| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(X) from table |
| Difficulty | Easy -1.2 This is a straightforward application of standard variance formula to a given probability distribution. Part (i) requires simple verification of E(X) using the definition, and part (ii) is a routine calculation of Var(X) = E(X²) - [E(X)]² with no conceptual challenges—purely mechanical arithmetic with a symmetric distribution that simplifies the work. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| \(r\) | 10 | 20 | 30 | 40 |
| \(\mathrm { P } ( X = r )\) | 0.2 | 0.3 | 0.3 | 0.2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 25\) because the distribution is symmetrical | E1 ANSWER GIVEN | 1 mark; allow correct calculation of \(\Sigma rp\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 10^2 \times 0.2 + 20^2 \times 0.3 + 30^2 \times 0.3 + 40^2 \times 0.2 = 730\) | M1 | for \(\Sigma r^2 p\) (at least 3 terms correct) |
| \(\text{Var}(X) = 730 - 25^2 = 105\) | M1dep, A1 CAO | dep for \(-25^2\); Total: 4 marks |
# Question 4:
## Part (i):
$E(X) = 25$ because the distribution is symmetrical | E1 ANSWER GIVEN | 1 mark; allow correct calculation of $\Sigma rp$
## Part (ii):
$E(X^2) = 10^2 \times 0.2 + 20^2 \times 0.3 + 30^2 \times 0.3 + 40^2 \times 0.2 = 730$ | M1 | for $\Sigma r^2 p$ (at least 3 terms correct)
$\text{Var}(X) = 730 - 25^2 = 105$ | M1dep, A1 CAO | dep for $-25^2$; **Total: 4 marks**
---4 The table shows the probability distribution of the random variable $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 10 & 20 & 30 & 40 \\
\hline
$\mathrm { P } ( X = r )$ & 0.2 & 0.3 & 0.3 & 0.2 \\
\hline
\end{tabular}
\end{center}
(i) Explain why $\mathrm { E } ( X ) = 25$.\\
(ii) Calculate $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2009 Q4 [4]}}