| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Easy -1.2 This is a straightforward S1 question testing basic discrete probability distribution concepts: finding the normalizing constant using ΣP(X=x)=1, calculating simple probabilities, and applying standard expectation formulas. All parts are routine recall and arithmetic with no problem-solving insight required, making it easier than average. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(k + 2k + 3k + 4k + 5k = 1\) | M1 | \(\sum P(X=x) = 1\) |
| \(15k = 1\) | ||
| \(k = \frac{1}{15}\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 4) = P(1) + P(2) + P(3) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15}\) | M1 | sum of 3 probabilities |
| \(= \frac{2}{5}\) | A1 | \(\frac{6}{15}\) or \(\frac{2}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 3 \times \frac{3}{15} + 4 \times \frac{4}{15} + 5 \times \frac{5}{15}\) | M1 | use of \(\sum xP(X=x)\) |
| \(= \frac{11}{3}\) | A1 | \(\frac{55}{15}\) or \(\frac{11}{3}\) or \(3\frac{2}{3}\) or \(3.\dot{6}\) or \(3.67\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(3X - 4) = 3E(X) - 4 = 11 - 4\) | M1 | \(3\times\text{theirs}-4\) |
| \(= 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(3X-4) = -1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 5 \times \frac{3}{15} + 8 \times \frac{4}{15} + 11 \times \frac{5}{15}\) | M1 | \(\sum(3x-4)kx\) |
| \(= 7\) | A1 | cao |
## Question 4:
**Part (a):**
$k + 2k + 3k + 4k + 5k = 1$ | M1 | $\sum P(X=x) = 1$
$15k = 1$ | |
$k = \frac{1}{15}$ | A1 | cso
**Part (b):**
$P(X < 4) = P(1) + P(2) + P(3) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15}$ | M1 | sum of 3 probabilities
$= \frac{2}{5}$ | A1 | $\frac{6}{15}$ or $\frac{2}{5}$
**Part (c):**
$E(X) = 1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 3 \times \frac{3}{15} + 4 \times \frac{4}{15} + 5 \times \frac{5}{15}$ | M1 | use of $\sum xP(X=x)$
$= \frac{11}{3}$ | A1 | $\frac{55}{15}$ or $\frac{11}{3}$ or $3\frac{2}{3}$ or $3.\dot{6}$ or $3.67$
**Part (d):**
$E(3X - 4) = 3E(X) - 4 = 11 - 4$ | M1 | $3\times\text{theirs}-4$
$= 7$ | A1 |
**OR:**
$E(3X-4) = -1 \times \frac{1}{15} + 2 \times \frac{2}{15} + 5 \times \frac{3}{15} + 8 \times \frac{4}{15} + 11 \times \frac{5}{15}$ | M1 | $\sum(3x-4)kx$
$= 7$ | A1 | cao
---4. The random variable $X$ has probability function
$$\mathrm { P } ( X = x ) = k x , \quad x = 1,2 , \ldots , 5$$
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 15 }$.
Find
\item $\mathrm { P } ( X < 4 )$,
\item $\mathrm { E } ( X )$,
\item $\mathrm { E } ( 3 X - 4 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2005 Q4 [8]}}