Question 9 - A-Level Maths

OCR MEI Further Statistics Major 2019 June — Question 9 15 marks

Exam Board	OCR MEI
Module	Further Statistics Major (Further Statistics Major)
Year	2019
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Uniform Random Variables
Type	Sum of independent uniforms
Difficulty	Moderate -0.5 This is a straightforward application of continuous uniform distribution properties and reading from a given PDF. Part (a) is trivial probability calculation (2/5), part (b) requires integration of a given piecewise function, and part (c) involves interpreting simulation data. All techniques are standard with no novel problem-solving required, making it slightly easier than average.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration

9 Every weekday Jonathan takes an underground train to work. On any weekday the time in minutes that he has to wait at the station for a train is modelled by the continuous uniform distribution over \([ 0,5 ]\).

Find the probability that Jonathan has to wait at least 3 minutes for a train. The total time that Jonathan has to wait on two days is modelled by the continuous random variable \(X\) with probability density function given by \(\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 25 } x & 0 \leqslant x \leqslant 5 , \\ \frac { 1 } { 25 } ( 10 - x ) & 5 < x \leqslant 10 , \\ 0 & \text { otherwise } . \end{cases}\)

Find the probability that Jonathan has to wait a total of at most 6 minutes on two days. Jonathan's friend suggests that the total waiting time for 5 days, \(T\) minutes, will almost certainly be less than 18 minutes. In order to investigate this suggestion, Jonathan constructs the simulation shown in Fig. 9. All of the numbers in the simulation have been rounded to 2 decimal places. \begin{table}[h]

	A	B	C	D	E	F
1	Mon	Tue	Wed	Thu	Fri	Total T
2	1.78	4.36	2.74	3.88	4.64	17.41
3	0.95	1.30	4.83	4.29	1.81	13.18
4	4.27	4.90	4.57	1.41	3.66	18.81
5	0.80	0.06	3.20	1.76	0.35	6.17
6	0.03	4.82	1.26	3.53	0.13	9.77
7	3.88	4.73	1.19	3.75	1.29	14.84
8	4.11	3.54	4.33	0.77	4.50	17.25
9	3.54	0.11	3.85	2.86	1.58	11.94
10	1.87	1.82	3.00	3.53	1.83	12.05
11	4.00	2.98	4.59	1.73	1.76	15.06
12	1.91	3.85	2.08	1.72	2.82	12.38
13	0.10	4.86	2.51	0.52	2.17	10.15
14	1.24	4.26	0.95	1.33	1.78	9.57
15	2.99	0.69	3.85	3.41	2.42	13.36
16	4.67	1.76	2.13	3.48	3.10	15.14
17	1.94	1.07	0.91	0.63	3.34	7.89
18	0.11	2.29	0.71	4.21	0.86	8.18
19	0.43	4.58	4.89	1.86	2.84	14.60
20	4.23	0.88	2.71	4.88	4.20	16.91
21	3.72	4.58	3.11	4.89	3.18	19.49

\captionsetup{labelformat=empty} \caption{Fig. 9}

\end{table}

Use the simulation to estimate \(\mathrm { P } ( T > 18 )\).
Explain how Jonathan could obtain a better estimate. Jonathan thinks that he can use the Central Limit Theorem to provide a very good approximation to the distribution of \(T\).
Find each of the following.
Jonathan travels to work on 200 days in a year.
Find the probability that the total waiting time for Jonathan in a year is more than 510 minutes.
[0pt] [3]

Show mark scheme Show mark scheme source

Question 9:

Answer	Marks	Guidance
9	(a)	P(at least 3 minutes) = 0.4

[1]

Answer	Marks	Guidance
Result	Res  33.5	Abs
value	Ran

k

Answer	Marks	Guidance
9	(b)	P(at most 6 minutes)

= 0.5

6  1 (10x)dx

5 25

Answer	Marks
= 0.5 + 0.18 = 0.68	M1

M1

A1

Answer	Marks
[3]	For addition of 0.5
oe use of area of trapezium or triangle	oe, using 1  P(>

6)

Answer	Marks	Guidance
9	(c)	Estimate of P(T 18) 2 0.1
20	B1

[1]

Answer	Marks	Guidance
9	(d)	Use more rows in the simulation
[1]	Allow ‘increase the sample size’, ‘use

more weeks’, etc

Answer	Marks	Guidance
9	(e)	E(T) = 12.5

Var(T)525

12 125 ( = 10.417)

Answer	Marks
12	B1

M1

A1

[3]

Answer	Marks	Guidance
9	(f)	 12.5,125

Total ⁓ N

12

Answer	Marks
Estimate of P(T > 18) = 0.0442	M1

A1

Answer	Marks
[2]	oe use of Mean for 5 days ~

N  2.5, 5 

12

Answer	Marks
BC	FT their 9(e) for

M1 only

Answer	Marks	Guidance
9	(g)	The sample size of 5 is small, so the CLT may not
give an accurate estimate	B1
[1]	Allow n is small, etc (but not 20 is

small)

Answer	Marks	Guidance
9	(h)	 2002.5,20025

Total ⁓ N

12  5000

so N 500, gives P(Total > 510) = 0.3121

Answer	Marks
12	B1

M1

A1

Answer	Marks
[3]	For Normal and correct mean (500)

For variance (416.67)

