Question 4 - A-Level Maths

OCR MEI Further Statistics B AS 2018 June — Question 4 15 marks

Exam Board	OCR MEI
Module	Further Statistics B AS (Further Statistics B AS)
Year	2018
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Uniform Random Variables
Type	Sample statistics from uniform
Difficulty	Easy -1.2 This question tests basic properties of continuous uniform distributions (probability, mean, variance) which are standard formula applications, followed by interpretation of a simulation spreadsheet. Parts (i) and (ii) require only direct recall and substitution into well-known formulas for U(0,10). The simulation context adds mild complexity but doesn't require sophisticated statistical reasoning—students likely just need to verify or interpret the given data. This is easier than average A-level material.
Spec	5.03a Continuous random variables: pdf and cdf 5.03b Solve problems: using pdf 5.03c Calculate mean/variance: by integration 5.05a Sample mean distribution: central limit theorem

4 The random variable \(X\) has a continuous uniform distribution on [ 0,10 ].

Find \(\mathrm { P } ( 3 < X < 6 )\).

Find each of the following.

\(\mathrm { E } ( X )\)
\(\operatorname { Var } ( X )\)

Marisa is investigating the sample mean, \(Y\), of 8 independent values of \(X\). She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of \(X\) together with the value of \(Y\). All of the values in the spreadsheet have been rounded to 2 decimal places. \begin{table}[h]

1	A	B	C	D	E	F	G	H	I	J
1	\(X _ { 1 }\)	\(X _ { 2 }\)	\(X _ { 3 }\)	\(X _ { 4 }\)	\(X _ { 5 }\)	\(X _ { 6 }\)	\(X _ { 7 }\)	\(X _ { 8 }\)	\(Y\)
2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
27

\captionsetup{labelformat=empty} \caption{Fig. 4.1}

\end{table}

Use the spreadsheet to estimate \(\mathrm { P } ( 3 < Y < 6 )\).
Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i). Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\).
Calculate an estimate of \(\mathrm { P } ( 3 < Y < 6 )\) using the Central Limit Theorem. A Normal probability plot of the 25 simulated values of \(Y\) is shown in Fig. 4.2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}
Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate \(\mathrm { P } ( 3 < Y < 6 )\). Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of \(Y\) between 3 and 6. This value is 808.
Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\). Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.
Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
4	(i)	P(3 < X < 6) = 0.3

[1]

Answer	Marks
(ii)	E(X) = 5

Var(X) = 25/

Answer	Marks
3	B1

B1

Answer	Marks
[2]	Allow 8.3 or better
(iii)	P(3 < Y < 6)

= 22/ or 0.88

Answer	Marks
25	M1

A1

Answer	Marks
[2]	For division by 25
(iv)	Because the mean value of Y is equal to the mean

value of X,

but the variance of Y is only one eighth of the

Answer	Marks
variance of X.	E1

E1

Answer	Marks
[2]	Or allow ‘as sample size increases,

variance decreases’

Answer	Marks
(v)	Using CLT, Y approx ~ N(5, 25/ )

24

Answer	Marks
P(3 < Y < 6) approx = 0.811	B1

M1

A1

Answer	Marks
[3]	For mean and Normal

For variance Can be implied by

correct answer

Answer	Marks
BC awrt 0.811	FT their Var(X) for M1 only
(vi)	Normal probability plot is very close to a straight

line

So this suggests that the distribution may be

Normal and that the Central Limit Theorem may

Answer	Marks
provide a good approximation.	E1*

*E1dep

[2]

Answer	Marks
(vii)	Yes because the estimated probability from the

mean of 808 is 0.808 which is close to the clt value

Answer	Marks	Guidance
of 0.811	B1
[1]	Must compare with 0.811, not 0.88	Do NOT FT their answer to part (v)
(viii)	You would expect there to be less variation with

1000 rows

since as the number of rows increases, the

estimated probability tends to get closer to the true

Answer	Marks
probability	E1*

*E1dep

Answer	Marks
[2]	Need more than ‘more values give a

better estimate’ or ‘or sample size is

bigger’ or similar for second E1.

Question 4:
4 | (i) | P(3 < X < 6) = 0.3 | B1
[1]
(ii) | E(X) = 5
Var(X) = 25/
3 | B1
B1
[2] | Allow 8.3 or better
(iii) | P(3 < Y < 6)
= 22/ or 0.88
25 | M1
A1
[2] | For division by 25
(iv) | Because the mean value of Y is equal to the mean
value of X,
but the variance of Y is only one eighth of the
variance of X. | E1
E1
[2] | Or allow ‘as sample size increases,
variance decreases’
(v) | Using CLT, Y approx ~ N(5, 25/ )
24
P(3 < Y < 6) approx = 0.811 | B1
M1
A1
[3] | For mean and Normal
For variance Can be implied by
correct answer
BC awrt 0.811 | FT their Var(X) for M1 only
(vi) | Normal probability plot is very close to a straight
line
So this suggests that the distribution may be
Normal and that the Central Limit Theorem may
provide a good approximation. | E1*
*E1dep
[2]
(vii) | Yes because the estimated probability from the
mean of 808 is 0.808 which is close to the clt value
of 0.811 | B1
[1] | Must compare with 0.811, not 0.88 | Do NOT FT their answer to part (v)
(viii) | You would expect there to be less variation with
1000 rows
since as the number of rows increases, the
estimated probability tends to get closer to the true
probability | E1*
*E1dep
[2] | Need more than ‘more values give a
better estimate’ or ‘or sample size is
bigger’ or similar for second E1.

