| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Direct probability from given distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability distribution concepts: reading from a table, solving simple inequalities, verifying variance using the standard formula, applying linear transformation properties (E(aX+b) and Var(aX+b)), and calculating probabilities by enumeration. All parts are routine textbook exercises requiring only direct application of standard formulas with no problem-solving insight needed. Part (d) requires some enumeration but is still mechanical. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(\boldsymbol { x }\) | 2 | 4 | 5 | 7 | 8 |
| \(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\) | 0.25 | 0.3 | 0.2 | 0.1 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(2X-3>5) = 0.45\) | B1 | 0.45 or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(X^2) = 2^2 \times 0.25 + 4^2 \times 0.3 + 5^2 \times 0.2 + 7^2 \times 0.1 + 8^2 \times 0.15 [= 25.3]\) | M1 | For correct method to find \(E(X^2)\); at least 3 terms correct and added; not implied by 25.3 alone |
| \(\text{Var}(X) = 2^2 \times 0.25 + 4^2 \times 0.3 + 5^2 \times 0.2 + 7^2 \times 0.1 + 8^2 \times 0.15 - 4.6^2\) | M1 | For use of correct equation ft their \(E(X^2)\) |
| \(= 25.3 - 4.6^2 = 4.14\) | A1* | For correct expression with all terms seen, leading to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([E(Y)=] 13.4 = a \times 4.6 - b\) | M1 | For writing or using correct equation for \(E(Y)\) |
| \([\text{Var}(Y)=] a^2 \times 4.14 = 66.24\) | M1 | For writing or using correct method for \(\text{Var}(Y)\) |
| \(a = 4\) | A1 | May be seen as part of expression \(4X \pm \ldots\) |
| \(b = 5\) | A1 | May be seen as part of expression \(\ldots X - 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sam throws 8 and Alex throws 2, 4 or 5; Sam throws 7 and Alex throws 2 or 4; Sam throws 5 or 4 or 2 and Alex throws 2 | M1 | For recognising all required combinations; implied by 3 correct probabilities; ignore repeats but not incorrect combinations |
| \(0.15 \times (0.25+0.3+0.2)\) or \(0.1 \times (0.25+0.3)\) or \((0.2+0.3+0.25) \times 0.25\) | M1 | For any one correct calculation from the 3 given |
| \(0.15\times(0.25+0.3+0.2) + 0.1\times(0.25+0.3) + (0.2+0.3+0.25)\times 0.25\) | M1 | For any 2 correct calculations from the 3 given |
| \(= 0.355 \left(= \frac{71}{200}\right)\) | A1 | 0.355 or equivalent |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(2X-3>5) = 0.45$ | B1 | 0.45 or equivalent |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X^2) = 2^2 \times 0.25 + 4^2 \times 0.3 + 5^2 \times 0.2 + 7^2 \times 0.1 + 8^2 \times 0.15 [= 25.3]$ | M1 | For correct method to find $E(X^2)$; at least 3 terms correct and added; not implied by 25.3 alone |
| $\text{Var}(X) = 2^2 \times 0.25 + 4^2 \times 0.3 + 5^2 \times 0.2 + 7^2 \times 0.1 + 8^2 \times 0.15 - 4.6^2$ | M1 | For use of correct equation ft their $E(X^2)$ |
| $= 25.3 - 4.6^2 = 4.14$ | A1* | For correct expression with all terms seen, leading to given answer |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[E(Y)=] 13.4 = a \times 4.6 - b$ | M1 | For writing or using correct equation for $E(Y)$ |
| $[\text{Var}(Y)=] a^2 \times 4.14 = 66.24$ | M1 | For writing or using correct method for $\text{Var}(Y)$ |
| $a = 4$ | A1 | May be seen as part of expression $4X \pm \ldots$ |
| $b = 5$ | A1 | May be seen as part of expression $\ldots X - 5$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sam throws 8 and Alex throws 2, 4 or 5; Sam throws 7 and Alex throws 2 or 4; Sam throws 5 or 4 or 2 and Alex throws 2 | M1 | For recognising all required combinations; implied by 3 correct probabilities; ignore repeats but not incorrect combinations |
| $0.15 \times (0.25+0.3+0.2)$ or $0.1 \times (0.25+0.3)$ or $(0.2+0.3+0.25) \times 0.25$ | M1 | For any one correct calculation from the 3 given |
| $0.15\times(0.25+0.3+0.2) + 0.1\times(0.25+0.3) + (0.2+0.3+0.25)\times 0.25$ | M1 | For any 2 correct calculations from the 3 given |
| $= 0.355 \left(= \frac{71}{200}\right)$ | A1 | 0.355 or equivalent |
---2. A spinner can land on the numbers $2,4,5,7$ or 8 only.
The random variable $X$ represents the number that this spinner lands on when it is spun once. The probability distribution of $X$ is given in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 2 & 4 & 5 & 7 & 8 \\
\hline
$\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )$ & 0.25 & 0.3 & 0.2 & 0.1 & 0.15 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( 2 X - 3 > 5 )$
Given that $\mathrm { E } ( X ) = 4.6$
\item show that $\operatorname { Var } ( X ) = 4.14$
The random variable $Y = a X - b$ where $a$ and $b$ are positive constants.\\
Given that
$$\mathrm { E } ( Y ) = 13.4 \quad \text { and } \quad \operatorname { Var } ( Y ) = 66.24$$
\item find the value of $a$ and the value of $b$
In a game Sam and Alex each spin the spinner once, landing on $X _ { 1 }$ and $X _ { 2 }$ respectively.\\
Sam's score is given by the random variable $S = X _ { 1 }$\\
Alex's score is given by the random variable $R = 2 X _ { 2 } - 3$\\
The person with the higher score wins the game. If the scores are the same it is a draw.
\item Find the probability that Sam wins the game.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2024 Q2 [12]}}