Answer	Marks
BC	oe, using the

distribution of the

mean time for a

year

Question 9:
9 | (a) | P(at least 3 minutes) = 0.4 | B1
[1]
Result | Res  33.5 | Abs
value | Ran
k
9 | (b) | P(at most 6 minutes)
= 0.5
6
 1 (10x)dx
5 25
= 0.5 + 0.18 = 0.68 | M1
M1
A1
[3] | For addition of 0.5
oe use of area of trapezium or triangle | oe, using 1  P(>
6)
9 | (c) | Estimate of P(T 18) 2 0.1
20 | B1
[1]
9 | (d) | Use more rows in the simulation | E1
[1] | Allow ‘increase the sample size’, ‘use
more weeks’, etc
9 | (e) | E(T) = 12.5
Var(T)525
12
125 ( = 10.417)
12 | B1
M1
A1
[3]
9 | (f) |  12.5,125
Total ⁓ N
12
Estimate of P(T > 18) = 0.0442 | M1
A1
[2] | oe use of Mean for 5 days ~
N  2.5, 5 
12
BC | FT their 9(e) for
M1 only
9 | (g) | The sample size of 5 is small, so the CLT may not
give an accurate estimate | B1
[1] | Allow n is small, etc (but not 20 is
small)
9 | (h) |  2002.5,20025
Total ⁓ N
12
 5000
so N 500, gives P(Total > 510) = 0.3121
12 | B1
M1
A1
[3] | For Normal and correct mean (500)
For variance (416.67)
BC | oe, using the
distribution of the
mean time for a
year

Show LaTeX source

9 Every weekday Jonathan takes an underground train to work. On any weekday the time in minutes that he has to wait at the station for a train is modelled by the continuous uniform distribution over $[ 0,5 ]$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Jonathan has to wait at least 3 minutes for a train.

The total time that Jonathan has to wait on two days is modelled by the continuous random variable $X$ with probability density function given by\\
$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 25 } x & 0 \leqslant x \leqslant 5 , \\ \frac { 1 } { 25 } ( 10 - x ) & 5 < x \leqslant 10 , \\ 0 & \text { otherwise } . \end{cases}$
\item Find the probability that Jonathan has to wait a total of at most 6 minutes on two days.

Jonathan's friend suggests that the total waiting time for 5 days, $T$ minutes, will almost certainly be less than 18 minutes. In order to investigate this suggestion, Jonathan constructs the simulation shown in Fig. 9. All of the numbers in the simulation have been rounded to 2 decimal places.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
 & A & B & C & D & E & F \\
\hline
1 & Mon & Tue & Wed & Thu & Fri & Total T \\
\hline
2 & 1.78 & 4.36 & 2.74 & 3.88 & 4.64 & 17.41 \\
\hline
3 & 0.95 & 1.30 & 4.83 & 4.29 & 1.81 & 13.18 \\
\hline
4 & 4.27 & 4.90 & 4.57 & 1.41 & 3.66 & 18.81 \\
\hline
5 & 0.80 & 0.06 & 3.20 & 1.76 & 0.35 & 6.17 \\
\hline
6 & 0.03 & 4.82 & 1.26 & 3.53 & 0.13 & 9.77 \\
\hline
7 & 3.88 & 4.73 & 1.19 & 3.75 & 1.29 & 14.84 \\
\hline
8 & 4.11 & 3.54 & 4.33 & 0.77 & 4.50 & 17.25 \\
\hline
9 & 3.54 & 0.11 & 3.85 & 2.86 & 1.58 & 11.94 \\
\hline
10 & 1.87 & 1.82 & 3.00 & 3.53 & 1.83 & 12.05 \\
\hline
11 & 4.00 & 2.98 & 4.59 & 1.73 & 1.76 & 15.06 \\
\hline
12 & 1.91 & 3.85 & 2.08 & 1.72 & 2.82 & 12.38 \\
\hline
13 & 0.10 & 4.86 & 2.51 & 0.52 & 2.17 & 10.15 \\
\hline
14 & 1.24 & 4.26 & 0.95 & 1.33 & 1.78 & 9.57 \\
\hline
15 & 2.99 & 0.69 & 3.85 & 3.41 & 2.42 & 13.36 \\
\hline
16 & 4.67 & 1.76 & 2.13 & 3.48 & 3.10 & 15.14 \\
\hline
17 & 1.94 & 1.07 & 0.91 & 0.63 & 3.34 & 7.89 \\
\hline
18 & 0.11 & 2.29 & 0.71 & 4.21 & 0.86 & 8.18 \\
\hline
19 & 0.43 & 4.58 & 4.89 & 1.86 & 2.84 & 14.60 \\
\hline
20 & 4.23 & 0.88 & 2.71 & 4.88 & 4.20 & 16.91 \\
\hline
21 & 3.72 & 4.58 & 3.11 & 4.89 & 3.18 & 19.49 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{table}
\item Use the simulation to estimate $\mathrm { P } ( T > 18 )$.
\item Explain how Jonathan could obtain a better estimate.

Jonathan thinks that he can use the Central Limit Theorem to provide a very good approximation to the distribution of $T$.
\item Find each of the following.

\begin{itemize}
  \item $\mathrm { E } ( T )$
  \item $\operatorname { Var } ( T )$
\item Use the Central Limit Theorem to estimate $\mathrm { P } ( T > 18 )$.
\item Comment briefly on the use of the Central Limit Theorem in this case.
\end{itemize}

Jonathan travels to work on 200 days in a year.
\item Find the probability that the total waiting time for Jonathan in a year is more than 510 minutes.\\[0pt]
[3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2019 Q9 [15]}}

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