| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate probability P(X in interval) |
| Difficulty | Standard +0.3 This is a standard continuous probability distribution question requiring routine integration techniques: finding k by integrating to 1, calculating P(X≥5) with a straightforward integral, and computing E(X) and Var(X) using standard formulas. While it involves multiple parts and careful algebraic manipulation, all steps follow textbook procedures 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 integration |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | 6 |
| Answer | Marks |
|---|---|
| 36 | M1 |
| A1 | 2.1 |
| 1.1 | Must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | M1 | M1 |
| Answer | Marks |
|---|---|
| k(108(cid:16)72)(cid:32)1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 36 | M | |
| [2] | I | |
| 3 | (ii) | P(Weight ≥ 1505) = P(X ≥ 5) |
| Answer | Marks |
|---|---|
| (cid:32)0.0741 | C |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | soi |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (iii) | Either |
| Answer | Marks |
|---|---|
| = 3 | M1 |
| A1 | 1.1 |
| 1.1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| OR | N | |
| Convincing argument about symmetry of distribution | Convincing argument about symmetry of distribution | M1 |
| Answer | Marks |
|---|---|
| Mean = 3 | A1 |
| Answer | Marks |
|---|---|
| Standard deviation = 1.34… | M1 |
| Answer | Marks |
|---|---|
| [6] | I1.1 |
| Answer | Marks |
|---|---|
| 1.1 | M |
Question 3:
3 | (i) | 6
k(cid:179) x(6(cid:16)x)dx(cid:32)1
0
k(cid:117)36(cid:32)1
1
k (cid:32) AG
36 | M1
A1 | 2.1
1.1 | Must be seen
BC
N
Alternative Method
6
(cid:170) 1 (cid:186)
k 3x2 (cid:16) x3 (cid:32)1
(cid:171) (cid:187)
(cid:172) 3 (cid:188)
0 | M1 | M1 | E
E
k(108(cid:16)72)(cid:32)1 | A1
1
k (cid:32) AG
36 | M
[2] | I
3 | (ii) | P(Weight ≥ 1505) = P(X ≥ 5)
E
1 6
(cid:32) (cid:179) x(6(cid:16)x)dx
36 5
(cid:32)0.0741 | C
B1
M1
A1
[3] | 3.4
1.1a
1.1 | soi
6
1 (cid:170) 1 (cid:186)
3x2(cid:16) x3 may be seen
(cid:171) (cid:187)
36(cid:172) 3 (cid:188)
5
BC
3 | (iii) | Either
6
Mean = E(X)(cid:32)(cid:179) x2(6(cid:16)x)dx
0
= 3 | M1
A1 | 1.1
1.1 | 6
(cid:170) 1 (cid:186)
2x3(cid:16) x4 may be seen
(cid:171) (cid:187)
(cid:172) 4 (cid:188)
0
BC Must be exact
OR | N
Convincing argument about symmetry of distribution | Convincing argument about symmetry of distribution | M1 | M1 | Sketch without comment
insufficient.
E
Mean = 3 | A1
6
E(X2)(cid:32)(cid:179) x3(6(cid:16)x)dx
0
=10.8
Var(X) = 10.8 – 32 = 1.8
Standard deviation = 1.34… | M1
C
A1
M1
A1
[6] | I1.1
1.1
1.1
1.1 | M
6
(cid:170)3 1 (cid:186)
x4(cid:16) x5 may be seen
(cid:171) (cid:187)
(cid:172)2 5 (cid:188)
0
BC3 At a factory, flour is packed into bags. A model for the mass in grams of flour packed into each bag is $1500 + X$, where $X$ is a continuous random variable with probability density function
$$f ( x ) = \left\{ \begin{array} { c c }
k x ( 6 - x ) & 0 \leq x \leq 6 \\
0 & \text { elsewhere, }
\end{array} \right.$$
where $k$ is a constant.\\
(i) Show that $k = \frac { 1 } { 36 }$.\\
(ii) Find the probability that a randomly selected bag of flour contains 1505 grams of flour or more.\\
(iii) Find
\begin{itemize}
\item the mean of $X$,
\item the standard deviation of $X$.
\end{itemize}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS Q3 [11]}}