| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.3 This is a systematic but straightforward problem requiring three probability equations (sum=1, E(X)=1.25, Var(X)=0.8875) to solve for three unknowns. Part (b) uses the standard variance transformation rule Var(aX+b)=a²Var(X), and part (c) recognizes a binomial distribution. All techniques are standard Further Statistics content with no novel insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( \mathrm { X } = \mathrm { x } )\) | \(a\) | \(b\) | \(c\) | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | b + 2c + 0.3 = 1.25 |
| Answer | Marks |
|---|---|
| b + 4c + 0.9 = 2.45 | M1 |
| M1 | 1.1 |
| 1.1 | M1 |
| M1 | Use Var(X) = x2p – 1.252 (not – 1.25 or +1.252) |
| Answer | Marks | Guidance |
|---|---|---|
| OR | (–1.25)2a + 0.252b + 0.752c + 0.1×1.752 = 0.8875 | M1 |
| 1.5626a + 0.0625b + 0.5625c = 0.58125 | 1.5626a + 0.0625b + 0.5625c = 0.58125 | A1 |
| Answer | Marks |
|---|---|
| a = 0.25 | A1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| 1.1 | A1 |
| Answer | Marks |
|---|---|
| [5] | Both correct, exact only (e.g. 72 , 31 ) |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 4Var(X) = 3.55 | B1 |
| [1] | 1.1 | B1 |
| [1] | 311 = 71 |
| Answer | Marks |
|---|---|
| (c) | B(20, 0.1) |
| npq = 1.8 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | M1 |
| Answer | Marks |
|---|---|
| [2] | Stated or implied as main method, e.g. by correct answer or |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 1:
1 | (a) | b + 2c + 0.3 = 1.25 | M1 | 1.1 | M1 | Use xp = 1.25, e.g. b + 2c = 0.95
E(X 2) = 0.8875 + 1.252 (= 2.45)
b + 4c + 0.9 = 2.45 | M1
M1 | 1.1
1.1 | M1
M1 | Use Var(X) = x2p – 1.252 (not – 1.25 or +1.252)
Use x2p = their 2.45 to obtain equation,
e.g. b + 4c = constant (1.55), allow from –1.25 or +1.252
OR | (–1.25)2a + 0.252b + 0.752c + 0.1×1.752 = 0.8875 | M1 | M1 | Clear use of Σ(x – 1.25)2p(x)
1.5626a + 0.0625b + 0.5625c = 0.58125 | 1.5626a + 0.0625b + 0.5625c = 0.58125 | A1 | A1 | A1 | A1 | 25a+ 1 b+ 9 c= 93
AEF, e.g. .
16 16 16 160
Can be implied by correct answers
b = 0.35, c = 0.3
a = 0.25 | A1
B1ft
[5] | 1.1
1.1 | A1
B1ft
[5] | Both correct, exact only (e.g. 72 , 31 )
0 0
0.9 – their (b + c), needs 0 a 0.9.
(b) | 4Var(X) = 3.55 | B1
[1] | 1.1 | B1
[1] | 311 = 71
3.55 or , exact
20 20
(c) | B(20, 0.1)
npq = 1.8 | M1
A1
[2] | 3.3
1.1 | M1
A1
[2] | Stated or implied as main method, e.g. by correct answer or
by np = 2 with no wrong working anywhere, not isw
1.8 or exact equivalent only, as final answer
(17.75 etc is M0A0 even if 0.1×20 = 2 seen)
Question | Answer | Marks | AO | Guidance1 A discrete random variable $X$ has the following distribution, where $a , b$ and $c$ are constants.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { x } )$ & $a$ & $b$ & $c$ & 0.1 \\
\hline
\end{tabular}
\end{center}
It is given that $\mathrm { E } ( X ) = 1.25$ and $\operatorname { Var } ( X ) = 0.8875$.
\begin{enumerate}[label=(\alph*)]
\item Determine the values of $a$, $b$ and $c$.
\item The random variable $Y$ is defined by $Y = 7 - 2 X$.
Write down the value of $\operatorname { Var } ( Y )$.
\item Twenty independent observations of $X$ are obtained. The number of those observations for which $X = 3$ is denoted by $T$.
Find the value of $\operatorname { Var } ( T )$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2024 Q1 [8]}}