| Exam Board | Edexcel |
|---|---|
| Module | FS1 (Further Statistics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.3 This is a standard Further Stats 1 question requiring systematic application of probability axioms and expectation properties. Part (a) involves solving simultaneous equations from given conditions (sum=1, E(4N+2)=14.8, conditional probability), then computing variance using standard formulas. Part (b) is straightforward expectation calculation with a piecewise fee structure. Part (c) tests conceptual understanding. While it has multiple parts and requires careful algebra, it follows predictable FS1 patterns with no novel problem-solving required. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p) |
| \(n\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( N = n )\) | \(a\) | 0.2 | 0.05 | 0.25 | \(b\) | \(c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4E(N) + 2 = 14.8\) or \(E(N) = 3.2\) | M1 | 3.1a - For using given information to find \(E(N)\). ALT: \(a+b+c=0.5\) oe |
| \(0.2+0.1+0.75+4b+5c = 3.2\) | M1 | 1.1b - For use of \(\sum nP(N=n) = 3.2\), at least 3 terms correct |
| \(\dfrac{c}{0.25+b+c} = 0.5\) or \(0.25 = c-b\) | M1 | 3.1a - Forming equation in \(b\) and \(c\) using conditional probability |
| \(b = 0.1\) and \(c = 0.35\) | ||
| \(E(N^2) = 1\times0.2 + 4\times0.05 + 9\times0.25 + 16\times0.1 + 25\times0.35\) [=13] | M1 | 1.1b - For using \(\sum n^2 P(N=n)\). Allow with letters \(b\) and \(c\) |
| \(\text{Var}(N) = 13 - 3.2^2\) | dM1 | 1.1b - Dependent on previous M mark. Correct method to find \(\text{Var}(N)\) |
| \(= 2.76\) | A1* | 2.1 - All previous marks must be awarded and 2.76 stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Fee table: \(0, 50, 70, 90, 100, 100\) with \(P(N=n)\): \(a, 0.2, 0.05, 0.25, b, c\) | M1 | 3.3 - Setting up new model with correct fees. At least 3 terms correct. Allow 0.5, 0.7, 0.9, 1 |
| \(50\times0.2 + 70\times0.05 + 90\times0.25 + 100\times0.1 + 100\times0.35\) | M1 | 1.1b - Correct method for calculating \(E(\text{fee})\), allow with letters \(b\) and \(c\) |
| \(= 81p\) | A1 | 1.1b - 81[p]. No units needed. Allow 0.81 if fees in pounds |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Poisson distribution will assign substantial probability to \(N > 5\) | B1 | 3.5b - A correct limitation |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4E(N) + 2 = 14.8$ or $E(N) = 3.2$ | M1 | 3.1a - For using given information to find $E(N)$. ALT: $a+b+c=0.5$ oe |
| $0.2+0.1+0.75+4b+5c = 3.2$ | M1 | 1.1b - For use of $\sum nP(N=n) = 3.2$, at least 3 terms correct |
| $\dfrac{c}{0.25+b+c} = 0.5$ or $0.25 = c-b$ | M1 | 3.1a - Forming equation in $b$ and $c$ using conditional probability |
| $b = 0.1$ and $c = 0.35$ | | |
| $E(N^2) = 1\times0.2 + 4\times0.05 + 9\times0.25 + 16\times0.1 + 25\times0.35$ [=13] | M1 | 1.1b - For using $\sum n^2 P(N=n)$. Allow with letters $b$ and $c$ |
| $\text{Var}(N) = 13 - 3.2^2$ | dM1 | 1.1b - Dependent on previous M mark. Correct method to find $\text{Var}(N)$ |
| $= 2.76$ | A1* | 2.1 - All previous marks must be awarded and 2.76 stated |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Fee table: $0, 50, 70, 90, 100, 100$ with $P(N=n)$: $a, 0.2, 0.05, 0.25, b, c$ | M1 | 3.3 - Setting up new model with correct fees. At least 3 terms correct. Allow 0.5, 0.7, 0.9, 1 |
| $50\times0.2 + 70\times0.05 + 90\times0.25 + 100\times0.1 + 100\times0.35$ | M1 | 1.1b - Correct method for calculating $E(\text{fee})$, allow with letters $b$ and $c$ |
| $= 81p$ | A1 | 1.1b - 81[p]. No units needed. Allow 0.81 if fees in pounds |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Poisson distribution will assign substantial probability to $N > 5$ | B1 | 3.5b - A correct limitation |
---\begin{enumerate}
\item Members of a photographic group may enter a maximum of 5 photographs into a members only competition.\\
Past experience has shown that the number of photographs, $N$, entered by a member follows the probability distribution shown below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$n$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( N = n )$ & $a$ & 0.2 & 0.05 & 0.25 & $b$ & $c$ \\
\hline
\end{tabular}
\end{center}
Given that $\mathrm { E } ( 4 N + 2 ) = 14.8$ and $\mathrm { P } ( N = 5 \mid N > 2 ) = \frac { 1 } { 2 }$\\
(a) show that $\operatorname { Var } ( N ) = 2.76$
The group decided to charge a 50p entry fee for the first photograph entered and then 20p for each extra photograph entered into the competition up to a maximum of $\pounds 1$ per person. Thus a member who enters 3 photographs pays 90 p and a member who enters 4 or 5 photographs just pays £1
Assuming that the probability distribution for the number of photographs entered by a member is unchanged,\\
(b) calculate the expected entry fee per member.
Bai suggests that, as the mean and variance are close, a Poisson distribution could be used to model the number of photographs entered by a member next year.\\
(c) State a limitation of the Poisson distribution in this case.
\hfill \mbox{\textit{Edexcel FS1 2021 Q4 [10]}}