| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Uniform Distribution |
| Type | Name the distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic discrete probability distributions. Part (a) requires simple recognition of a discrete uniform distribution, parts (b) and (c) involve routine calculations of expectation and variance using standard formulas, part (d) applies linear transformation properties (E(aX+b) and Var(aX+b)), and part (e) requires finding when transformed values match. All techniques are standard bookwork with no problem-solving insight required, making it easier than average but not trivial due to the multi-part nature and arithmetic involved. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(y\) | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( Y = y )\) | \(\frac { 4 } { 30 }\) | \(\frac { 9 } { 30 }\) | \(\frac { 6 } { 30 }\) | \(\frac { 5 } { 30 }\) | \(\frac { 6 } { 30 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (Discrete) uniform (distribution) | B1 | B0 if "continuous uniform" stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([\) By symmetry \(]\ E(X) = \mathbf{13}\) | B1 | B1 for 13 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{10^2+12^2+14^2+16^2}{4} - 13^2\) or \(\frac{696}{4}-169\) or \(174-169\) | M1 | Fully correct expression; may use \(E(X-\mu)^2 = \frac{3^2\times2+1^2\times2}{4}\) |
| \(= \mathbf{5}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(Y) = \frac{1}{30}(1\times4 + 2\times9 + 3\times6 + 4\times5 + 5\times6) = \frac{90}{30} = \mathbf{3}\) | M1;A1 | M1 for attempt at \(E(Y)\) with at least 3 correct products |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(Y^2) = \frac{1}{30}(1^2\times4 + 2^2\times9 + 3^2\times6 + 4^2\times5 + 5^2\times6) = \left[\frac{324}{30} \text{ or } 10.8\right]\) | M1 | 1st M1 for attempt at \(E(Y^2)\) with at least 3 correct products or 10.8 o.e. |
| \(\text{Var}(Y) = \text{"10.8"} - \text{"[3]"}^2 = \mathbf{1.8}\) | M1;A1 | 2nd M1 for correct \(\text{Var}(Y)\) expression (ft 10.8 and 3). NB \(\text{Var}(Y)=...=10.8\) scores M1M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(W)=E(Y) \implies aE(X)+b = E(W)\) or \(E(Y)\) or "3"; i.e. \(\text{"13"}a+b=\text{"3"}\) | M1;A1ft | 1st M1 for correct use of \(E(aX+b)\) formula. 1st A1ft for correct equation in \(a\) and \(b\) |
| \(\text{Var}(W)=\text{Var}(Y) \implies a^2\times\text{"5"}=\text{"1.8"}\); so \(a=\frac{3}{5}\) or 0.6 | M1;A1 | 2nd M1 for correct use of \(\text{Var}(aX+b)\) formula. 2nd A1 for \(a=0.6\) |
| \(b = \mathbf{-4.8}\) | A1 | 3rd A1 for \(b=-4.8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Values of \(w\) are: \(10\times\text{"0.6"}-\text{"4.8"} = 1.2\) or \(2.4\) or \(3.6\) or \(4.8\), i.e. all non integers | M1 | M1 for clear attempt to find all possible values of \(w\) (ft their \(a\) and \(b\)); \(w\) values needn't be correct, or state no integer values for \(w\) |
| \([\text{So no cases are possible when } W=Y \text{ so}]\ P(W=Y) = \mathbf{0}\) | A1 | A1 for answer of 0 provided it's true for their \(a\) and \(b\) |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (Discrete) uniform (distribution) | B1 | B0 if "continuous uniform" stated |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[$ By symmetry $]\ E(X) = \mathbf{13}$ | B1 | B1 for 13 |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{10^2+12^2+14^2+16^2}{4} - 13^2$ or $\frac{696}{4}-169$ or $174-169$ | M1 | Fully correct expression; may use $E(X-\mu)^2 = \frac{3^2\times2+1^2\times2}{4}$ |
| $= \mathbf{5}$ | A1 | |
## Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(Y) = \frac{1}{30}(1\times4 + 2\times9 + 3\times6 + 4\times5 + 5\times6) = \frac{90}{30} = \mathbf{3}$ | M1;A1 | M1 for attempt at $E(Y)$ with at least 3 correct products |
## Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(Y^2) = \frac{1}{30}(1^2\times4 + 2^2\times9 + 3^2\times6 + 4^2\times5 + 5^2\times6) = \left[\frac{324}{30} \text{ or } 10.8\right]$ | M1 | 1st M1 for attempt at $E(Y^2)$ with at least 3 correct products or 10.8 o.e. |
| $\text{Var}(Y) = \text{"10.8"} - \text{"[3]"}^2 = \mathbf{1.8}$ | M1;A1 | 2nd M1 for correct $\text{Var}(Y)$ expression (ft 10.8 and 3). NB $\text{Var}(Y)=...=10.8$ scores M1M0 |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(W)=E(Y) \implies aE(X)+b = E(W)$ or $E(Y)$ or "3"; i.e. $\text{"13"}a+b=\text{"3"}$ | M1;A1ft | 1st M1 for correct use of $E(aX+b)$ formula. 1st A1ft for correct equation in $a$ and $b$ |
| $\text{Var}(W)=\text{Var}(Y) \implies a^2\times\text{"5"}=\text{"1.8"}$; so $a=\frac{3}{5}$ or **0.6** | M1;A1 | 2nd M1 for correct use of $\text{Var}(aX+b)$ formula. 2nd A1 for $a=0.6$ |
| $b = \mathbf{-4.8}$ | A1 | 3rd A1 for $b=-4.8$ |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Values of $w$ are: $10\times\text{"0.6"}-\text{"4.8"} = 1.2$ or $2.4$ or $3.6$ or $4.8$, i.e. all non integers | M1 | M1 for clear attempt to find all possible values of $w$ (ft their $a$ and $b$); $w$ values needn't be correct, or state no integer values for $w$ |
| $[\text{So no cases are possible when } W=Y \text{ so}]\ P(W=Y) = \mathbf{0}$ | A1 | A1 for answer of 0 provided it's true for their $a$ and $b$ |
---4. A spinner can land on the numbers $10,12,14$ and 16 only and the probability of the spinner landing on each number is the same.\\
The random variable $X$ represents the number that the spinner lands on when it is spun once.
\begin{enumerate}[label=(\alph*)]
\item State the name of the probability distribution of $X$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the value of $\mathrm { E } ( X )$
\item Find $\operatorname { Var } ( X )$
A second spinner can land on the numbers 1, 2, 3, 4 and 5 only. The random variable $Y$ represents the number that this spinner lands on when it is spun once. The probability distribution of $Y$ is given in the table below
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$y$ & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( Y = y )$ & $\frac { 4 } { 30 }$ & $\frac { 9 } { 30 }$ & $\frac { 6 } { 30 }$ & $\frac { 5 } { 30 }$ & $\frac { 6 } { 30 }$ \\
\hline
\end{tabular}
\end{center}
\end{enumerate}\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( Y )$
\item $\operatorname { Var } ( Y )$
The random variable $W = a X + b$, where $a$ and $b$ are constants and $a > 0$ Given that $\mathrm { E } ( W ) = \mathrm { E } ( Y )$ and $\operatorname { Var } ( W ) = \operatorname { Var } ( Y )$
\end{enumerate}\item find the value of $a$ and the value of $b$.
Each of the two spinners is spun once.
\item Find $\mathrm { P } ( W = Y )$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q4 [16]}}