| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Moderate -0.8 This is a straightforward discrete probability distribution question requiring only routine calculations: substituting values into a given formula, using the fact that probabilities sum to 1 to find k, calculating E(X) using the standard formula, and reading off a probability from the table. All steps are mechanical with no problem-solving or insight required, making it easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(6 k\) | \(10 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r\): 0, 1, 2, 3, 4; \(P(X=r)\): \(6k\), \(10k\), \(12k\), \(12k\), \(10k\) | B1, B1 | 1 value correct; all 3 correct |
| \(50k = 1 \Rightarrow k = \frac{1}{50}\) | M1 | sum of 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = 110k = 2.2\) | M1, A1 | sum of \(rp\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X > 2.2) = 22k = 0.44\) | B1 |
## Question 4:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $r$: 0, 1, 2, 3, 4; $P(X=r)$: $6k$, $10k$, $12k$, $12k$, $10k$ | B1, B1 | 1 value correct; all 3 correct |
| $50k = 1 \Rightarrow k = \frac{1}{50}$ | M1 | sum of 1 |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = 110k = 2.2$ | M1, A1 | sum of $rp$; cao |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X > 2.2) = 22k = 0.44$ | B1 | |
---4 The number, $X$, of children per family in a certain city is modelled by the probability distribution $\mathrm { P } ( X = r ) = k ( 6 - r ) ( 1 + r )$ for $r = 0,1,2,3,4$.\\
(i) Copy and complete the following table and hence show that the value of $k$ is $\frac { 1 } { 50 }$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $6 k$ & $10 k$ & & & \\
\hline
\end{tabular}
\end{center}
(ii) Calculate $\mathrm { E } ( X )$.\\
(iii) Hence write down the probability that a randomly selected family in this city has more than the mean number of children.
\hfill \mbox{\textit{OCR MEI S1 2005 Q4 [6]}}