| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring only routine calculations: normalizing probabilities by summing to 1 (finding k), then applying standard expectation and variance formulas. All steps are mechanical with no conceptual challenges or problem-solving insight needed—easier than average A-level. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r\): 2, 3, 4, 5; \(P(X=r)\): \(k\), \(8k\), \(15k\), \(24k\) | B1 | For correct table (to \(k\) or correct probabilities 0.06, 0.16, 0.30, 0.48) |
| \(3k + 8k + 15k + 24k = 1\) | M1 | For their four multiples of \(k\) added and \(=1\) — Allow M1A1 even if done in part (ii) |
| \(k = 0.02\) | A1 | or \(k = \frac{1}{50}\) (with or without working) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (2\times0.06)+(3\times0.16)+(4\times0.30)+(5\times0.48) = 4.2\) or \(\frac{21}{5}\) | M1 | For \(\Sigma rp\) (at least 3 terms correct), provided 4 reasonable probabilities seen — If probs wrong but sum \(=1\) allow full marks; NB \(E(X)=210k\), \(E(X^2)=924k\) gets M1A0M1M1A0 |
| A1 | cao | |
| \(E(X^2) = (4\times0.06)+(9\times0.16)+(16\times0.30)+(25\times0.48) = 18.48\) | M1 | For \(\Sigma r^2 p\) (at least 3 terms correct) — Use of \(E(X-\mu)^2\) gets M1; should see \((-2.2)^2\), \((-1.2)^2\), \((-0.2)^2\), \(0.8^2\) |
| \(\text{Var}(X) = 18.48 - 4.2^2\) | M1 | dep for \(-\) their \(E(X)^2\) — FT their \(E(X)\) provided \(\text{Var}(X)>0\) |
| \(= 0.84 = \frac{21}{25}\) | A1 | FT their \(E(X)\) provided \(\text{Var}(X)>0\) and \(E(X^2)\) is correct — Division by 4 gives max M1A1M1M1A0 |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r$: 2, 3, 4, 5; $P(X=r)$: $k$, $8k$, $15k$, $24k$ | B1 | For correct table (to $k$ or correct probabilities 0.06, 0.16, 0.30, 0.48) |
| $3k + 8k + 15k + 24k = 1$ | M1 | For their four multiples of $k$ added and $=1$ — Allow M1A1 even if done in part (ii) |
| $k = 0.02$ | A1 | or $k = \frac{1}{50}$ (with or without working) |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = (2\times0.06)+(3\times0.16)+(4\times0.30)+(5\times0.48) = 4.2$ or $\frac{21}{5}$ | M1 | For $\Sigma rp$ (at least 3 terms correct), provided 4 reasonable probabilities seen — If probs wrong but sum $=1$ allow full marks; NB $E(X)=210k$, $E(X^2)=924k$ gets M1A0M1M1A0 |
| | A1 | cao |
| $E(X^2) = (4\times0.06)+(9\times0.16)+(16\times0.30)+(25\times0.48) = 18.48$ | M1 | For $\Sigma r^2 p$ (at least 3 terms correct) — Use of $E(X-\mu)^2$ gets M1; should see $(-2.2)^2$, $(-1.2)^2$, $(-0.2)^2$, $0.8^2$ |
| $\text{Var}(X) = 18.48 - 4.2^2$ | M1 | dep for $-$ their $E(X)^2$ — FT their $E(X)$ provided $\text{Var}(X)>0$ |
| $= 0.84 = \frac{21}{25}$ | A1 | FT their $E(X)$ provided $\text{Var}(X)>0$ and $E(X^2)$ is correct — Division by 4 gives max M1A1M1M1A0 |4 The probability distribution of the random variable $X$ is given by the formula
$$\mathrm { P } ( X = r ) = k \left( r ^ { 2 } - 1 \right) \text { for } r = 2,3,4,5 .$$
(i) Show the probability distribution in a table, and find the value of $k$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 Q4 [8]}}