| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from given formula |
| Difficulty | Easy -1.2 This is a straightforward question requiring only direct substitution into a given formula (part i), standard expectation and variance calculations using given probabilities (part ii), and a simple application of expectation (part iii). All parts are routine bookwork with no problem-solving or insight required—significantly easier than average A-level questions. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=4) = \frac{1}{40}(4)(5) = \frac{1}{2}\) (Answer given) | B1 | Calculation must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = (2+12+36+80)\frac{1}{40}\), so \(E(X) = 3.25\) | M1 A1 cao | Sum of \(rp\) |
| \(\text{Var}(X) = (2+24+108+320)\frac{1}{40} - 3.25^2\) | M1 | Sum of \(r^2p\) |
| \(= 11.35 - 10.5625\) | M1 dep | \(-3.25^2\) |
| \(= 0.7875\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Expected number of weeks \(= \frac{6}{40} \times 45 = 6.75\) weeks | M1 A1 | Use of \(np\) |
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=4) = \frac{1}{40}(4)(5) = \frac{1}{2}$ (Answer given) | B1 | Calculation must be seen |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = (2+12+36+80)\frac{1}{40}$, so $E(X) = 3.25$ | M1 A1 cao | Sum of $rp$ |
| $\text{Var}(X) = (2+24+108+320)\frac{1}{40} - 3.25^2$ | M1 | Sum of $r^2p$ |
| $= 11.35 - 10.5625$ | M1 dep | $-3.25^2$ |
| $= 0.7875$ | A1 cao | |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Expected number of weeks $= \frac{6}{40} \times 45 = 6.75$ weeks | M1 A1 | Use of $np$ |
---3 Jeremy is a computing consultant who sometimes works at home. The number, $X$, of days that Jeremy works at home in any given week is modelled by the probability distribution
$$\mathrm { P } ( X = r ) = \frac { 1 } { 40 } r ( r + 1 ) \quad \text { for } r = 1,2,3,4 .$$
(i) Verify that $\mathrm { P } ( X = 4 ) = \frac { 1 } { 2 }$.\\
(ii) Calculate $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
(iii) Jeremy works for 45 weeks each year. Find the expected number of weeks during which he works at home for exactly 2 days.
\hfill \mbox{\textit{OCR MEI S1 2005 Q3 [8]}}