| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Total journey time probabilities |
| Difficulty | Moderate -0.3 This is a straightforward application of normal distribution properties and linear combinations. Parts (a-c) involve standard calculations: single normal probability, sum of independent normals, and scaling a normal distribution. Part (d) requires a brief comment on independence. The question is slightly easier than average because it clearly signposts the required techniques, involves routine calculations with no conceptual surprises, and the independence assumption is explicitly stated, making it a textbook-style exercise rather than requiring problem-solving insight. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions |
| \multirow{2}{*}{} | Journey time (minutes) | |
| \cline { 2 - 3 } | Mean | Standard deviation |
| A to B | 21 | 3 |
| B to C | 29 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | P(< 23) = 0.748 (0.74750...) |
| [1] | BC | |
| (b) | A to C distribution X : N(50, 25) + 1 minute |
| Answer | Marks |
|---|---|
| = 0.421 (0.42074...) | B1 |
| Answer | Marks |
|---|---|
| [4] | For Normal and mean |
| Answer | Marks |
|---|---|
| (c) | Total time : N(145, 80) |
| P( > 150) = 0.288 (0.28807...) | B1 |
| Answer | Marks |
|---|---|
| [3] | For mean |
| Answer | Marks |
|---|---|
| BC | Or mean = 5×29 |
| Answer | Marks |
|---|---|
| (d) | It seems likely that the assumption is valid, since |
| Answer | Marks |
|---|---|
| another day. | B1 |
| [1] | Allow sensible alternative such as |
| Answer | Marks |
|---|---|
| the assumption is not valid.’ | Or eg ‘Not independent |
Question 3:
3 | (a) | P(< 23) = 0.748 (0.74750...) | B1
[1] | BC
(b) | A to C distribution X : N(50, 25) + 1 minute
stopped
P(Total Journey < 50) = P(X < 49)
= 0.421 (0.42074...) | B1
B1
M1
A1
[4] | For Normal and mean
For variance
oe, using N(51, 25) and P(< 50)
BC
(c) | Total time : N(145, 80)
P( > 150) = 0.288 (0.28807...) | B1
B1
B1
[3] | For mean
For variance
BC | Or mean = 5×29
variance = 5×42
(d) | It seems likely that the assumption is valid, since
it is unlikely that a delay on one day would affect
another day. | B1
[1] | Allow sensible alternative such as
‘There may be bad weather for a
particular week which might make
journeys slower and mean that the
assumption is not valid.’
‘There may be roadworks for a
period of some days which would
affect journey times and mean that
the assumption is not valid.’ | Or eg ‘Not independent
because if buses run late on
one day, it may affect the
service on the next day’3 A bus runs from point A on the outskirts of a city, stops at point B outside the rail station, and continues to point C in the city centre.\\
The journey times for the sections A to B and B to C vary according to traffic conditions, and are modelled by independent Normal distributions with means and standard deviations as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
\multirow{2}{*}{} & \multicolumn{2}{|c|}{Journey time (minutes)} \\
\cline { 2 - 3 }
& Mean & Standard deviation \\
\hline
A to B & 21 & 3 \\
\hline
B to C & 29 & 4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find the probability that a randomly chosen journey from A to B takes less than the scheduled time of 23 minutes.
For every journey, the bus stops for 1 minute when it reaches B to drop off and pick up passengers.
\item Find the probability that a randomly chosen journey from A to C takes less than the scheduled time of 50 minutes.
Mary travels on the bus from the station at B to her workplace at C every working day. You should assume that times for her bus journeys on different days are independent.
\item Find the probability that the total time taken for her five journeys on the bus in a randomly chosen week is at least $2 \frac { 1 } { 2 }$ hours.
\item Comment on the assumption that times on different days are independent.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2019 Q3 [9]}}