Question 2 - A-Level Maths

OCR MEI Further Statistics B AS 2021 November — Question 2 10 marks

Exam Board	OCR MEI
Module	Further Statistics B AS (Further Statistics B AS)
Year	2021
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear combinations of normal random variables
Type	Pure expectation and variance calculation
Difficulty	Moderate -0.3 This is a straightforward application of standard results for linear combinations of binomial distributions and the Central Limit Theorem. Parts (a)-(c) require only reading from a spreadsheet and basic reasoning about binomial distributions. Part (d) is a routine CLT application requiring calculation of E(Z) and Var(Z) using linearity properties, then standardizing—all standard techniques with no novel insight required. Slightly easier than average due to the guided structure.
Spec	2.03a Mutually exclusive and independent events 2.04b Binomial distribution: as model B(n,p)

2 Natasha is investigating binomial distributions. She constructs the spreadsheet in Fig. 2 which shows the first 3 and last 4 rows of a simulation involving two independent variables, \(X\) and \(Y\), with distributions \(\mathrm { B } ( 10,0.3 )\) and \(\mathrm { B } ( 50,0.3 )\) respectively. The spreadsheet also shows the corresponding value of the random variable \(Z\), defined by \(Z = 5 X - Y\), for each pair of values of \(X\) and \(Y\). There are 100 simulated values of each of \(X , Y\) and \(Z\). The spreadsheet also shows whether each value of \(Z\) is greater than 6, and cells D103 and D104 show the number of values of \(Z\) which are greater than 6 and not greater than 6 respectively. \begin{table}[h]

1	A	B	C	D	E
1	X	Y	\(\mathbf { Z } = \mathbf { 5 } \mathbf { X } - \mathbf { Y }\)	\(\mathbf { Z } > \mathbf { 6 }\)
2	4	13	7	Y
3	4	17	3	N
4	3	21	-6	N
5
6
98	1	14	-9	N
99	5	12	13	Y
100	3	18	-3	N
101	3	15	0	N
102
103			Number of Y	19
104			Number of N	81
105

\captionsetup{labelformat=empty} \caption{Fig. 2}

\end{table}

Use the information in the spreadsheet to write down an estimate of \(\mathrm { P } ( Z > 6 )\).
Explain how a more reliable estimate of \(\mathrm { P } ( Z > 6 )\) could be obtained.
1. State the greatest possible value of \(Z\).
2. Explain why it is very unlikely that \(Z\) would have this value.
Use the Central Limit Theorem to calculate an estimate of the probability that the mean of 20 independent values of \(Z\) is greater than 2 .

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	Estimate =

19 = 0.19

Answer	Marks	Guidance
100	B1
[1]	1.1
2	(b)	By using more rows in the spreadsheet.
[1]	1.1
2	(c)	(i)
[1]	1.1
2	(c)	(ii)
very very small.	E1
[1]	3.2b
2	(d)	E(Z) = 0

Var(Z) = 52 × 2.1 + 10.5

= 63

So mean of 20 values of Z ~ approx. N(0, 63/20)

P(Mean > 2) = P(Normal > 2.025)

Answer	Marks
= 0.127	B1

M1

A1

M1

B1

A1

Answer	Marks
[6]	1.1

1.1a

1.1

3.3

3.4

Answer	Marks	Guidance
1.1	For continuity correction	If no continuity correction

answer is 0.130 which can

score max 5/6

Question 2:
2 | (a) | Estimate =
19
= 0.19
100 | B1
[1] | 1.1
2 | (b) | By using more rows in the spreadsheet. | E1
[1] | 1.1
2 | (c) | (i) | Max value = 50 | E1
[1] | 1.1
2 | (c) | (ii) | Very unlikely since Y has to be 0 and P(Y = 0) is
very very small. | E1
[1] | 3.2b
2 | (d) | E(Z) = 0
Var(Z) = 52 × 2.1 + 10.5
= 63
So mean of 20 values of Z ~ approx. N(0, 63/20)
P(Mean > 2) = P(Normal > 2.025)
= 0.127 | B1
M1
A1
M1
B1
A1
[6] | 1.1
1.1a
1.1
3.3
3.4
1.1 | For continuity correction | If no continuity correction
answer is 0.130 which can
score max 5/6

Show LaTeX source

2 Natasha is investigating binomial distributions. She constructs the spreadsheet in Fig. 2 which shows the first 3 and last 4 rows of a simulation involving two independent variables, $X$ and $Y$, with distributions $\mathrm { B } ( 10,0.3 )$ and $\mathrm { B } ( 50,0.3 )$ respectively. The spreadsheet also shows the corresponding value of the random variable $Z$, defined by $Z = 5 X - Y$, for each pair of values of $X$ and $Y$.

There are 100 simulated values of each of $X , Y$ and $Z$. The spreadsheet also shows whether each value of $Z$ is greater than 6, and cells D103 and D104 show the number of values of $Z$ which are greater than 6 and not greater than 6 respectively.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
1 & A & B & C & D & E \\
\hline
1 & X & Y & $\mathbf { Z } = \mathbf { 5 } \mathbf { X } - \mathbf { Y }$ & $\mathbf { Z } > \mathbf { 6 }$ &  \\
\hline
2 & 4 & 13 & 7 & Y &  \\
\hline
3 & 4 & 17 & 3 & N &  \\
\hline
4 & 3 & 21 & -6 & N &  \\
\hline
5 &  &  &  &  &  \\
\hline
6 &  &  &  &  &  \\
\hline
98 & 1 & 14 & -9 & N &  \\
\hline
99 & 5 & 12 & 13 & Y &  \\
\hline
100 & 3 & 18 & -3 & N &  \\
\hline
101 & 3 & 15 & 0 & N &  \\
\hline
102 &  &  &  &  &  \\
\hline
103 &  &  & Number of Y & 19 &  \\
\hline
104 &  &  & Number of N & 81 &  \\
\hline
105 &  &  &  &  &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the information in the spreadsheet to write down an estimate of $\mathrm { P } ( Z > 6 )$.
\item Explain how a more reliable estimate of $\mathrm { P } ( Z > 6 )$ could be obtained.
\item \begin{enumerate}[label=(\roman*)]
\item State the greatest possible value of $Z$.
\item Explain why it is very unlikely that $Z$ would have this value.
\end{enumerate}\item Use the Central Limit Theorem to calculate an estimate of the probability that the mean of 20 independent values of $Z$ is greater than 2 .
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2021 Q2 [10]}}

This paper (6 questions)

View full paper

Q1 9 Q2 10 Q3 12 Q4 9 Q5 9 Q6 11