| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Find median or percentiles |
| Difficulty | Challenging +1.2 This is a multi-part CDF question requiring standard techniques: finding the median involves solving F(x)=0.5 (a quadratic equation), and part (b) requires finding k, then the mean and variance through integration of the pdf, followed by a probability calculation. While it involves several steps and integration, these are routine procedures for Further Statistics students with no novel insight required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (a) | k(a×a−0.5a 2 )=1 |
| Answer | Marks |
|---|---|
| 2 2 | M1 |
| Answer | Marks |
|---|---|
| [7] | 3.1a |
| Answer | Marks |
|---|---|
| 3.2a | Do not allow any credit if no attempt to find k |
| Answer | Marks | Guidance |
|---|---|---|
| 12 | (b) | 1 1 1 |
| Answer | Marks |
|---|---|
| = 0.81426… – 0.18573… = 0.6285 | B1 |
| Answer | Marks |
|---|---|
| [7] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | No credit if f(x) not found Can be in terms of k in place of |
Question 12:
12 | (a) | k(a×a−0.5a 2 )=1
2
k =
a2
2
(am−0.5m2)=0.5
a2
(am−0.5m 2 )=0.25a 2 [ 2m2−4am+a2 =0]
⇒
2(m−a) 2−a 2 =0
2 2± 2
m=a± 0.5a=a± a=a
2 2
2 2− 2
m=a− 0.5a=a− a=a
2 2 | M1
A1
B1
M1
M1
M1
A1
[7] | 3.1a
1.1
1.1
1.1
2.1
1.1
3.2a | Do not allow any credit if no attempt to find k
For attempt to solve by formula or completing the square
For at least one correct solution
For choosing correct answer
12 | (b) | 1 1 1
fE(𝑥𝑥()X=)=∫ 1(0 10 1 −(1𝑥𝑥0)x=−x2−)dx 𝑥𝑥
50 5 50
0 50
= 3.333
10
E(X2)=∫ 10 1 =(130x2 −x3)dx[ = 16.667]
0 50
Var(X)=16.667−3.333 2 =5.556
Standard deviation = 2.357 50
= 9
P(within 1 sd of mean) = P(0.97631 < X < 5.69036)
=0.02(10×5.69036−0.5×5.69036 2
−0.02(10×0.97631−0.5×0.97631 2 )
= 0.81426… – 0.18573… = 0.6285 | B1
M1
A1
M1
A1
M1
A1
[7] | 3.1a
1.1
1.1
1.1
1.1
3.3
1.1 | No credit if f(x) not found Can be in terms of k in place of
1
With their two term f(x) Can be in terms of k in place of f5o0r M1
1
50
BC
With their two term f(x) Can be in terms of k in place of for
M1 1
50
Dep on mean and sd correct.
Or
10−5√2 10+5√2
P� 3 < 𝑋𝑋 < 3 �
Allow exact answer 4 2
9
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 continuous random variable $X$ has cumulative distribution function given by
$$F ( x ) = \begin{cases} 0 & x < 0 \\ k \left( a x - 0.5 x ^ { 2 } \right) & 0 \leqslant x \leqslant a \\ 1 & x > a \end{cases}$$
where $a$ and $k$ are positive constants.
\begin{enumerate}[label=(\alph*)]
\item Determine the median of $X$ in terms of $a$.
\item Given that $a = 10$, determine the probability that $X$ is within one standard deviation of its mean.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2022 Q12 [14]}}