| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.8 This is a Further Maths statistics question requiring students to set up and solve simultaneous equations using the constraint that probabilities sum to 1, plus the condition E(V) = Var(V). While the individual calculations are standard, combining these constraints to find two unknowns requires careful algebraic manipulation and understanding of variance formula, making it moderately challenging but still within typical Further Maths scope. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(v\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( \mathrm { V } = \mathrm { v } )\) | \(p\) | \(q\) | 0.12 | 0.2 |
| Answer | Marks |
|---|---|
| 3 | DR E(V) = q + 0.24 + 0.6 |
| Answer | Marks |
|---|---|
| p = 1 – 0.32 – q = 0.32 | B1 |
| Answer | Marks |
|---|---|
| [8] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | Use E(V 2) – [E(V)]2 |
| Answer | Marks |
|---|---|
| SC: E(V 2) – E(V): (B1B1) q = 0.6 B1, p = 0.08 B1 | Or: E(V – E(V))2: M1 |
Question 3:
3 | DR E(V) = q + 0.24 + 0.6
E(V 2) = q + 0.48 + 1.8
q + 0.84 = q + 2.28 – (q + 0.84)2
(q + 0.84)2 = 1.44
so q = –0.84 1.2
Reject q = –2.04 as negative
q = 0.36
p = 1 – 0.32 – q = 0.32 | B1
B1
M1
A1
M1
B1
A1
B1ft
[8] | 3.1a
1.1
2.1
1.1
1.1
2.3
2.2a
1.1 | Use E(V 2) – [E(V)]2
Correct simplified quad, eg q2 + 1.68q – 0.7344 = 0
Solve quadratic, valid method seen or implied
Explicitly reject invalid solution, with reason
0.36 or exact equivalent seen, e.g. 9/25
p = 0.68 – their q. Withhold if extra solutions seen
SC: E(V 2) – E(V): (B1B1) q = 0.6 B1, p = 0.08 B1 | Or: E(V – E(V))2: M1
then eliminate p: M1
Or 625q2 + 1050q – 459 = 0
Can be implied by one correct root
Or reason to take + root of 1.44
E.g. 8/25
Use Poisson: 0/83 In this question you must show detailed reasoning.
A discrete random variable $V$ has the following probability distribution, where $p$ and $q$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$v$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( \mathrm { V } = \mathrm { v } )$ & $p$ & $q$ & 0.12 & 0.2 \\
\hline
\end{tabular}
\end{center}
It is given that $\mathrm { E } ( V ) = \operatorname { Var } ( V )$.
Determine the value of $p$ and the value of $q$.
\hfill \mbox{\textit{OCR Further Statistics 2022 Q3 [8]}}