| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct distribution then calculate probability |
| Difficulty | Standard +0.3 This is a straightforward S1 question requiring standard techniques: calculating P(X=6) using complementary probability (all dice ≤6 minus all dice ≤5), then applying expectation and variance formulas to a given distribution. The probability distribution is provided, making part (ii) purely computational with no problem-solving required. Slightly easier than average due to the routine nature of both parts. |
| Spec | 2.04a Discrete probability distributions |
| \(r\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 1 } { 216 }\) | \(\frac { 7 } { 216 }\) | \(\frac { 19 } { 216 }\) | \(\frac { 37 } { 216 }\) | \(\frac { 61 } { 216 }\) | \(\frac { 91 } { 216 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=6) = 1 - P(X < 6) = 1 - \left(\frac{5}{6}\right)^3\) | M1 | For \(\left(\frac{5}{6}\right)^3\) |
| M1 | For \(1 - \left(\frac{5}{6}\right)^3\) | |
| \(= \frac{91}{216}\) | A1 | NB ANSWER GIVEN |
| OR: \(= \left(\frac{1}{6}\right)^3 + 3\times\left(\frac{5}{6}\right)\times\left(\frac{1}{6}\right)^2 + 3\times\left(\frac{5}{6}\right)^2\times\left(\frac{1}{6}\right)\) | M1 | For second or third product term. Correct including \(\times 3\) or probabilities seen on correct tree diagram |
| M1 | For attempt at three terms. With no extras, but allow omission of \(\times 3\). NB Zero for \(1-(\text{sum of probs given in part (ii)})\) | |
| \(= \frac{91}{216}\) | A1 | NB ANSWER GIVEN |
| OR: \(1 + 15 + 75\), \(= \frac{1+15+75}{216}\) | M1 M1 | For 15 or 75 seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \left(1\times\frac{1}{216}\right)+\left(2\times\frac{7}{216}\right)+\left(3\times\frac{19}{216}\right)+\left(4\times\frac{37}{216}\right)+\left(5\times\frac{61}{216}\right)+\left(6\times\frac{91}{216}\right)\) | M1 | For \(\Sigma rp\) (at least 3 terms correct) |
| \(=\frac{1071}{216}=\frac{119}{24}=4.96\) (exact answer 4.9583333) | A1 | CAO; accept fractional answers; do not allow answer of 5 unless more accurate answer given first |
| \(E(X^2)=\left(1\times\frac{1}{216}\right)+\left(4\times\frac{7}{216}\right)+\left(9\times\frac{19}{216}\right)+\left(16\times\frac{37}{216}\right)+\left(25\times\frac{61}{216}\right)+\left(36\times\frac{91}{216}\right)\) | M1* | For \(\Sigma r^2p\) (at least 3 terms correct); use of \(E(X-\mu)^2\) gets M1 for attempt |
| \(=\frac{5593}{216}=25.89\) | M1* dep | For \(-\) their \((E(X))^2\) |
| \(\text{Var}(X)=25.89\ldots - 4.958\ldots^2\) | ||
| \(=1.31\); accept answers in range 1.28 to 1.31 with correct working (exact answer 1.308449...) | A1 | FT their \(E(X)\) provided \(\text{Var}(X)>0\); deduct at most 1 mark for over-specification |
| [5] |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=6) = 1 - P(X < 6) = 1 - \left(\frac{5}{6}\right)^3$ | M1 | For $\left(\frac{5}{6}\right)^3$ |
| | M1 | For $1 - \left(\frac{5}{6}\right)^3$ |
| $= \frac{91}{216}$ | A1 | **NB ANSWER GIVEN** |
| **OR:** $= \left(\frac{1}{6}\right)^3 + 3\times\left(\frac{5}{6}\right)\times\left(\frac{1}{6}\right)^2 + 3\times\left(\frac{5}{6}\right)^2\times\left(\frac{1}{6}\right)$ | M1 | For second or third product term. Correct including $\times 3$ or probabilities seen on correct tree diagram |
| | M1 | For attempt at three terms. With no extras, but allow omission of $\times 3$. NB Zero for $1-(\text{sum of probs given in part (ii)})$ |
| $= \frac{91}{216}$ | A1 | **NB ANSWER GIVEN** |
| **OR:** $1 + 15 + 75$, $= \frac{1+15+75}{216}$ | M1 M1 | For 15 or 75 seen |
# Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \left(1\times\frac{1}{216}\right)+\left(2\times\frac{7}{216}\right)+\left(3\times\frac{19}{216}\right)+\left(4\times\frac{37}{216}\right)+\left(5\times\frac{61}{216}\right)+\left(6\times\frac{91}{216}\right)$ | M1 | For $\Sigma rp$ (at least 3 terms correct) |
| $=\frac{1071}{216}=\frac{119}{24}=4.96$ (exact answer 4.9583333) | A1 | CAO; accept fractional answers; do not allow answer of 5 unless more accurate answer given first |
| $E(X^2)=\left(1\times\frac{1}{216}\right)+\left(4\times\frac{7}{216}\right)+\left(9\times\frac{19}{216}\right)+\left(16\times\frac{37}{216}\right)+\left(25\times\frac{61}{216}\right)+\left(36\times\frac{91}{216}\right)$ | M1* | For $\Sigma r^2p$ (at least 3 terms correct); use of $E(X-\mu)^2$ gets M1 for attempt |
| $=\frac{5593}{216}=25.89$ | M1* dep | For $-$ their $(E(X))^2$ |
| $\text{Var}(X)=25.89\ldots - 4.958\ldots^2$ | | |
| $=1.31$; accept answers in range 1.28 to 1.31 with correct working (exact answer 1.308449...) | A1 | FT their $E(X)$ provided $\text{Var}(X)>0$; deduct at most 1 mark for over-specification |
| **[5]** | | |
---6 Three fair six-sided dice are thrown. The random variable $X$ represents the highest of the three scores on the dice.\\
(i) Show that $\mathrm { P } ( X = 6 ) = \frac { 91 } { 216 }$.
The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 1 } { 216 }$ & $\frac { 7 } { 216 }$ & $\frac { 19 } { 216 }$ & $\frac { 37 } { 216 }$ & $\frac { 61 } { 216 }$ & $\frac { 91 } { 216 }$ \\
\hline
\end{tabular}
\end{center}
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{OCR MEI S1 2015 Q6 [8]}}