| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.3 This is a standard S1 question requiring systematic application of expectation and variance formulas with symmetry recognition. The algebra is straightforward (linear equations from E(X) and Var(X)), and parts (c)-(d) use routine transformations. Slightly above average due to multiple unknowns and several parts, but all techniques are textbook exercises with no novel insight required. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | - 2 | - 1 | 0 | 1 | 4 |
| \(\mathrm { P } ( X = x )\) | \(a\) | \(b\) | \(c\) | \(b\) | \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{E(X) =\}\ -2a - b + 0\times c + b + 4a\) or \(2a\) | M1 | For any correct expression for \(E(X)\) in terms of \(a\) (or \(a, b, c\)) |
| \(\{2a = 0.5\text{ so}\}\ \mathbf{a = 0.25}\) | A1 | For \(a = 0.25\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{E(X^2) =\}\ (-2)^2\times a + (-1)^2\times b + 0 + 1^2\times b + 4^2\times a\) or \(20a + 2b\) | M1 | For attempt at an expression for \(E(X^2)\) with at least 3 correct non-zero terms |
| \(\{\text{Var}(X) =\}\ \text{"20}a + 2b\text{"} - 0.5^2\) | M1 | For a correct expression for \(\text{Var}(X)\) eg "18\(a-c+1\)"\(-0.5^2\). Allow with their value of \(a\) substituted |
| \(20a + 2b - 0.25 = 5.01\) eg "4.75" \(+ 2b = 5.01\) | A1 | For a correct equation for \(b\) (or possibly \(c\)) eg "18\(a-c+1\)"\(-0.5^2 = 5.01\). Allow with their value of \(a\) substituted |
| \(\{2b = 0.26\text{ so}\}\ \mathbf{b = 0.13}\) | A1 | For either \(b = 0.13\) or \(c = 0.24\) |
| \(\{\)Use of sum of probs \(= 1\) to calculate a 2nd value\(\}\ \mathbf{c = 0.24}\) | A1ft | For using \(c = 1 - 2\times\text{"0.25"} - 2\times\text{"0.13"}\) or \(b = (1-2\times\text{"0.25"}-\text{"0.24"})\div 2\) to gain the correct ft answer for their 2nd value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{E(Y) = 5 - 8\times 0.5\} = \mathbf{1}\) | B1 | For \(\{E(Y) =\}\ 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\text{Var}(Y) =\}\ (-8)^2\times 5.01\) | M1 | For correct use of \(\text{Var}(aX+b) = a^2\,\text{Var}(X)\) |
| \(= 320.64\) awrt \(\mathbf{321}\) | A1 | For awrt 321 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4X^2 > 5 - 8X\) | M1 | For correct quadratic inequality (may be inside prob statement) or table of values |
| \((2X-1)(2X+5) > 0\ \Rightarrow\ X > 0.5\) | M1A1 | 2nd M1 for attempt to solve or identifying correct \(X\) values. A1 for \(X > 0.5\) [may also have \(X < -2.5\)] |
| So need \(X = 1\) or \(4\) or probability of \(a + b\) | M1 | For realising need \(X = 1\) and 4 only or answer of their \((a+b)\) |
| \(= \mathbf{0.38}\) | A1 | For 0.38 (or exact equivalent) only (correct answer only 5/5) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{E(X) =\}\ -2a - b + 0\times c + b + 4a$ or $2a$ | M1 | For any correct expression for $E(X)$ in terms of $a$ (or $a, b, c$) |
| $\{2a = 0.5\text{ so}\}\ \mathbf{a = 0.25}$ | A1 | For $a = 0.25$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{E(X^2) =\}\ (-2)^2\times a + (-1)^2\times b + 0 + 1^2\times b + 4^2\times a$ or $20a + 2b$ | M1 | For attempt at an expression for $E(X^2)$ with at least 3 correct non-zero terms |
| $\{\text{Var}(X) =\}\ \text{"20}a + 2b\text{"} - 0.5^2$ | M1 | For a correct expression for $\text{Var}(X)$ eg "18$a-c+1$"$-0.5^2$. Allow with their value of $a$ substituted |
| $20a + 2b - 0.25 = 5.01$ eg "4.75" $+ 2b = 5.01$ | A1 | For a correct equation for $b$ (or possibly $c$) eg "18$a-c+1$"$-0.5^2 = 5.01$. Allow with their value of $a$ substituted |
| $\{2b = 0.26\text{ so}\}\ \mathbf{b = 0.13}$ | A1 | For either $b = 0.13$ or $c = 0.24$ |
| $\{$Use of sum of probs $= 1$ to calculate a 2nd value$\}\ \mathbf{c = 0.24}$ | A1ft | For using $c = 1 - 2\times\text{"0.25"} - 2\times\text{"0.13"}$ or $b = (1-2\times\text{"0.25"}-\text{"0.24"})\div 2$ to gain the correct ft answer for their 2nd value |
## Part (c)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{E(Y) = 5 - 8\times 0.5\} = \mathbf{1}$ | B1 | For $\{E(Y) =\}\ 1$ |
## Part (c)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\text{Var}(Y) =\}\ (-8)^2\times 5.01$ | M1 | For correct use of $\text{Var}(aX+b) = a^2\,\text{Var}(X)$ |
| $= 320.64$ awrt $\mathbf{321}$ | A1 | For awrt 321 |
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4X^2 > 5 - 8X$ | M1 | For correct quadratic inequality (may be inside prob statement) or table of values |
| $(2X-1)(2X+5) > 0\ \Rightarrow\ X > 0.5$ | M1A1 | 2nd M1 for attempt to solve or identifying correct $X$ values. A1 for $X > 0.5$ [may also have $X < -2.5$] |
| So need $X = 1$ or $4$ or probability of $a + b$ | M1 | For realising need $X = 1$ and 4 only or answer of their $(a+b)$ |
| $= \mathbf{0.38}$ | A1 | For 0.38 (or exact equivalent) only (correct answer only 5/5) |\begin{enumerate}
\item The discrete random variable $X$ has the following probability distribution
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 2 & - 1 & 0 & 1 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $a$ & $b$ & $c$ & $b$ & $a$ \\
\hline
\end{tabular}
\end{center}
Given that $\mathrm { E } ( X ) = 0.5$\\
(a) find the value of $a$.
Given also that $\operatorname { Var } ( X ) = 5.01$\\
(b) find the value of $b$ and the value of $c$.
The random variable $Y = 5 - 8 X$\\
(c) Find\\
(i) $\mathrm { E } ( Y )$\\
(ii) $\operatorname { Var } ( Y )$\\
(d) Find $\mathrm { P } \left( 4 X ^ { 2 } > Y \right)$
\hfill \mbox{\textit{Edexcel S1 2021 Q5 [15]}}