| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Find cumulative distribution F(x) |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing standard probability distribution calculations. All parts require direct application of formulas (cumulative distribution, expectation, variance, linear transformations) with no problem-solving insight needed. The arithmetic involves simple fractions, and part (d)(iii) requires only substitution and comparison of values already computed. |
| Spec | 5.02a 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 | 1 | 3 | 4 | 6 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 1 } { 4 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 3 }\) | \(\frac { 1 } { 12 }\) | \(\frac { 1 } { 6 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F(5) = \dfrac{5}{6}\) | B1 | For \(\dfrac{5}{6}\) or exact equivalents e.g. \(\dfrac{10}{12}\) or \(0.8\overline{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = -2\times\dfrac{1}{4}+1\times\dfrac{1}{6}+3\times\dfrac{1}{3}+4\times\dfrac{1}{12}+6\times\dfrac{1}{6}\) or \(\dfrac{1}{12}(-6+2+12+4+12)\) | M1 | For attempt at \(E(X)\) with at least 3 correct products seen. Answer only M1A1. Full method leading to answer e.g. \(\div\) by 5 (or \(n\)) is usually M0 |
| \(= 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X^2) = (-2)^2\times\dfrac{1}{4}+1\times\dfrac{1}{6}+3^2\times\dfrac{1}{3}+4^2\times\dfrac{1}{12}+6^2\times\dfrac{1}{6}\) | M1 | 1st M1: attempt at \(E(X^2)\) with at least 3 correct products seen (even if labelled \(\text{Var}(X)\)). Condone \(-2^2\) if it later becomes \(+4\) but only 3 correct products needed for M1 |
| \(\dfrac{1}{12}(12+2+36+16+72)\) or \(\dfrac{138}{12}\) or \(\dfrac{23}{2}\) | ||
| \(\text{Var}(X) = \dfrac{23}{2} - 2^2\) | M1 | 2nd M1: correct numerical expression for \(\text{Var}(X)\) ft their \(E(X)\) and \(E(X^2)\) |
| \(= 7.5\) | A1 | For 7.5 or exact equivalent e.g. \(\dfrac{15}{2}\). Answer only M1M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(Y) = 7 - 2E(X) = 3\) | B1 | For 3 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Var}(Y) = (-2)^2\,\text{Var}(X)\) or \(4\,\text{Var}(X)\) | M1 | Correct use of \(\text{Var}(aX+b)\) formula. ft their value of \(\text{Var}(X)\) even if \(< 0\) |
| \(= 30\) | A1 | For 30 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(7 - 2X > X \Rightarrow 7 > 3X\) | M1 | For attempt at solving a correct inequality as far as \(a > bx\) (\(a\) and \(b\) both \(> 0\)) |
| so \(X = 1\) or \(X = -2\) | A1 | For identifying \(X=1\) and \(X=-2\) as required values (or \(Y=11\) and \(5\)) |
| \(P(Y > X) = \dfrac{5}{12}\) | A1 | For \(\dfrac{5}{12}\) or exact equivalent |
# Question 1:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $F(5) = \dfrac{5}{6}$ | B1 | For $\dfrac{5}{6}$ or exact equivalents e.g. $\dfrac{10}{12}$ or $0.8\overline{3}$ |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = -2\times\dfrac{1}{4}+1\times\dfrac{1}{6}+3\times\dfrac{1}{3}+4\times\dfrac{1}{12}+6\times\dfrac{1}{6}$ or $\dfrac{1}{12}(-6+2+12+4+12)$ | M1 | For attempt at $E(X)$ with at least 3 correct products seen. Answer only M1A1. Full method leading to answer e.g. $\div$ by 5 (or $n$) is usually M0 |
| $= 2$ | A1 | |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X^2) = (-2)^2\times\dfrac{1}{4}+1\times\dfrac{1}{6}+3^2\times\dfrac{1}{3}+4^2\times\dfrac{1}{12}+6^2\times\dfrac{1}{6}$ | M1 | 1st M1: attempt at $E(X^2)$ with at least 3 correct products seen (even if labelled $\text{Var}(X)$). Condone $-2^2$ if it later becomes $+4$ but only 3 correct products needed for M1 |
| $\dfrac{1}{12}(12+2+36+16+72)$ or $\dfrac{138}{12}$ or $\dfrac{23}{2}$ | | |
| $\text{Var}(X) = \dfrac{23}{2} - 2^2$ | M1 | 2nd M1: correct numerical expression for $\text{Var}(X)$ ft their $E(X)$ and $E(X^2)$ |
| $= 7.5$ | A1 | For 7.5 or exact equivalent e.g. $\dfrac{15}{2}$. Answer only M1M1A1 |
## Part (d)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(Y) = 7 - 2E(X) = 3$ | B1 | For 3 only |
## Part (d)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Var}(Y) = (-2)^2\,\text{Var}(X)$ or $4\,\text{Var}(X)$ | M1 | Correct use of $\text{Var}(aX+b)$ formula. ft their value of $\text{Var}(X)$ even if $< 0$ |
| $= 30$ | A1 | For 30 only |
## Part (d)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $7 - 2X > X \Rightarrow 7 > 3X$ | M1 | For attempt at solving a correct inequality as far as $a > bx$ ($a$ and $b$ both $> 0$) |
| so $X = 1$ or $X = -2$ | A1 | For identifying $X=1$ and $X=-2$ as required values (or $Y=11$ and $5$) |
| $P(Y > X) = \dfrac{5}{12}$ | A1 | For $\dfrac{5}{12}$ or exact equivalent |\begin{enumerate}
\item The discrete random variable $X$ has the probability distribution given in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 2 & 1 & 3 & 4 & 6 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 1 } { 4 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 3 }$ & $\frac { 1 } { 12 }$ & $\frac { 1 } { 6 }$ \\
\hline
\end{tabular}
\end{center}
(a) Write down the value of $\mathrm { F } ( 5 )$\\
(b) Find $\mathrm { E } ( X )$\\
(c) Find $\operatorname { Var } ( X )$
The random variable $Y = 7 - 2 X$\\
(d) Find\\
(i) $\mathrm { E } ( Y )$\\
(ii) $\operatorname { Var } ( Y )$\\
(iii) $\mathrm { P } ( Y > X )$
\includegraphics[max width=\textwidth, alt={}, center]{70137e9a-0a6b-48b5-8dd4-c436cb063351-03_2261_47_313_37}\\
\hfill \mbox{\textit{Edexcel S1 2016 Q1 [12]}}