| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing standard probability distribution calculations. Part (a) uses the fact probabilities sum to 1 (routine). Parts (b)-(d) apply standard formulas for E(X), Var(X), and Var(aX+b) directly from the table. Part (e) requires solving an inequality but is still mechanical. All steps are textbook exercises with no problem-solving insight required, making it easier than average. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | - 1 | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 5 }\) | \(a\) | \(\frac { 1 } { 10 }\) | \(a\) | \(\frac { 1 } { 5 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2a + \frac{2}{5} + \frac{1}{10} = 1\), \(a = \frac{1}{4}\) or 0.25 | M1 A1 | M1 for clear attempt to use \(\sum P(X=x)=1\). NB Division by 5 scores 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1\) | B1 | B1 for 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 1\times\frac{1}{5} + 1\times\frac{1}{10} + 4\times\frac{1}{4} + 9\times\frac{1}{5}\) (= 3.1) | M1 | M1 for attempting \(\sum x^2 P(X=x)\), at least two terms correct |
| \(\text{Var}(X) = 3.1 - 1^2 = 2.1\) or \(\frac{21}{10}\) | M1 A1 | 2nd M1 for \(E(X^2)-[E(X)]^2\). Correct answer only 3/3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(Y) = (-2)^2 \text{Var}(X) = 8.4\) or \(\frac{42}{5}\) | M1 A1 | M1 for \((-2)^2\text{Var}(X)\) or \(4\text{Var}(X)\). Correct answer only 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \geq Y\) when \(X = 3\) or \(2\), probability \(= \frac{1}{4} + \frac{1}{5} = \frac{9}{20}\) | M1 A1ft A1 | M1 for identifying \(X=2,3\); A1ft for \(P(X=2)+P(X=3)\); 2nd A1 for \(\frac{9}{20}\) or 0.45 |
# Question 3:
## Part (a)
| $2a + \frac{2}{5} + \frac{1}{10} = 1$, $a = \frac{1}{4}$ or 0.25 | M1 A1 | M1 for clear attempt to use $\sum P(X=x)=1$. NB Division by 5 scores 0 |
## Part (b)
| $E(X) = 1$ | B1 | B1 for 1 |
## Part (c)
| $E(X^2) = 1\times\frac{1}{5} + 1\times\frac{1}{10} + 4\times\frac{1}{4} + 9\times\frac{1}{5}$ (= 3.1) | M1 | M1 for attempting $\sum x^2 P(X=x)$, at least two terms correct |
|---|---|---|
| $\text{Var}(X) = 3.1 - 1^2 = 2.1$ or $\frac{21}{10}$ | M1 A1 | 2nd M1 for $E(X^2)-[E(X)]^2$. Correct answer only 3/3 |
## Part (d)
| $\text{Var}(Y) = (-2)^2 \text{Var}(X) = 8.4$ or $\frac{42}{5}$ | M1 A1 | M1 for $(-2)^2\text{Var}(X)$ or $4\text{Var}(X)$. Correct answer only 2/2 |
## Part (e)
| $X \geq Y$ when $X = 3$ or $2$, probability $= \frac{1}{4} + \frac{1}{5} = \frac{9}{20}$ | M1 A1ft A1 | M1 for identifying $X=2,3$; A1ft for $P(X=2)+P(X=3)$; 2nd A1 for $\frac{9}{20}$ or 0.45 |
---\begin{enumerate}
\item The discrete random variable $X$ has probability distribution given by
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 1 & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 5 }$ & $a$ & $\frac { 1 } { 10 }$ & $a$ & $\frac { 1 } { 5 }$ \\
\hline
\end{tabular}
\end{center}
where $a$ is a constant.\\
(a) Find the value of $a$.\\
(b) Write down $\mathrm { E } ( X )$.\\
(c) Find $\operatorname { Var } ( X )$.
The random variable $Y = 6 - 2 X$\\
(d) Find $\operatorname { Var } ( Y )$.\\
(e) Calculate $\mathrm { P } ( X \geqslant Y )$.\\
\hfill \mbox{\textit{Edexcel S1 Q3 [11]}}