| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Piecewise PDF with k |
| Difficulty | Standard +0.3 This is a standard Further Statistics piecewise PDF question requiring integration to find the constant, probability calculations, and setting up a median equation. All techniques are routine for this level—integrating polynomials, using the total probability condition, and applying definitions of mean and median. The algebra is straightforward with no novel problem-solving required, making it slightly easier than average. |
| 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 |
|---|---|---|
| 2 | (i) | (A) |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Shape of each part separately, |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | (B) |
| Answer | Marks |
|---|---|
| 3 3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.2 |
| Answer | Marks |
|---|---|
| 2.2a | M |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (ii) | (A) |
| Answer | Marks |
|---|---|
| 24 | E |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | Attempt to find area from –1 to |
| Answer | Marks |
|---|---|
| (B) | 0 1(cid:11) 1x(cid:14)x3(cid:12) |
| Answer | Marks |
|---|---|
| 4 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | Must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (iii) | area from –1 to 0 is 1, so require |
| Answer | Marks |
|---|---|
| 3 3 6 | E1 |
| Answer | Marks |
|---|---|
| [3] | 2.1 |
| Answer | Marks |
|---|---|
| 1.1 | N |
Question 2:
2 | (i) | (A) | B1
B1
[2] | 1.1
1.1 | Shape of each part separately,
domain correct
All correctN, including y-
intercept, which may be labelled
a or 1
E3
2 | (i) | (B) | Total area = 1
1(cid:11) a(cid:14)x2(cid:12)
Area = a(cid:14)(cid:179) dx
0
1 1
So 2a(cid:14) (cid:32)1 (cid:159)a(cid:32) AG
3 3 | B1
M1
A1
[3] | 1.2
I2.1
C
2.2a | M
Use of this principle somewhere
in solution
Attempt at two (or more) areas
including a correct integral
2 | (ii) | (A) | 1 1(cid:167)1 (cid:183)
(cid:14)(cid:179)2 (cid:168) (cid:14)x2 (cid:184)dx
3 0 (cid:169)3 (cid:185) P
13
(cid:32) (cid:32)0.5417 S
24 | E
M1
A1
[2] | 1.1a
1.1 | Attempt to find area from –1 to
1
2
Must be seen
BC
As fraction, or given correct to
3 or 4 d.p.
(B) | 0 1(cid:11) 1x(cid:14)x3(cid:12)
E(X)(cid:32)(cid:179) 1xdx(cid:14)(cid:179) dx
(cid:16)13 0 3
1
(cid:32)
4 | M1
A1
[2] | 1.1a
1.1 | Must be seen
BC
2 | (iii) | area from –1 to 0 is 1, so require
3
m(cid:11) 1(cid:14)x2(cid:12)
(cid:179) dx(cid:32) 1
0 3 6
m
(cid:170)1x(cid:14)1x3(cid:186) (cid:32) 1
(cid:172)3 3 (cid:188) 0 6
1m(cid:14)1m3 (cid:32) 1 (cid:159)2m3(cid:14)2m(cid:16)1(cid:32)0 AG
3 3 6 | E1
M1
A1
[3] | 2.1
1.1a
1.1 | N2 The continuous random variable $X$ takes values in the interval $- 1 \leq x \leq 1$ and has probability density function
$$f ( x ) = \left\{ \begin{array} { l r }
a & - 1 \leq x < 0 \\
a + x ^ { 2 } & 0 \leq x \leq 1
\end{array} \right.$$
where $a$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item (A) Sketch the probability density function.\\
(B) Show that $a = \frac { 1 } { 3 }$.
\item Find\\
(A) $\mathrm { P } \left( X < \frac { 1 } { 2 } \right)$,\\
(B) the mean of $X$.
\item Show that the median of $X$ satisfies the equation $2 m ^ { 3 } + 2 m - 1 = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major Q2 [12]}}