| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Probability distribution from formula |
| Difficulty | Moderate -0.8 This is a straightforward S1 question testing basic probability distribution properties. Part (i) requires summing probabilities to equal 1 (routine algebra), and part (ii) involves standard application of expectation and variance formulas with simple arithmetic. No problem-solving insight needed, just methodical calculation following learned procedures. |
| Spec | 2.04a Discrete probability distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k + 0.01 + k + 0.04 + k + 0.09 + k + 0.16 + k + 0.25 = 1\), \(5k + 0.55 = 1\) | M1 | For equation in \(k\) — Allow substitution of \(k=0.09\) to show probabilities add to 1 with convincing working |
| \(k = 0.09\) | A1 | NB Answer Given |
| \(r\): 1, 2, 3, 4, 5; \(P(X=r)\): 0.1, 0.13, 0.18, 0.25, 0.34 | B1 | Complete correct table — Must tabulate probabilities, though may be seen in part (ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (1\times0.1)+(2\times0.13)+(3\times0.18)+(4\times0.25)+(5\times0.34)\) | M1 | For \(\Sigma rp\) (at least 3 terms correct), provided 5 reasonable probabilities seen — If probs wrong but sum \(=1\) allow max M1A0M1M1A1 |
| \(= 3.6\) | A1 | CAO |
| \(E(X^2) = (1\times0.1)+(4\times0.13)+(9\times0.18)+(16\times0.25)+(25\times0.34) = 14.74\) | M1* | For \(\Sigma r^2 p\) (at least 3 terms correct) — Use of \(E(X-\mu)^2\) gets M1; should see \((-2.6)^2\), \((-1.6)^2\), \((-0.6)^2\), \(0.4^2\), \(1.4^2\) |
| \(\text{Var}(X) = 14.74 - 3.6^2\) | M1* dep | For \(-\) their \((E[X])^2\), FT their \(E(X)\) provided \(\text{Var}(X)>0\) |
| \(= 1.78\) | A1 | CAO — Division by 5 or other spurious value gives max M1A1M1M1A0 |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k + 0.01 + k + 0.04 + k + 0.09 + k + 0.16 + k + 0.25 = 1$, $5k + 0.55 = 1$ | M1 | For equation in $k$ — Allow substitution of $k=0.09$ to show probabilities add to 1 with convincing working |
| $k = 0.09$ | A1 | **NB Answer Given** |
| $r$: 1, 2, 3, 4, 5; $P(X=r)$: 0.1, 0.13, 0.18, 0.25, 0.34 | B1 | Complete correct table — Must tabulate probabilities, though may be seen in part (ii) |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = (1\times0.1)+(2\times0.13)+(3\times0.18)+(4\times0.25)+(5\times0.34)$ | M1 | For $\Sigma rp$ (at least 3 terms correct), provided 5 reasonable probabilities seen — If probs wrong but sum $=1$ allow max M1A0M1M1A1 |
| $= 3.6$ | A1 | CAO |
| $E(X^2) = (1\times0.1)+(4\times0.13)+(9\times0.18)+(16\times0.25)+(25\times0.34) = 14.74$ | M1* | For $\Sigma r^2 p$ (at least 3 terms correct) — Use of $E(X-\mu)^2$ gets M1; should see $(-2.6)^2$, $(-1.6)^2$, $(-0.6)^2$, $0.4^2$, $1.4^2$ |
| $\text{Var}(X) = 14.74 - 3.6^2$ | M1* dep | For $-$ their $(E[X])^2$, FT their $E(X)$ provided $\text{Var}(X)>0$ |
| $= 1.78$ | A1 | CAO — Division by 5 or other spurious value gives max M1A1M1M1A0 |
---3 The probability distribution of the random variable $X$ is given by the formula
$$\mathrm { P } ( X = r ) = k + 0.01 r ^ { 2 } \text { for } r = 1,2,3,4,5 .$$
(i) Show that $k = 0.09$. Using this value of $k$, display the probability distribution of $X$ in a table.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 Q3 [8]}}