| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question requiring routine calculations: substituting values into a given formula, using the sum-to-1 property to find k, calculating E(X) and Var(X) using standard formulas. While it involves multiple parts and some arithmetic, all techniques are standard textbook exercises with no problem-solving insight required. The formula is provided rather than derived, making this slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(r\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm { P } ( X = r )\) | \(91 k\) | \(61 k\) | \(37 k\) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | r |
| Answer | Marks | Guidance |
|---|---|---|
| P(X = r) | 91k | 61k |
| 1 | (b) | 91k + 61k + 37k + 19k + 7k + k = 1 |
| Answer | Marks |
|---|---|
| 216 | M1 |
| Answer | Marks |
|---|---|
| [2] | For equation |
| AG | Zero if only write |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | 0.50 |
| Answer | Marks |
|---|---|
| r | B1 |
| Answer | Marks |
|---|---|
| [2] | For heights by eye |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | The distribution has (strong) positive skew |
| [1] | Allow ‘The distribution is J-shaped’ |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (e) | DR |
| Answer | Marks |
|---|---|
| 1728 | 1 M1 |
| Answer | Marks |
|---|---|
| [5] | FT their probabilities for all M marks |
Question 1:
1 | (a) | r | 1 | 2 | 3 | 4 | 5 | 6 | B1
[1]
P(X = r) | 91k | 61k | 37k | 19k | 7k | k
1 | (b) | 91k + 61k + 37k + 19k + 7k + k = 1
216k = 1
So k 1
216 | M1
A1
[2] | For equation
AG | Zero if only write
216k = 1 so
k 1
216
1 | (c) | 0.50
0.40
)r
0.30 =X(P
0.20
0.10
0.00
1 2 3 4 5 6
r | B1
B1
[2] | For heights by eye
For axes (including scales) and labels
Should be a probability scale rather 91,
61 etc so 91k is ok
1 | (d) | The distribution has (strong) positive skew | B1
[1] | Allow ‘The distribution is J-shaped’
Do not allow ‘Decreasing distribution’
1 | (e) | DR
E(X)1 91 2 61 3 37 4 19 5 7 6
216 216 216 216 216 2
= 49 = 2.0417
24
E(X2)12 91 22 61 32 37 42 19
216 216 216 216
52 7 62 1 1183 5.4769
216 216 216
Var(X) = 5.4769 – (2.0417)2
= 1.308 or 2261
1728 | 1 M1
16
A1
M1*
*DM1
A1
[5] | FT their probabilities for all M marks
(provided they sum to 1)
Allow fraction or decimal form
Dep on valid attempt at E(X2) – at least
four correct terms but could be ito k1 A fair six-sided dice is rolled three times.\\
The random variable $X$ represents the lowest of the three scores.\\
The probability distribution of $X$ is given by the formula\\
$\mathrm { P } ( X = r ) = k \left( 127 - 39 r + 3 r ^ { 2 } \right)$ for $r = 1,2,3,4,5,6$.
\begin{enumerate}[label=(\alph*)]
\item Complete the copy of the table in the Printed Answer Booklet.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm { P } ( X = r )$ & $91 k$ & $61 k$ & $37 k$ & & & \\
\hline
\end{tabular}
\end{center}
\item Show that $k = \frac { 1 } { 216 }$.
\item Draw a graph to illustrate the distribution.
\item Comment briefly on the shape of the distribution.
\item In this question you must show detailed reasoning.
Find each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q1 [11]}}