| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| 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 probability distribution question requiring only standard techniques: summing probabilities to equal 1 to find k, then applying textbook formulas for E(X) and Var(X). The arithmetic is simple with small integer values, and no problem-solving insight is needed—just methodical application of basic definitions. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k + 0.01 + k + 0.04 + k + 0.09 + k + 0.16 + k + 0.25 = 1\) | M1 | For equation in \(k\) |
| \(5k + 0.55 = 1\) | ||
| \(k = 0.09\) | A1 | NB Answer Given. Allow substitution of \(k = 0.09\) to show probabilities add to 1 with convincing working |
| B1 | Complete correct table. Must tabulate probabilities, though may be seen in part(ii) | |
| r | 1 | 2 |
| \(P(X = r)\) | 0.1 | 0.13 |
| Total | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (1 \times 0.1) + (2 \times 0.13) + (3 \times 0.18) + (4 \times 0.25) + (5 \times 0.34)\) | M1 | For \(\Sigma rp\) (at least 3 terms correct) Provided 5 reasonable probabilities seen. If probs wrong but sum = 1 allow max M1A0M1M1A0 (provided all probabilities \(\geq 0\) and \(< 1\)). No marks if all probs =0.2 |
| \(= 3.6\) | A1 | CAO |
| \(E(X^2) = (1 \times 0.1) + (4 \times 0.13) + (9 \times 0.18) + (16 \times 0.25) + (25 \times 0.34) = 14.74\) | M1* | For \(\Sigma r^2 p\) (at least 3 terms correct) for – their \([E[X]]^2\) FT their \(E(X)\) provided \(\text{Var}(X) > 0\) |
| \(\text{Var}(X) = 14.74 - 3.6^2\) | M1* dep | |
| \(= 1.78\) | A1 | CAO. Use of \(E(X - \mu)^2\) gets M1 for attempt at \(\sum(x - \mu)^2\) should see (-2.6)², (-1.6)², (-0.6)², 0.4², 1.4², (if \(E(X)\) wrong FT their \(E(X)\)) (all 5 correct for M1), then M1 for \(\Sigma p(x - \mu)^2\) (at least 3 terms correct with their probabilities). Division by 5 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0, or M1A0M1M1A0 if \(E(X)\) also divided by 5. Unsupported correct answers get 5 marks (Probably from calculator) |
| Total | [5] |
## (i)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $k + 0.01 + k + 0.04 + k + 0.09 + k + 0.16 + k + 0.25 = 1$ | M1 | For equation in $k$ |
| $5k + 0.55 = 1$ | | |
| $k = 0.09$ | A1 | NB Answer Given. Allow substitution of $k = 0.09$ to show probabilities add to 1 with convincing working |
| | B1 | Complete correct table. Must tabulate probabilities, though may be seen in part(ii) |
| | | |
| | r | 1 | 2 | 3 | 4 | 5 |
| | $P(X = r)$ | 0.1 | 0.13 | 0.18 | 0.25 | 0.34 |
| **Total** | **[3]** | |
## (ii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $E(X) = (1 \times 0.1) + (2 \times 0.13) + (3 \times 0.18) + (4 \times 0.25) + (5 \times 0.34)$ | M1 | For $\Sigma rp$ (at least 3 terms correct) Provided 5 reasonable probabilities seen. If probs wrong but sum = 1 allow max M1A0M1M1A0 (provided all probabilities $\geq 0$ and $< 1$). No marks if all probs =0.2 |
| $= 3.6$ | A1 | CAO |
| $E(X^2) = (1 \times 0.1) + (4 \times 0.13) + (9 \times 0.18) + (16 \times 0.25) + (25 \times 0.34) = 14.74$ | M1* | For $\Sigma r^2 p$ (at least 3 terms correct) for – their $[E[X]]^2$ FT their $E(X)$ provided $\text{Var}(X) > 0$ |
| $\text{Var}(X) = 14.74 - 3.6^2$ | M1* dep | |
| $= 1.78$ | A1 | CAO. Use of $E(X - \mu)^2$ gets M1 for attempt at $\sum(x - \mu)^2$ should see (-2.6)², (-1.6)², (-0.6)², 0.4², 1.4², (if $E(X)$ wrong FT their $E(X)$) (all 5 correct for M1), then M1 for $\Sigma p(x - \mu)^2$ (at least 3 terms correct with their probabilities). Division by 5 or other spurious value at end and/or rooting final answer gives max M1A1M1M1A0, or M1A0M1M1A0 if $E(X)$ also divided by 5. Unsupported correct answers get 5 marks (Probably from calculator) |
| **Total** | **[5]** | |
---The probability distribution of the random variable $X$ is given by the formula
$$\mathrm{P}(X = r) = k + 0.01r^2 \text{ for } r = 1, 2, 3, 4, 5.$$
\begin{enumerate}[label=(\roman*)]
\item Show that $k = 0.09$. Using this value of $k$, display the probability distribution of $X$ in a table. [3]
\item Find $\mathrm{E}(X)$ and $\mathrm{Var}(X)$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2014 Q5 [8]}}