiring standard binomial probability calculations and expectation/variance formulas. While it has multiple parts, each step follows routine procedures: identifying binomial outcomes (3C1, 3C2 combinations), applying probability formulas, and using E(X) and Var(X) definitions. The bonus round part (f) tests understanding of linear transformations but remains mechanical. Slightly easier than average due to the structured table guidance and lack of conceptual challenges. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 30 | 15 | 0 | - 15 |
| \(\mathrm { P } ( X = x )\) | 0.216 | 0.064 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| To score 15 points, 2 correct and 1 not correct: \([0.6\times0.6\times0.4]+[0.6\times0.4\times0.6]+[0.4\times0.6\times0.6]\) or \(3\times(0.6\times0.6\times0.4)=0.432\) | M1, A1cso | M1 for \(0.6^2\times0.4\) may imply tree diagram; just \(3\times0.144\) or \(2\times0.216\) is M0; A1 cso for \(3\times0.6^2\times0.4\) seen and no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1-(0.216+0.432+0.064)=\mathbf{0.288}\) or \(3\times0.6\times(0.4)^2\) | B1 | Answer may be seen in table; fractions: \(\frac{27}{125},\frac{54}{125},\frac{36}{125},\frac{8}{125}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([(30,0),(0,30)\) or \((15,15)]\): \(0.216\times'0.288'+0.288'\times0.216+0.432\times0.432\) awrt \(\mathbf{0.311}\) | M1 A1ft, A1 | M1 for either \(0.216\times'0.288'=(0.062208)\) or \(0.432\times0.432=0.186624\); 1st A1ft for fully correct expression; 2nd A1 for awrt 0.311 or \(\frac{972}{3125}\) |
| SC: 6 questions 4 correct: Award M1 & 1st A1 for \(6C4\times0.6^4\times0.4^2\) or \(15\times0.6^4\times0.4^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X)=[30\times0.216]+[15\times0.432]+[0\times0.288]+[(-15)\times0.064]\) | M1 | M1 for correct expression for \(E(X)\) (0 term not required, ft their (b)) |
| \(E(X)=12\) | A1 | 12 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X^2)=30^2\times0.216+15^2\times0.432+0^2\times0.288+(-15)^2\times0.064\ (=306)\) | M1 | 1st M1 for correct expression for \(E(X^2)\) (0 term not required, ft their (b)); condone \(-15^2\) |
| \(\text{Var}(X)=E(X^2)-[E(X)]^2='306'-'12'^2=\mathbf{162}\) | M1, A1 | 2nd M1 for correct expression for \(\text{Var}(X)\); ignore label so \(\text{Var}(X)=[E(X^2)]=306\) can score M1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(Y=\) number of points scored in bonus round; distribution table with \([y]\): 60, 35, 10, \(-15\) and \([P(Y=y)]\): 0.216, 0.432, 0.288, 0.064 | M1 | 1st M1 for correct distribution for \(Y\) (ft(b)) or \(20\times0.6+(-5)\times0.4\) or \(Y=\frac{5}{3}X+10\) |
| \(E(Y)=60\times0.216+35\times0.432+10\times0.288+(-15)\times0.064\) | dM1 | 2nd dM1 for correct expression for \(E(Y)\) or \(3\times(20\times0.6+(-5)\times0.4)\) or \(E(Y)=\frac{5}{3}E(X)+10\) |
| \(=\mathbf{30}\) | A1 | A1 for 30 with at least 1 M mark scored; answer only is 0/3 but 30 after M1 is 3/3 |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| To score 15 points, 2 correct and 1 not correct: $[0.6\times0.6\times0.4]+[0.6\times0.4\times0.6]+[0.4\times0.6\times0.6]$ or $3\times(0.6\times0.6\times0.4)=0.432$ | M1, A1cso | M1 for $0.6^2\times0.4$ may imply tree diagram; just $3\times0.144$ or $2\times0.216$ is M0; A1 cso for $3\times0.6^2\times0.4$ seen and no incorrect working |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1-(0.216+0.432+0.064)=\mathbf{0.288}$ or $3\times0.6\times(0.4)^2$ | B1 | Answer may be seen in table; fractions: $\frac{27}{125},\frac{54}{125},\frac{36}{125},\frac{8}{125}$ |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[(30,0),(0,30)$ or $(15,15)]$: $0.216\times'0.288'+0.288'\times0.216+0.432\times0.432$ awrt $\mathbf{0.311}$ | M1 A1ft, A1 | M1 for either $0.216\times'0.288'=(0.062208)$ or $0.432\times0.432=0.186624$; 1st A1ft for fully correct expression; 2nd A1 for awrt 0.311 or $\frac{972}{3125}$ |
| **SC**: 6 questions 4 correct: Award M1 & 1st A1 for $6C4\times0.6^4\times0.4^2$ or $15\times0.6^4\times0.4^2$ | | |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X)=[30\times0.216]+[15\times0.432]+[0\times0.288]+[(-15)\times0.064]$ | M1 | M1 for correct expression for $E(X)$ (0 term not required, ft their (b)) |
| $E(X)=12$ | A1 | **12** only |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2)=30^2\times0.216+15^2\times0.432+0^2\times0.288+(-15)^2\times0.064\ (=306)$ | M1 | 1st M1 for correct expression for $E(X^2)$ (0 term not required, ft their (b)); condone $-15^2$ |
| $\text{Var}(X)=E(X^2)-[E(X)]^2='306'-'12'^2=\mathbf{162}$ | M1, A1 | 2nd M1 for correct expression for $\text{Var}(X)$; ignore label so $\text{Var}(X)=[E(X^2)]=306$ can score M1M0A0 |
## Part (f)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $Y=$ number of points scored in bonus round; distribution table with $[y]$: 60, 35, 10, $-15$ and $[P(Y=y)]$: 0.216, 0.432, 0.288, 0.064 | M1 | 1st M1 for correct distribution for $Y$ (ft(b)) or $20\times0.6+(-5)\times0.4$ or $Y=\frac{5}{3}X+10$ |
| $E(Y)=60\times0.216+35\times0.432+10\times0.288+(-15)\times0.064$ | dM1 | 2nd dM1 for correct expression for $E(Y)$ or $3\times(20\times0.6+(-5)\times0.4)$ or $E(Y)=\frac{5}{3}E(X)+10$ |
| $=\mathbf{30}$ | A1 | A1 for 30 with at least 1 M mark scored; answer only is 0/3 but 30 after M1 is 3/3 |
---\begin{enumerate}
\item In a quiz, a team gains 10 points for every question it answers correctly and loses 5 points for every question it does not answer correctly. The probability of answering a question correctly is 0.6 for each question. One round of the quiz consists of 3 questions.
\end{enumerate}
The discrete random variable $X$ represents the total number of points scored in one round. The table shows the incomplete probability distribution of $X$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 30 & 15 & 0 & - 15 \\
\hline
$\mathrm { P } ( X = x )$ & 0.216 & & & 0.064 \\
\hline
\end{tabular}
\end{center}
(a) Show that the probability of scoring 15 points in a round is 0.432\\
(b) Find the probability of scoring 0 points in a round.\\
(c) Find the probability of scoring a total of 30 points in 2 rounds.\\
(d) Find $\mathrm { E } ( X )$\\
(e) Find $\operatorname { Var } ( X )$
In a bonus round of 3 questions, a team gains 20 points for every question it answers correctly and loses 5 points for every question it does not answer correctly.\\
(f) Find the expected number of points scored in the bonus round.\\
\hfill \mbox{\textit{Edexcel S1 2015 Q5 [14]}}