| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic discrete probability distribution properties. Part (a) uses ΣP(X=x)=1 to find k, parts (b-d) apply standard formulas for probability, expectation and variance with simple arithmetic, and part (e) uses E(Y²)=Var(Y)+[E(Y)]² with linear transformation rules. All steps are routine applications of bookwork formulas with no problem-solving or insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1^2}{k}+\frac{2^2}{k}+\frac{3^2}{k}+\frac{4^2}{k}=1\); \(\frac{30}{k}=1\), so \(k=30\) | M1, A1cso | M1 for clear use of sum of probs \(=1\); A1 for correct conclusion with no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| \(1-P(X=4) = 1-\frac{16}{30} = \frac{7}{15}\) | M1, A1 | M1 for \(1-P(X=4)\) or \(P(X=1)+P(X=2)+P(X=3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1\times\frac{1}{30}+2\times\frac{4}{30}+3\times\frac{9}{30}+4\times\frac{16}{30} = \frac{10}{3}\) | M1, A1 | M1 for at least 3 correct products |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 1^2\times\frac{1}{30}+2^2\times\frac{4}{30}+3^2\times\frac{9}{30}+4^2\times\frac{16}{30} = \frac{354}{30}\); \(\text{Var}(X)=\frac{354}{30}-\left(\frac{100}{30}\right)^2 = \frac{31}{45}\) | M1 A1, M1, A1 | 1st M1 for \(E(X^2)\) attempt; 2nd M1 for \(\text{Var}(X)=E(X^2)-[E(X)]^2\) |
| Answer | Marks |
|---|---|
| \(E(Y)=3E(X)-1=9\); \(\text{Var}(Y)=3^2\text{Var}(X)=6.2\); \(E(Y^2)=\text{Var}(Y)+[E(Y)]^2=6.2+81=\mathbf{87.2}\) | M1, M1, M1 A1 |
# Question 6:
## Part (a)
| $\frac{1^2}{k}+\frac{2^2}{k}+\frac{3^2}{k}+\frac{4^2}{k}=1$; $\frac{30}{k}=1$, so $k=30$ | M1, A1cso | M1 for clear use of sum of probs $=1$; A1 for correct conclusion with no incorrect working |
## Part (b)
| $1-P(X=4) = 1-\frac{16}{30} = \frac{7}{15}$ | M1, A1 | M1 for $1-P(X=4)$ or $P(X=1)+P(X=2)+P(X=3)$ |
## Part (c)
| $E(X) = 1\times\frac{1}{30}+2\times\frac{4}{30}+3\times\frac{9}{30}+4\times\frac{16}{30} = \frac{10}{3}$ | M1, A1 | M1 for at least 3 correct products |
## Part (d)
| $E(X^2) = 1^2\times\frac{1}{30}+2^2\times\frac{4}{30}+3^2\times\frac{9}{30}+4^2\times\frac{16}{30} = \frac{354}{30}$; $\text{Var}(X)=\frac{354}{30}-\left(\frac{100}{30}\right)^2 = \frac{31}{45}$ | M1 A1, M1, A1 | 1st M1 for $E(X^2)$ attempt; 2nd M1 for $\text{Var}(X)=E(X^2)-[E(X)]^2$ |
## Part (e)
| $E(Y)=3E(X)-1=9$; $\text{Var}(Y)=3^2\text{Var}(X)=6.2$; $E(Y^2)=\text{Var}(Y)+[E(Y)]^2=6.2+81=\mathbf{87.2}$ | M1, M1, M1 A1 | |
---\begin{enumerate}
\item The random variable $X$ has probability function
\end{enumerate}
$$\mathrm { P } ( X = x ) = \frac { x ^ { 2 } } { k } \quad x = 1,2,3,4$$
(a) Show that $k = 30$\\
(b) Find $\mathrm { P } ( X \neq 4 )$\\
(c) Find the exact value of $\mathrm { E } ( X )$\\
(d) Find the exact value of $\operatorname { Var } ( X )$
Given that $Y = 3 X - 1$\\
(e) find $\mathrm { E } \left( Y ^ { 2 } \right)$\\
\hfill \mbox{\textit{Edexcel S1 2015 Q6 [14]}}