| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Single-piece PDF with k |
| Difficulty | Standard +0.3 This is a standard continuous probability distribution question requiring routine integration techniques: finding k by integrating to 1, calculating E(X), and evaluating cumulative probabilities. While it involves a polynomial requiring careful algebra, these are well-practiced A-level Further Maths techniques with no novel problem-solving or conceptual insight required. Slightly above average difficulty due to the algebraic manipulation needed. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | k 1 (5x4−16x2+11x)dx=1 |
| Answer | Marks |
|---|---|
| k 76 = 1 k = 67 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.1a | |
| 1.1 | Integration BC Ignore any working for the integration | |
| 7 | (b) | E ( X ) = 1 67 ( 5 x 5 − 1 6 x 3 + 1 1 x 2 ) d x |
| Answer | Marks |
|---|---|
| = 0.5184 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | BC Ignore any working for the integration |
| Answer | Marks | Guidance |
|---|---|---|
| OR | (c) | 0 .4 1 7 67 ( 5 x 4 − 1 6 x 2 + 1 1 x ) d x [ = 0.499] |
| Answer | Marks |
|---|---|
| So median must be between 0.417 and 0.418 | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1a |
| Answer | Marks |
|---|---|
| 2.2a | BC |
Question 7:
7 | (a) | k 1 (5x4−16x2+11x)dx=1
0
k 76 = 1 k = 67 | M1
A1
[2] | 3.1a
1.1 | Integration BC Ignore any working for the integration
7 | (b) | E ( X ) = 1 67 ( 5 x 5 − 1 6 x 3 + 1 1 x 2 ) d x
0
= 3 =0.4286
7
P ( X E ( X ) ) = 0 .4 2 8 6 67 ( 5 x 4 − 1 6 x 2 + 1 1 x ) d x
0
= 0.5184 | M1
A1
M1
A1
[4] | 3.1a
1.1
3.1a
1.1 | BC Ignore any working for the integration
BC Allow 0.5185 Ignore any working for the
integration
7
OR | (c) | 0 .4 1 7 67 ( 5 x 4 − 1 6 x 2 + 1 1 x ) d x [ = 0.499]
0
0 .4 1 8 67 ( 5 x 4 − 1 6 x 2 + 1 1 x ) d x [ = 0.501]
0
At the median, m, F(m) = 0.5 so the median must be
between 0.417 and 0.418
6 16 11
(𝑚5− 𝑚3+ 𝑚2)−0.5 = 0
7 3 2
0.417 gives −0.00091 0.418 gives 0.00076
So median must be between 0.417 and 0.418 | M1
M1
A1
[3]
M1
M1
A1 | 1.1a
1.1
2.2a | BC
BC
Both of the values 0.499 and 0.501 must be seen for this
final mark
Oe Must include k Allow x in place of m.
Sub 0.417 or 0.418
Both values substituted for final mark
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.7 Many cars have pollen filters to try to remove as much pollen as possible from the passenger compartment. In a test car, the amount of pollen is regularly monitored. The amount of pollen is measured using a scale from 0 to 1 , and is modelled by the continuous random variable $X$ with probability density function given by\\
$f ( x ) = \begin{cases} k \left( 5 x ^ { 4 } - 16 x ^ { 2 } + 11 x \right) & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise } , \end{cases}$\\
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 6 } { 7 }$.
\item Determine $\mathrm { P } ( X < \mathrm { E } ( X ) )$.
\item Verify that the median amount of pollen according to the model lies between 0.417 and 0.418.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2022 Q7 [9]}}