Show LaTeX source

4 The random variable $X$ has a continuous uniform distribution on [ 0,10 ].\\
(i) Find $\mathrm { P } ( 3 < X < 6 )$.\\
(ii) Find each of the following.

\begin{itemize}
  \item $\mathrm { E } ( X )$
  \item $\operatorname { Var } ( X )$
\end{itemize}

Marisa is investigating the sample mean, $Y$, of 8 independent values of $X$. She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of $X$ together with the value of $Y$. All of the values in the spreadsheet have been rounded to 2 decimal places.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline
1 & A & B & C & D & E & F & G & H & I & J \\
\hline
1 & $X _ { 1 }$ & $X _ { 2 }$ & $X _ { 3 }$ & $X _ { 4 }$ & $X _ { 5 }$ & $X _ { 6 }$ & $X _ { 7 }$ & $X _ { 8 }$ & $Y$ &  \\
\hline
2 & 6.31 & 2.45 & 3.27 & 3.06 & 4.16 & 1.53 & 0.43 & 7.99 & 3.65 &  \\
\hline
3 & 1.70 & 1.52 & 7.10 & 8.93 & 6.44 & 2.70 & 9.96 & 7.83 & 5.77 &  \\
\hline
4 & 9.15 & 0.52 & 4.95 & 6.99 & 6.52 & 3.15 & 0.81 & 5.35 & 4.68 &  \\
\hline
5 & 0.65 & 2.71 & 7.92 & 9.65 & 0.50 & 4.87 & 6.46 & 2.67 & 4.43 &  \\
\hline
6 & 3.09 & 6.11 & 3.96 & 0.09 & 0.18 & 4.67 & 0.67 & 6.20 & 3.12 &  \\
\hline
7 & 7.06 & 5.84 & 1.97 & 3.60 & 9.36 & 1.97 & 4.48 & 3.47 & 4.72 &  \\
\hline
8 & 1.46 & 1.57 & 5.45 & 0.37 & 3.76 & 7.56 & 8.48 & 9.12 & 4.72 &  \\
\hline
9 & 9.42 & 1.85 & 4.91 & 1.61 & 1.94 & 8.00 & 1.77 & 5.34 & 4.36 &  \\
\hline
10 & 2.98 & 5.32 & 2.91 & 4.12 & 9.16 & 1.76 & 9.97 & 6.88 & 5.39 &  \\
\hline
11 & 2.83 & 3.44 & 3.28 & 7.85 & 1.00 & 0.93 & 8.77 & 4.03 & 4.01 &  \\
\hline
12 & 4.51 & 0.59 & 5.84 & 9.87 & 8.65 & 3.94 & 7.18 & 0.23 & 5.10 &  \\
\hline
13 & 4.49 & 0.69 & 3.65 & 8.78 & 4.96 & 8.96 & 3.77 & 1.43 & 4.59 &  \\
\hline
14 & 6.57 & 8.08 & 4.85 & 6.75 & 7.92 & 0.27 & 9.69 & 4.04 & 6.02 &  \\
\hline
15 & 8.35 & 1.09 & 8.63 & 8.04 & 7.23 & 2.12 & 2.57 & 9.59 & 5.95 &  \\
\hline
16 & 5.24 & 9.53 & 6.08 & 8.21 & 3.61 & 7.07 & 6.65 & 7.63 & 6.75 &  \\
\hline
17 & 7.89 & 5.50 & 3.09 & 0.71 & 6.47 & 5.49 & 6.47 & 4.95 & 5.07 &  \\
\hline
18 & 8.36 & 7.27 & 2.35 & 9.04 & 0.58 & 2.26 & 3.01 & 7.90 & 5.10 &  \\
\hline
19 & 3.76 & 1.01 & 9.61 & 9.65 & 7.89 & 9.98 & 6.28 & 4.34 & 6.56 &  \\
\hline
20 & 9.94 & 6.84 & 3.38 & 5.53 & 0.26 & 8.53 & 5.72 & 5.12 & 5.66 &  \\
\hline
21 & 7.25 & 9.10 & 0.34 & 2.88 & 4.66 & 2.65 & 6.37 & 7.63 & 5.11 &  \\
\hline
22 & 7.18 & 7.14 & 5.38 & 0.04 & 4.09 & 6.47 & 4.96 & 4.23 & 4.94 &  \\
\hline
23 & 8.69 & 5.04 & 4.90 & 2.94 & 2.00 & 4.23 & 4.13 & 0.97 & 4.11 &  \\
\hline
24 & 3.46 & 6.33 & 0.48 & 9.35 & 0.23 & 1.18 & 7.97 & 6.37 & 4.42 &  \\
\hline
25 & 2.37 & 7.26 & 7.16 & 1.24 & 5.26 & 2.80 & 3.55 & 3.84 & 4.19 &  \\
\hline
26 & 2.16 & 8.30 & 7.17 & 3.32 & 2.96 & 1.30 & 9.11 & 0.31 & 4.33 &  \\
\hline
27 &  &  &  &  &  &  &  &  &  &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{table}

(iii) Use the spreadsheet to estimate $\mathrm { P } ( 3 < Y < 6 )$.\\
(iv) Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i).

Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to $\mathrm { P } ( 3 < Y < 6 )$.\\
(v) Calculate an estimate of $\mathrm { P } ( 3 < Y < 6 )$ using the Central Limit Theorem.

A Normal probability plot of the 25 simulated values of $Y$ is shown in Fig. 4.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}

(vi) Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate $\mathrm { P } ( 3 < Y < 6 )$.

Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of $Y$ between 3 and 6. This value is 808.\\
(vii) Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to $\mathrm { P } ( 3 < Y < 6 )$.

Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.\\
(viii) Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.

\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2018 Q4 [15]}}

This paper (6 questions)

View full paper

Q1 6 Q2 9 Q3 10 Q4 15 Q5 10 Q6 10