| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Probability Distributions and Random Variables |
| Type | Calculate probability P(X in interval) |
| Difficulty | Standard +0.8 This is a substantial Further Maths statistics question requiring multiple techniques: finding constants using continuity conditions, probability calculations from CDFs, solving cubic equations for the median, computing mean and variance by integration, and comparing distributions conceptually. While each individual step uses standard methods, the length (7 parts), the need to work with both CDF and PDF, and the requirement for integration and cubic equation solving make this moderately challenging for Further Maths level. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (a) | F(4) = 1 so k(8×42 – 43 – 24) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 40 | B1 | |
| [1] | 2.1 | AG |
| 11 | (b) | P( 2.5 ≤ T ≤ 3.5) = F(3.5) – F(2.5) |
| Answer | Marks |
|---|---|
| = 0.51875 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (c) | 2 3 |
| Answer | Marks |
|---|---|
| 2 3 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 1.1 | Condoned additional substitution of 2 in equation since this gives |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (d) | (8𝑚𝑚 – 𝑚𝑚 – 24)=20 |
| Answer | Marks |
|---|---|
| So median is 2.95 to 2 dp | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | For either |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (e) | f(t)= 1 (16t−3t2) |
| Answer | Marks |
|---|---|
| = F(3.531) – F(2.402) = 0.586 | B1 |
| Answer | Marks |
|---|---|
| [7] | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | f | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | For main part of graph |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | g | The probability will be less than the Normal probability |
| Answer | Marks |
|---|---|
| is involved | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.2a | For partial explanation. |
Question 11:
11 | (a) | F(4) = 1 so k(8×42 – 43 – 24) = 1
So 40k = 1 so k = 1
40 | B1
[1] | 2.1 | AG
11 | (b) | P( 2.5 ≤ T ≤ 3.5) = F(3.5) – F(2.5)
= 1 (8×3.52 −3.53−24)− 1 (8×2.52 −2.53−24)
40 40
= 0.51875 | M1
A1
[2] | 1.1a
1.1
11 | (c) | 2 3
0.02 5(8𝑚𝑚 – 𝑚𝑚 – 2 4)=0.5
2 3 | M1
A1
[2] | 2.1
1.1 | Condoned additional substitution of 2 in equation since this gives
zero
AG
11 | (d) | (8𝑚𝑚 – 𝑚𝑚 – 24)=20
F(32.945) =2 0.496
𝑚𝑚 −8𝑚𝑚 +44=0
F(2.955) = 0.501
So median is 2.95 to 2 dp | B1
E1
[2] | 3.4
1.1 | For either
OR
Calculation of
For 2.945 (= 0.1538) 2
For 2.955 (= –0𝑚𝑚.05−3)8 𝑚𝑚 +44
So change of sign
11 | (e) | f(t)= 1 (16t−3t2)
40
E(T) =
4
2
= 22.9 6 7
∫ 0 . 0 25𝑡𝑡(16𝑡𝑡 – 3𝑡𝑡 )𝑑𝑑𝑡𝑡
E(T2) =
= 9.12 4 2 2
∫2 0.025𝑡𝑡 (16𝑡𝑡 – 3𝑡𝑡 )𝑑𝑑𝑡𝑡
Var(T)=9.12−2.9672 =0.3189
SD = 0.5647
P( µ – σ < T < µ + σ) = P(2.402 < T < 3.531).
= F(3.531) – F(2.402) = 0.586 | B1
M1
A1
M1
A1
M1
A1
[7] | 3.1a
1.1
1.1
1.1
1.1
3.3
1.1 | BC
BC
11 | f | B1
B1
[2] | 1.1
1.1 | For main part of graph
For axes and for part where f(t) = 0
Numbers not required on F(t) axis but
2 and 4 required on t-axis
11 | g | The probability will be less than the Normal probability
because the graph of the Normal distribution is more
peaked and only the central section of the Normal curve
is involved | E1
E1
[2] | 2.2a
2.2a | For partial explanation.
For full explanation
Must clearly state which is greater to get any marks at all.
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For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored11 The length of time in minutes for which a particular geyser erupts is modelled by the continuous random variable $T$ with cumulative distribution function given by\\
$\mathrm { F } ( t ) = \begin{cases} 0 & t \leqslant 2 , \\ k \left( 8 t ^ { 2 } - t ^ { 3 } - 24 \right) & 2 < t < 4 , \\ 1 & t \geqslant 4 , \end{cases}$\\
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 1 } { 40 }$.
\item Find the probability that a randomly selected eruption time lies between 2.5 and 3.5 minutes.
\item Show that the median $m$ of the distribution satisfies the equation $m ^ { 3 } - 8 m ^ { 2 } + 44 = 0$.
\item Verify that the median eruption time is 2.95 minutes, correct to 2 decimal places.
The mean and standard deviation of $T$ are denoted by $\mu$ and $\sigma$ respectively.
\item Find $\mathrm { P } ( \mu - \sigma < T < \mu + \sigma )$.
\item Sketch the graph of the probability density function of $T$.
\item A Normally distributed random variable $X$ has the same mean and standard deviation as $T$.
By considering the shape of the Normal distribution, and without doing any calculations, explain whether $\mathrm { P } ( \mu - \sigma < X < \mu + \sigma )$ will be greater than, equal to or less than the probability that you calculated in part (e).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q11 [18]}}