| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Moderate -0.8 This is a straightforward S1 question on discrete probability distributions. Part (i) requires simple logical reasoning (if 3 are correct, the 4th must be too), part (ii) is basic probability calculation (1/4! = 1/24), and part (iii) involves standard application of E(X) and Var(X) formulas with given probabilities. All techniques are routine for S1 with no problem-solving insight required. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | |||
| P(X = r) | \(\frac{3}{8}\) | \(\frac{1}{3}\) | \(\frac{1}{4}\) | 0 | \(\frac{1}{24}\) |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Impossible because if 3 letters are correct, the | |
| fourth must be also. | E1 | 1 |
| (ii) | There is only one way to place letters correctly. |
| Answer | Marks |
|---|---|
| 4 3 2 | E1 |
| Answer | Marks |
|---|---|
| 4 3 2 | 2 |
| (iii) | 1 1 1 |
| Answer | Marks |
|---|---|
| = 1 | ∑ |
| Answer | Marks |
|---|---|
| Var( X ) > 0 | 5 |
| TOTAL | 8 |
Question 1:
--- 1
(i) ---
1
(i) | Impossible because if 3 letters are correct, the
fourth must be also. | E1 | 1
(ii) | There is only one way to place letters correctly.
There are 4! = 24 ways to arrange 4 letters.
OR:
1 1 1
x x NOTE: ANSWER GIVEN
4 3 2 | E1
E1
1 1 1
B1 for x B1 for x
4 3 2 | 2
(iii) | 1 1 1
E( X ) = 1 x + 2 x + 4 x = 1
3 4 24
1 1 1
E( X² ) = 1 x + 4 x + 16 x = 2
3 4 24
So Var( X ) = 2 – 12
= 1 | ∑
M1 For xp(at least 2 non-
zero terms correct)
A1 CAO
M1 for ∑x2p (at least 2 non-
zero terms correct)
M1dep for – their E( X )²
A1 FT their E(X) provided
Var( X ) > 0 | 5
TOTAL | 8Four letters are taken out of their envelopes for signing. Unfortunately they are replaced randomly, one in each envelope.
The probability distribution for the number of letters, $X$, which are now in the correct envelope is given in the following table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$r$ & 0 & 1 & & & \\
\hline
P(X = r) & $\frac{3}{8}$ & $\frac{1}{3}$ & $\frac{1}{4}$ & 0 & $\frac{1}{24}$ \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Explain why the case $X = 3$ is impossible. [1]
\item Explain why P($X = 4$) = $\frac{1}{24}$. [2]
\item Calculate E($X$) and Var($X$). [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q1 [8]}}