| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Find cumulative distribution F(x) |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic discrete uniform distribution properties. Parts (a)-(c) involve simple recall and calculation with 4 equally likely values. Parts (d)-(f) apply standard linear transformation formulas for expectation and variance. All steps are routine textbook exercises requiring no problem-solving insight. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F(3) = \frac{3}{4}\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 2.5\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 1^2 \times \frac{1}{4} + 2^2 \times \frac{1}{4} + 3^2 \times \frac{1}{4} + 4^2 \times \frac{1}{4} = \frac{15}{2}\) | M1 | Correct expression for \(E(X^2)\); \(\frac{15}{2}\) on its own does not imply this mark |
| \(\text{Var}(X) = \frac{15}{2} - \left(\frac{5}{2}\right)^2 = \frac{5}{4}\) | M1A1 cso | 2nd M1 for correct \(\text{Var}(X)\) expression using their \(E(X^2)\) and \(E(X)\); A1 for \(\frac{5}{4}\) cso |
| Alt (c): \(\text{Var}(X) = \frac{n^2-1}{12}\) | M1 | May be implied by 2nd M1 |
| \(\frac{4^2-1}{12}\) or \(\frac{(4+1)(4-1)}{12}\) | M1 | 15/12 on its own does not score this mark |
| \(\frac{5}{4}\) | A1 cso | Dependent on both M marks and no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y=y) = \frac{1}{4}\) | B1 | May be in a table, but must be \(\frac{1}{4}\) for each probability |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(Y) = \text{Var}(kX+c) = k^2\text{Var}(X)\) | M1 | M1 for \(k^2\text{Var}(X)\) |
| \(= \frac{5}{4}k^2\) | A1 | oe |
| Answer | Marks |
|---|---|
| \(c = 3-k\) | B1 |
## Question 6:
### Part (a):
$F(3) = \frac{3}{4}$ | B1 | oe
### Part (b):
$E(X) = 2.5$ | B1 | oe
### Part (c):
$E(X^2) = 1^2 \times \frac{1}{4} + 2^2 \times \frac{1}{4} + 3^2 \times \frac{1}{4} + 4^2 \times \frac{1}{4} = \frac{15}{2}$ | M1 | Correct expression for $E(X^2)$; $\frac{15}{2}$ on its own does not imply this mark
$\text{Var}(X) = \frac{15}{2} - \left(\frac{5}{2}\right)^2 = \frac{5}{4}$ | M1A1 cso | 2nd M1 for correct $\text{Var}(X)$ expression using their $E(X^2)$ and $E(X)$; A1 for $\frac{5}{4}$ cso
**Alt (c):** $\text{Var}(X) = \frac{n^2-1}{12}$ | M1 | May be implied by 2nd M1
$\frac{4^2-1}{12}$ or $\frac{(4+1)(4-1)}{12}$ | M1 | 15/12 on its own does not score this mark
$\frac{5}{4}$ | A1 cso | Dependent on both M marks and no incorrect working
### Part (d):
$P(Y=y) = \frac{1}{4}$ | B1 | May be in a table, but must be $\frac{1}{4}$ for each probability
### Part (e):
$\text{Var}(Y) = \text{Var}(kX+c) = k^2\text{Var}(X)$ | M1 | M1 for $k^2\text{Var}(X)$
$= \frac{5}{4}k^2$ | A1 | oe
Note: $\text{Var}(Y) = \frac{3^2+(3+k)^2+(3+2k)^2+(3+3k)^2}{4} - \left(\frac{3+(3+3k)}{2}\right)^2$ scores M1A1 if correct expression seen
### Part (f):
$c = 3-k$ | B1 |
---\begin{enumerate}
\item The random variable $X$ has a discrete uniform distribution and takes the values $1,2,3,4$ Find\\
(a) $\mathrm { F } ( 3 )$, where $\mathrm { F } ( x )$ is the cumulative distribution function of $X$,\\
(b) $\mathrm { E } ( X )$.\\
(c) Show that $\operatorname { Var } ( X ) = \frac { 5 } { 4 }$
\end{enumerate}
The random variable $Y$ has a discrete uniform distribution and takes the values
$$3,3 + k , 3 + 2 k , 3 + 3 k$$
where $k$ is a constant.\\
(d) Write down $\mathrm { P } ( Y = y )$ for $y = 3,3 + k , 3 + 2 k , 3 + 3 k$
The relationship between $X$ and $Y$ may be written in the form $Y = k X + c$ where $c$ is a constant.\\
(e) Find $\operatorname { Var } ( Y )$ in terms of $k$.\\
(f) Express $c$ in terms of $k$.\\
\hfill \mbox{\textit{Edexcel S1 2015 Q6 [9]}}