| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probabilities in table form with k |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability distribution properties. Part (a) uses the fact that probabilities sum to 1 (simple algebra), followed by routine calculations of cumulative probability, expectation, and variance using standard formulas. All steps are mechanical applications of definitions with no problem-solving or insight required. |
| Spec | 2.04a Discrete probability distributions5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.03a Continuous random variables: pdf and cdf |
| \(x\) | 2 | 3 | 4 | 6 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 5 } { 21 }\) | \(\frac { 2 k } { 21 }\) | \(\frac { 7 } { 21 }\) | \(\frac { k } { 21 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{5}{21} + \frac{2k}{21} + \frac{7}{21} + \frac{k}{21} = 1\) | M1 | M1 Award for verification. Sub in \(k=3\) and show \(\sum x P(X = x) = 1\). Require at least three correct terms seen or line 2 of scheme. |
| \(\frac{12 + 3k}{21} = 1\) | ||
| \(k = 3\) * AG | A1 | A1 Correct solution only including verification. |
| (2 marks) | ||
| (b) \(\frac{11}{21}\) | B1 | B1 Award for exact equivalent. |
| (1 mark) | ||
| (c) \(E(X) = 2 \times \frac{5}{21} + 3 \times \frac{6}{21} + 4 \times \frac{7}{21} + 6 \times \frac{1}{7}\) | M1 | M1 At least two correct terms required for method, follow through 'their \(k\)' for method. Correct answer only, award M1 A1. |
| \(= 3\frac{11}{21}\) or \(\frac{74}{21}\) or awrt 3.52 | A1 | A1 for \(3\frac{11}{21}\) or \(\frac{74}{21}\) or awrt 3.52 |
| (2 marks) | ||
| (d) \(E(X^2) = 2^2 \times \frac{5}{21} + 3^2 \times \frac{6}{21} + 4^2 \times \frac{7}{21} + 6^2 \times \frac{1}{7}\) | M1 | M1 At least two correct terms required for method. M0 if probability is squared. |
| \(= 14\) | A1 | Correct answer only, award M1 A1. Accept exact equivalent of 14 for A1. |
| (2 marks) | ||
| (e) \(\text{Var}(X) = 14 - \left(3\frac{11}{21}\right)^2\) | M1 | M1 for use of correct formula in both. |
| \(= \frac{257}{441}\) or \(\frac{698}{441}\) or awrt 1.6 | A1 | A1 for \(1.6\) can be implied by correct final answer. Working needs to be clearly labelled to award first method mark without second stage of calculation. |
| \(\text{Var}(7X - 5) = 7^2 \text{Var}(X)\) | M1 | |
| \(= 77\frac{5}{9}\) or \(\frac{698}{9}\) or awrt 77.6 | A1 | |
| (4 marks) |
**(a)** $\frac{5}{21} + \frac{2k}{21} + \frac{7}{21} + \frac{k}{21} = 1$ | M1 | M1 Award for verification. Sub in $k=3$ and show $\sum x P(X = x) = 1$. Require at least three correct terms seen or line 2 of scheme.
| | |
| $\frac{12 + 3k}{21} = 1$ | | |
| | |
| $k = 3$ * AG | A1 | A1 Correct solution only including verification.
| | (2 marks) |
**(b)** $\frac{11}{21}$ | B1 | B1 Award for exact equivalent.
| | (1 mark) |
**(c)** $E(X) = 2 \times \frac{5}{21} + 3 \times \frac{6}{21} + 4 \times \frac{7}{21} + 6 \times \frac{1}{7}$ | M1 | M1 At least two correct terms required for method, follow through 'their $k$' for method. Correct answer only, award M1 A1.
| | |
| $= 3\frac{11}{21}$ or $\frac{74}{21}$ or awrt 3.52 | A1 | A1 for $3\frac{11}{21}$ or $\frac{74}{21}$ or awrt 3.52
| | (2 marks) |
**(d)** $E(X^2) = 2^2 \times \frac{5}{21} + 3^2 \times \frac{6}{21} + 4^2 \times \frac{7}{21} + 6^2 \times \frac{1}{7}$ | M1 | M1 At least two correct terms required for method. M0 if probability is squared.
| | |
| $= 14$ | A1 | Correct answer only, award M1 A1. Accept exact equivalent of 14 for A1.
| | (2 marks) |
**(e)** $\text{Var}(X) = 14 - \left(3\frac{11}{21}\right)^2$ | M1 | M1 for use of correct formula in both.
| | |
| $= \frac{257}{441}$ or $\frac{698}{441}$ or awrt 1.6 | A1 | A1 for $1.6$ can be implied by correct final answer. Working needs to be clearly labelled to award first method mark without second stage of calculation.
| | |
| $\text{Var}(7X - 5) = 7^2 \text{Var}(X)$ | M1 | |
| | |
| $= 77\frac{5}{9}$ or $\frac{698}{9}$ or awrt 77.6 | A1 | |
| | (4 marks) |
**Total: 11 marks**
---3. The discrete random variable $X$ can take only the values $2,3,4$ or 6 . For these values the probability distribution function is given by
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 2 & 3 & 4 & 6 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 5 } { 21 }$ & $\frac { 2 k } { 21 }$ & $\frac { 7 } { 21 }$ & $\frac { k } { 21 }$ \\
\hline
\end{tabular}
\end{center}
where $k$ is a positive integer.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 3$
Find
\item $\mathrm { F } ( 3 )$
\item $\mathrm { E } ( X )$
\item $\mathrm { E } \left( X ^ { 2 } \right)$
\item $\operatorname { Var } ( 7 X - 5 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2012 Q3 [11]}}