| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Moderate -0.3 This is a standard S1 question testing routine calculations with discrete probability distributions. Parts (a)-(d) involve straightforward E(X) and Var(X) formulas with simple arithmetic. Parts (e)-(g) require basic normal distribution understanding and probability calculations, but follow predictable patterns. The multi-part structure adds length rather than conceptual difficulty, making it slightly easier than average overall. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution |
| \(a\) | 1 | 4 | 5 | 7 |
| \(\mathrm { P } ( A = a )\) | 0.40 | 0.20 | 0.25 | 0.15 |
| \(b\) | 1 | 3 | 4 | \(k\) |
| \(\mathrm { P } ( B = b )\) | 0.25 | 0.25 | 0.25 | 0.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(E(A) = 1\times0.4 + 4\times0.2 + 5\times0.25 + 7\times0.15\) | M1 | At least 3 correct products |
| \(= \mathbf{3.5}\) | A1cso | 3.5 with no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(E(A^2) = 1\times0.4 + 16\times0.2 + 25\times0.25 + 49\times0.15\ [=17.2]\) | M1 | At least 3 correct products |
| \(\text{Var}(A) = E(A^2) - [E(A)]^2 = 17.2 - 3.5^2\) | M1 | Use of \(E(A^2)-[E(A)]^2\) ft their \(E(A^2)\) |
| \(= \mathbf{4.95}\) | A1 | Or exact equivalent e.g. \(\frac{99}{20}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| (Discrete) uniform distribution | B1 | Continuous uniform is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| By symmetry \(k=6\) | B1 | Stating \(k=6\) with suitable reason e.g. symmetry or full calculation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Sam has \(Z = \frac{3.5-4}{3} = -\frac{1}{6}\) and Tim needs \(\frac{3.5-A}{4} < -\frac{1}{6}\) so \(A > 4.166...\) | M1 | Suitable calculation for \(A\); or stating \(A=5\) or \(7\) or \(A >\) awrt 4.2 |
| So prob \(= 0.25 + 0.15 = \mathbf{0.4}\) | A1 | Must be based on correct calculation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Largest possible \(\mu=7\) and smallest possible \(\sigma=1\) | B1, B1 | Each may be implied from standardisation with 3.5 |
| \(P(X>3.5)\) is then \(P\left(Z > \frac{3.5-7}{1}\right) = P(Z > -3.5)\) | M1 | Attempting correct probability; ft standardisation using 3.5, \(\mu\neq4\), \(\sigma\neq3\) |
| \(= \mathbf{0.9998}\) (tables) or \(0.999767...\) (calc) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(P(A=7)\times P(B=1) = ``0.15"\times0.25\) | M1 | For \(``0.15"\times0.25\) ft their value of \(A\) from (f) |
| \(= \mathbf{0.0375}\) | A1cso | \(= \frac{3}{80}\); must clearly come from \(A=7\) and \(B=1\) in (f) |
# Question 6:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $E(A) = 1\times0.4 + 4\times0.2 + 5\times0.25 + 7\times0.15$ | M1 | At least 3 correct products |
| $= \mathbf{3.5}$ | A1cso | 3.5 with no incorrect working |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $E(A^2) = 1\times0.4 + 16\times0.2 + 25\times0.25 + 49\times0.15\ [=17.2]$ | M1 | At least 3 correct products |
| $\text{Var}(A) = E(A^2) - [E(A)]^2 = 17.2 - 3.5^2$ | M1 | Use of $E(A^2)-[E(A)]^2$ ft their $E(A^2)$ |
| $= \mathbf{4.95}$ | A1 | Or exact equivalent e.g. $\frac{99}{20}$ |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| (Discrete) uniform distribution | B1 | Continuous uniform is B0 |
## Part (d):
| Working | Marks | Notes |
|---------|-------|-------|
| By symmetry $k=6$ | B1 | Stating $k=6$ with suitable reason e.g. symmetry or full calculation |
## Part (e):
| Working | Marks | Notes |
|---------|-------|-------|
| Sam has $Z = \frac{3.5-4}{3} = -\frac{1}{6}$ and Tim needs $\frac{3.5-A}{4} < -\frac{1}{6}$ so $A > 4.166...$ | M1 | Suitable calculation for $A$; or stating $A=5$ or $7$ or $A >$ awrt 4.2 |
| So prob $= 0.25 + 0.15 = \mathbf{0.4}$ | A1 | Must be based on correct calculation |
## Part (f):
| Working | Marks | Notes |
|---------|-------|-------|
| Largest possible $\mu=7$ and smallest possible $\sigma=1$ | B1, B1 | Each may be implied from standardisation with 3.5 |
| $P(X>3.5)$ is then $P\left(Z > \frac{3.5-7}{1}\right) = P(Z > -3.5)$ | M1 | Attempting correct probability; ft standardisation using 3.5, $\mu\neq4$, $\sigma\neq3$ |
| $= \mathbf{0.9998}$ (tables) or $0.999767...$ (calc) | A1 | |
## Part (g):
| Working | Marks | Notes |
|---------|-------|-------|
| $P(A=7)\times P(B=1) = ``0.15"\times0.25$ | M1 | For $``0.15"\times0.25$ ft their value of $A$ from (f) |
| $= \mathbf{0.0375}$ | A1cso | $= \frac{3}{80}$; must clearly come from $A=7$ and $B=1$ in (f) |6. The random variable $A$ represents the score when a spinner is spun. The probability distribution for $A$ is given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$a$ & 1 & 4 & 5 & 7 \\
\hline
$\mathrm { P } ( A = a )$ & 0.40 & 0.20 & 0.25 & 0.15 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { E } ( A ) = 3.5$
\item Find $\operatorname { Var } ( A )$
The random variable $B$ represents the score on a 4 -sided die. The probability distribution for $B$ is given in the following table where $k$ is a positive integer.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$b$ & 1 & 3 & 4 & $k$ \\
\hline
$\mathrm { P } ( B = b )$ & 0.25 & 0.25 & 0.25 & 0.25 \\
\hline
\end{tabular}
\end{center}
\item Write down the name of the probability distribution of $B$.
\item Given that $\mathrm { E } ( B ) = \mathrm { E } ( A )$ state, giving a reason, the value of $k$.
The random variable $X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)$
Sam and Tim are playing a game with the spinner and the die.
They each spin the spinner once to obtain their value of $A$ and each roll the die once to obtain their value of $B$.\\
Their value of $A$ is taken as their value of $\mu$ and their value of $B$ is taken as their value of $\sigma$. The person with the larger value of $\mathrm { P } ( X > 3.5 )$ is the winner.
\item Given that Sam obtained values of $a = 4$ and $b = 3$ and Tim obtained $b = 4$ find, giving a reason, the probability that Tim wins.
\item Find the largest value of $\mathrm { P } ( X > 3.5 )$ achievable in this game.
\item Find the probability of achieving this value.
\includegraphics[max width=\textwidth, alt={}, center]{81d5e460-9559-4d25-aa08-6440559aec83-21_2255_50_314_34}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2020 Q6 [15]}}