| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for expectation and variance given a discrete probability distribution. Part (iv) requires only routine calculation of E(X) = Σr·P(X=r) and Var(X) using the standard formula. The conceptual parts (i-ii) involve basic reasoning about permutations, and parts (v-vi) are simple probability calculations. While there are multiple parts, each is a standard textbook exercise requiring no problem-solving insight. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = r )\) | \(\frac { 11 } { 30 }\) | \(\frac { 3 } { 8 }\) | \(\frac { 1 } { 6 }\) | \(\frac { 1 } { 12 }\) | 0 | \(\frac { 1 } { 120 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | Because if 4 are matched correctly then the fifth |
| must also be matched correctly | B1 | |
| [1] | 2.2a | Do not allow 1 - ∑others |
| (ii) | AG | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 2.4 | Or: there are 5! = 120 possible |
| Answer | Marks |
|---|---|
| correct. | Do not allow 1 - ∑others |
| Answer | Marks |
|---|---|
| (iii) | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | For line graph with axes labelled. |
| Answer | Marks |
|---|---|
| heights of lines/points | B0 if tops of lines/points |
| Answer | Marks |
|---|---|
| (iv) | E(X) = 1 |
| Var(X) = 1 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| (v) | oe | B1 |
| [1] | 1.1 | Accept 0.258 or better |
| (vi) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | Accept 0.096, 0.09554 |
Question 2:
2 | (i) | Because if 4 are matched correctly then the fifth
must also be matched correctly | B1
[1] | 2.2a | Do not allow 1 - ∑others
(ii) | AG | M1
E1
[2] | 1.1a
2.4 | Or: there are 5! = 120 possible
arrangements of which only one is
correct. | Do not allow 1 - ∑others
Accept 1/5!
(iii) | B1
B1
[2] | 1.1
1.1 | For line graph with axes labelled.
Correct linear scale used for
probabilities and visibly correct
heights of lines/points | B0 if tops of lines/points
joined.
(iv) | E(X) = 1
Var(X) = 1 | B1
B1
[2] | 1.1a
1.1
(v) | oe | B1
[1] | 1.1 | Accept 0.258 or better
(vi) | M1
A1
[2] | 1.1a
1.1 | Accept 0.096, 0.095542 In a quiz, competitors have to match 5 landmarks to the 5 British counties which the landmarks are in. The random variable $X$ represents the number of correct matches that a competitor gets, assuming that the competitor guesses randomly. The probability distribution of $X$ is given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = r )$ & $\frac { 11 } { 30 }$ & $\frac { 3 } { 8 }$ & $\frac { 1 } { 6 }$ & $\frac { 1 } { 12 }$ & 0 & $\frac { 1 } { 120 }$ \\
\hline
\end{tabular}
\end{center}
(i) Explain why $\mathrm { P } ( X = 4 )$ must be 0 .\\
(ii) Explain how the value $\frac { 1 } { 120 }$ for $\mathrm { P } ( X = 5 )$ is calculated.\\
(iii) Draw a graph to illustrate the distribution.\\
(iv) Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$\\
(v) Find $\mathrm { P } ( X > \mathrm { E } ( X ) )$.\\
(vi) There are 12 competitors in the quiz. Assuming that they all guess randomly, find the probability that at least one of them gets all five matches correct.
\end{itemize}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2018 Q2 [10]}}