| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Multi-part piecewise CDF |
| Difficulty | Challenging +1.2 This is a Further Maths statistics question requiring understanding of CDF properties (continuity, F(20)=0, F(30)=1) to set up a system of equations, then using given probability information to solve for constants. While it involves multiple steps and algebraic manipulation, the approach is methodical and the techniques are standard for Further Statisticsβharder than typical A-level but not requiring exceptional insight. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (a) | 11 |
| Answer | Marks |
|---|---|
| 200 | B1 |
| Answer | Marks |
|---|---|
| A1 | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | Soi For at least 2 of the three |
| Answer | Marks |
|---|---|
| Please annotate page 19 | B1 |
| Answer | Marks |
|---|---|
| [7] | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (b) | 1 (π2+10πβ600) = 0.9 |
| Answer | Marks |
|---|---|
| π = 29.1 (29.132β¦) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | For equation using their non-zero values of a, b and c |
Question 12:
12 | (a) | 11
F(20) = 0,F(30) = 1,F(25) =
24
π(400+20π+π) = 0
π(900+30π+π) = 1
11
π(625+25π+π) =
24
1
π = ,π = 10,π = β600
600
P(X > 27) = 1β 1 (272+10Γ27β600)
600
67
P(X > 27) = 0.335 or
200 | B1
M1
M1
M1
A1
M1
A1 | 2.2a
2.2a
1.1
1.1
1.1
3.1a
1.1 | Soi For at least 2 of the three
For F(20) = 0
For F(30) = 1
11
For F(25) =
24
For solving either by elimination or by calculator
For calculation using their non-zero values of a, b and c. oe
Alternative method
F(30)βF(20) = 1
π(900+30π+π)βπ(400+20π+π) = 1
11
F(25)βF(20) =
24
11
π(625+25π+π)β π(400+20π+π) =
24
1
π = ,π = 10
600
P(X > 27) = 1 (302+10Γ30+π)β 1 (272+10Γ27+π)
600 600
67
P(X > 27) = 0.335 or
200
Please annotate page 19 | B1
M1
B1
M1
A1
M1
A1
[7] | 2.2a
2.2a
1.1
1.1
1.1
3.1a
1.1 | soi
soi
For solving either by elimination or by calculator
For calculation using their non-zero values of a, b and c
12 | (b) | 1 (π2+10πβ600) = 0.9
600
Or 1 (π2+10π+π)β 1 (202+10Γ20+π) = 0.9
600 600
π = 29.1 (29.132β¦) | M1
A1
[2] | 3.1a
1.1 | For equation using their non-zero values of a, b and c
Or For equation using their non-zero values of a and b
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.12 The cumulative distribution function of the continuous random variable $X$ is given by\\
$F ( x ) = \begin{cases} 0 & x < 20 , \\ a \left( x ^ { 2 } + b x + c \right) & 20 \leqslant x \leqslant 30 , \\ 1 & x > 30 , \end{cases}$\\
where $a$, $b$ and $c$ are constants.\\
You are given that $\mathrm { P } ( X < 25 ) = \frac { 11 } { 24 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X > 27 )$.
\item Find the 90th percentile of $X$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q12 [9]}}