| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring routine application of standard formulas for discrete probability distributions. Students substitute values into the given probability function, sum probabilities over an interval, calculate E(X) and Var(X) using standard formulas, then apply the variance transformation rule. All steps are mechanical with no problem-solving or insight required—easier than average A-level. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) N.B. Part (a) doesn't have to be in a table, could be a list \(P(X = 1) = \ldots\) etc | B1, B1, B1 | |
| \(x\) | 1 | 2 |
| \(P(X = x)\) | \(\frac{1}{36}\) | \(\frac{3}{36}\) |
| 0.0278, 0.0833, 0.139, 0.194, 0.25, 0.306 | (Accept awrt 3 s.f) | (3 marks) |
| (b) \(P(3) + P(4) + P(5) = \frac{21}{36}\) or \(\frac{7}{12}\) or awrt 0.583 | M1, A1 | (2 marks) |
| (c) \(E(X) = 1 \times \frac{1}{36} + 2 \times \frac{3}{36} + ... = \frac{161}{36}\) or 4.472 or \(4\frac{17}{36}\) | M1, A1 | (2 marks) |
| (d) \(E(X^2) = \frac{1}{36} + 2^2 \times \frac{3}{36} + ... = \frac{791}{36}\) or full expression or \(21\frac{35}{36}\) or awrt 21.97 | M1, A1 | |
| \(\text{Var}(X) = \frac{791}{36} - \left(\frac{161}{36}\right)^2 = \mathbf{1.9714...} *\) | M1, A1 c.s.o. | (4 marks) |
| (e) \(\text{Var}(2 - 3X) = 9 \times 1.97\) or \((-3)^2 \times 1.97 = 17.73\) | M1, A1 | awrt 17.7 or \(\frac{2555}{144}\) |
**(a)** N.B. Part (a) doesn't have to be in a table, could be a list $P(X = 1) = \ldots$ etc | B1, B1, B1 |
| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X = x)$ | $\frac{1}{36}$ | $\frac{3}{36}$ | $\frac{5}{36}$ | $\frac{7}{36}$ | $\frac{9}{36}$ | $\frac{11}{36}$ |
0.0278, 0.0833, 0.139, 0.194, 0.25, 0.306 | (Accept awrt 3 s.f) | (3 marks)
**(b)** $P(3) + P(4) + P(5) = \frac{21}{36}$ or $\frac{7}{12}$ or awrt 0.583 | M1, A1 | (2 marks)
**(c)** $E(X) = 1 \times \frac{1}{36} + 2 \times \frac{3}{36} + ... = \frac{161}{36}$ or 4.472 or $4\frac{17}{36}$ | M1, A1 | (2 marks)
**(d)** $E(X^2) = \frac{1}{36} + 2^2 \times \frac{3}{36} + ... = \frac{791}{36}$ or full expression or $21\frac{35}{36}$ or awrt 21.97 | M1, A1 |
$\text{Var}(X) = \frac{791}{36} - \left(\frac{161}{36}\right)^2 = \mathbf{1.9714...} *$ | M1, A1 c.s.o. | (4 marks)
**(e)** $\text{Var}(2 - 3X) = 9 \times 1.97$ or $(-3)^2 \times 1.97 = 17.73$ | M1, A1 | awrt **17.7** or $\frac{2555}{144}$ | (2 marks)
**Total: 13 marks**
---\begin{enumerate}
\item The random variable $X$ has probability function
\end{enumerate}
$$\mathrm { P } ( X = x ) = \frac { ( 2 x - 1 ) } { 36 } \quad x = 1,2,3,4,5,6$$
\begin{enumerate}[label=(\alph*)]
\item Construct a table giving the probability distribution of $X$.
Find
\item $\mathrm { P } ( 2 < X \leqslant 5 )$,
\item the exact value of $\mathrm { E } ( X )$.
\item Show that $\operatorname { Var } ( X ) = 1.97$ to 3 significant figures.
\item Find $\operatorname { Var } ( 2 - 3 X )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2007 Q3 [13]}}