| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Continuous CDF with polynomial pieces |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring multiple techniques: finding k using continuity conditions, probability calculations using the CDF, solving a quadratic equation for the median, and comparing measures of central tendency by finding the PDF (via differentiation) and calculating the mean (via integration). The multi-part nature, requirement to differentiate/integrate the CDF, and conceptual understanding of mode/mean/median relationships make this moderately challenging, though each individual step follows standard procedures. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | [F(2) = 1 so] k(12×2 – 22) = 1 |
| Answer | Marks |
|---|---|
| 20 | M1 |
| Answer | Marks |
|---|---|
| (b) | P(1 ≤ X ≤ 1.5) = F(1.5) – F(1) |
| Answer | Marks |
|---|---|
| = 0.2375 | M1 |
| Answer | Marks |
|---|---|
| (c) | 0.05(12m – m2) = 0.5 |
| Answer | Marks |
|---|---|
| m = 0.901 (reject m = 11.099) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [3] | BC | Accept 6 – √26 (0.9009) |
| (d) | f(x) = 0.05(12 – 2x) |
| Answer | Marks |
|---|---|
| The mean is the largest. | M1 |
| Answer | Marks |
|---|---|
| [5] | For differentiation |
| Answer | Marks |
|---|---|
| obtained | Max M1M1A1M0A0 if the |
Question 4:
4 | (a) | [F(2) = 1 so] k(12×2 – 22) = 1
So 20k = 1 or k = 1/ so k = 0.05
20 | M1
A1
[2]
(b) | P(1 ≤ X ≤ 1.5) = F(1.5) – F(1)
= 0.05(12 1.5 1.52) 0.05(12 1 12)
= 0.2375 | M1
A1
[2]
(c) | 0.05(12m – m2) = 0.5
0.05m2 – 0.6m + 0.5 = 0
m = 0.901 (reject m = 11.099) | M1
M1
A1
[3] | BC | Accept 6 – √26 (0.9009)
(d) | f(x) = 0.05(12 – 2x)
2 2 3 1
E(X) 0.05x(122x)dxor x( x)dx
0 0 5 10
= 0.933
Maximum value (in range of x-values) = 0 so
mode = 0
The mean is the largest. | M1
M1
A1
M1
A1
[5] | For differentiation
BC
For using range
Dep on all 3 values correctly
obtained | Max M1M1A1M0A0 if the
mode is not found
Must justify the value of the
mode4 The cumulative distribution function of the continuous random variable $X$ is given by\\
$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 , \\ k \left( 12 x - x ^ { 2 } \right) & 0 \leqslant x \leqslant 2 , \\ 1 & x > 2 , \end{cases}$ where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 0.05$.
\item Find $\mathrm { P } ( 1 \leqslant X \leqslant 1.5 )$.
\item Find the median of $X$, correct to 3 significant figures.
\item Find which of the median, mean and mode of $X$ is the largest of the three measures of central tendency.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2019 Q4 [12]}}