| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring basic algebraic manipulation. Part (i) uses the fundamental property that probabilities sum to 1, then sets up a simple simultaneous equation system from the expectation formula. Part (ii) applies the standard variance formula. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = r )\) | 0.5 | 0.35 | \(p\) | \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| (A) \(0.5 + 0.35 + p + q = 1\) so \(p + q = 0.15\) | M1, A1 | B1 \(p + q\) in a correct equation before they reach \(p + q = 0.15\) |
| (B) \(0 \times 0.5 + 1 \times 0.35 + 2p + 3q = 0.67\) so \(2p + 3q = 0.32\) | M1, A1 | B1 \(2p + 3q\) in a correct equation before they reach \(2p + 3q = 0.32\) |
| (C) From above: \(2p + 2q = 0.30\) so \(q = 0.02, p = 0.13\) | M1, A1 | (B1) for any 1 correct answer; B2 for both correct answers |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X^2) = 0 \times 0.5 + 1 \times 0.35 + 4 \times 0.13 + 9 \times 0.02 = 1.05\) | M1 | \(\Sigma x^p\) (at least 2 non zero terms correct); M1dep for \((\neg 0.67^2)\); provided \(\text{Var}(X) > 0\) |
| \(\text{Var}(X) = \text{'their 1.05'} - 0.67^2 = 0.6011\) (awrt 0.6) | A1 | A1cao (No n or n-1 divisors) |
## (i)
**(A)** $0.5 + 0.35 + p + q = 1$ so $p + q = 0.15$ | M1, A1 | B1 $p + q$ in a correct equation before they reach $p + q = 0.15$
**(B)** $0 \times 0.5 + 1 \times 0.35 + 2p + 3q = 0.67$ so $2p + 3q = 0.32$ | M1, A1 | B1 $2p + 3q$ in a correct equation before they reach $2p + 3q = 0.32$
**(C)** From above: $2p + 2q = 0.30$ so $q = 0.02, p = 0.13$ | M1, A1 | (B1) for any 1 correct answer; B2 for both correct answers | M1 for 1 correct answer; M1 for correct sum of all 3 (or all 5) with no extras; A1cao (condone 0.96 www) | 2 marks total
## (ii)
$E(X^2) = 0 \times 0.5 + 1 \times 0.35 + 4 \times 0.13 + 9 \times 0.02 = 1.05$ | M1 | $\Sigma x^p$ (at least 2 non zero terms correct); M1dep for $(\neg 0.67^2)$; provided $\text{Var}(X) > 0$
$\text{Var}(X) = \text{'their 1.05'} - 0.67^2 = 0.6011$ (awrt 0.6) | A1 | A1cao (No n or n-1 divisors) | 3 marks total
---3 In a game of darts, a player throws three darts. Let $X$ represent the number of darts which hit the bull's-eye. The probability distribution of $X$ is shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = r )$ & 0.5 & 0.35 & $p$ & $q$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item (A) Show that $p + q = 0.15$.\\
(B) Given that the expectation of $X$ is 0.67 , show that $2 p + 3 q = 0.32$.\\
(C) Find the values of $p$ and $q$.
\item Find the variance of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2008 Q3 [7]}}