| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2004 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Easy -1.3 This is a straightforward S1 question requiring only direct application of standard formulas for discrete probability distributions. All parts involve routine calculations: summing probabilities, applying E(X) = Σxp(x), using linearity of expectation E(aX+b) = aE(X)+b, and the variance formula. No problem-solving insight or novel reasoning is required—purely mechanical application of memorized techniques. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 3 }\) | \(\frac { 1 } { 2 }\) | \(\frac { 1 } { 12 }\) | \(\frac { 1 } { 12 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(1 < X \leq 3) = P(X=2) + P(X=3)\) | M1 | |
| \(= \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}\) | A1 | \(\frac{2}{12}\); \(\frac{1}{6}\); \(0.167\); \(0.166\overline{6}\); \(0.16\overline{6}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F(2.6) = P(X \leq 2) = 1 - P(X=3) = 1 - \frac{1}{12} = \frac{11}{12}\); \(0.917\); \(0.91\overline{6}\) | B1 | (or: \(P(X \leq 2) = \frac{1}{3} + \frac{1}{2} + \frac{1}{12} = \frac{11}{12}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \left(0 \times \frac{1}{3}\right) + \ldots + \left(3 \times \frac{1}{12}\right) = \frac{11}{12}\) | M1 | Use of \(\sum x P(X=x)\) |
| A1 | \(\frac{11}{12}\); AWRT \(0.917\) | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(2X-3) = 2E(X) - 3\) | M1 | Use of \(E(ax+b)\) |
| \(= 2 \times \frac{11}{12} - 3 = -\frac{14}{12} = -\frac{7}{6}\) | A1 | \(-\frac{7}{6}\); \(-1\frac{1}{6}\); AWRT \(-1.17\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = 1^2 \times \frac{1}{2} + \ldots + 3^2 \times \frac{1}{12} - \left(\frac{11}{12}\right)^2\) | M1 | Use of \(E(X^2) - \{E(X)\}^2\) |
| A1 | Correct substitution | |
| \(= \frac{107}{144}\) | A1 | \(\frac{107}{144}\); AWRT \(0.743\) |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(1 < X \leq 3) = P(X=2) + P(X=3)$ | M1 | |
| $= \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$ | A1 | $\frac{2}{12}$; $\frac{1}{6}$; $0.167$; $0.166\overline{6}$; $0.16\overline{6}$ | **(2)** |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F(2.6) = P(X \leq 2) = 1 - P(X=3) = 1 - \frac{1}{12} = \frac{11}{12}$; $0.917$; $0.91\overline{6}$ | B1 | (or: $P(X \leq 2) = \frac{1}{3} + \frac{1}{2} + \frac{1}{12} = \frac{11}{12}$) | **(1)** |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \left(0 \times \frac{1}{3}\right) + \ldots + \left(3 \times \frac{1}{12}\right) = \frac{11}{12}$ | M1 | Use of $\sum x P(X=x)$ |
| | A1 | $\frac{11}{12}$; AWRT $0.917$ | **(2)** |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(2X-3) = 2E(X) - 3$ | M1 | Use of $E(ax+b)$ |
| $= 2 \times \frac{11}{12} - 3 = -\frac{14}{12} = -\frac{7}{6}$ | A1 | $-\frac{7}{6}$; $-1\frac{1}{6}$; AWRT $-1.17$ | **(2)** |
### Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = 1^2 \times \frac{1}{2} + \ldots + 3^2 \times \frac{1}{12} - \left(\frac{11}{12}\right)^2$ | M1 | Use of $E(X^2) - \{E(X)\}^2$ |
| | A1 | Correct substitution |
| $= \frac{107}{144}$ | A1 | $\frac{107}{144}$; AWRT $0.743$ | **(3)** |
---3. A discrete random variable $X$ has the probability function shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 2 }$ & $\frac { 1 } { 12 }$ & $\frac { 1 } { 12 }$ \\
\hline
\end{tabular}
\end{center}
Find
\begin{enumerate}[label=(\alph*)]
\item $\mathrm { P } ( 1 < X \leq 3 )$,
\item $\mathrm { F } ( 2.6 )$,
\item $\mathrm { E } ( X )$,
\item $\mathrm { E } ( 2 X - 3 )$,
\item $\operatorname { Var } ( X )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2004 Q3 [10]}}