| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward discrete probability distribution question requiring only basic probability axioms (probabilities sum to 1) and standard expectation/variance formulas. Part (i) is trivial arithmetic, part (ii) is routine calculation from definitions, and part (iii) applies basic independence. No problem-solving insight needed, just mechanical application of standard S1 techniques. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(r\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = r )\) | \(p\) | 0.1 | 0.05 | 0.05 | 0.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = 0.55\) | B1 cao | |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = 0 \times 0.55 + 1 \times 0.1 + 2 \times 0.05 + 3 \times 0.05 + 4 \times 0.25 = 1.35\) | M1 | For \(\Sigma rp\) (at least 3 non zero terms correct) |
| A1 CAO | No 'n' or 'n-1' divisors | |
| \(E(X^2) = 0 \times 0.55 + 1 \times 0.1 + 4 \times 0.05 + 9 \times 0.05 + 16 \times 0.25 = 4.75\) | M1 | For \(\Sigma r^2p\) (at least 3 non zero terms correct) |
| \(\text{Var}(X) = 4.75 - 1.35^2 = 2.9275\) awfw (2.9275 – 2.93) | M1dep | For their \(E(X)^2\) provided \(\text{Var}(X) > 0\) |
| A1 cao | No 'n' or 'n-1' divisors | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{at least 2 both times}) = (0.05 + 0.05 + 0.25)^2 = 0.1225\) | M1 | For \((0.05+0.05+0.25)^2\) or \(0.35^2\) seen |
| A1 cao | awfw (0.1225 – 0.123) or 49/400 | |
| TOTAL | 8 |
## Question 3:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = 0.55$ | B1 cao | |
| **Total** | **1** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 0 \times 0.55 + 1 \times 0.1 + 2 \times 0.05 + 3 \times 0.05 + 4 \times 0.25 = 1.35$ | M1 | For $\Sigma rp$ (at least 3 non zero terms correct) |
| | A1 CAO | No 'n' or 'n-1' divisors |
| $E(X^2) = 0 \times 0.55 + 1 \times 0.1 + 4 \times 0.05 + 9 \times 0.05 + 16 \times 0.25 = 4.75$ | M1 | For $\Sigma r^2p$ (at least 3 non zero terms correct) |
| $\text{Var}(X) = 4.75 - 1.35^2 = 2.9275$ awfw (2.9275 – 2.93) | M1dep | For their $E(X)^2$ provided $\text{Var}(X) > 0$ |
| | A1 cao | No 'n' or 'n-1' divisors |
| **Total** | **5** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{at least 2 both times}) = (0.05 + 0.05 + 0.25)^2 = 0.1225$ | M1 | For $(0.05+0.05+0.25)^2$ or $0.35^2$ seen |
| | A1 cao | awfw (0.1225 – 0.123) or 49/400 |
| **TOTAL** | **8** | |
---3 A zoologist is studying the feeding behaviour of a group of 4 gorillas. The random variable $X$ represents the number of gorillas that are feeding at a randomly chosen moment. The probability distribution of $X$ is shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = r )$ & $p$ & 0.1 & 0.05 & 0.05 & 0.25 \\
\hline
\end{tabular}
\end{center}
(i) Find the value of $p$.\\
(ii) Find the expectation and variance of $X$.\\
(iii) The zoologist observes the gorillas on two further occasions. Find the probability that there are at least two gorillas feeding on both occasions.
\hfill \mbox{\textit{OCR MEI S1 2009 Q3 [8]}}