Question 5 - A-Level Maths

OCR MEI Further Statistics Major 2021 November — Question 5 17 marks

Exam Board	OCR MEI
Module	Further Statistics Major (Further Statistics Major)
Year	2021
Session	November
Marks	17
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Z-tests (known variance)
Type	One-tail z-test (upper tail)
Difficulty	Standard +0.3 This is a straightforward Further Maths statistics question requiring standard techniques: (a) and (b) involve routine linear combinations of normal distributions, while (c) is a textbook one-sample t-test with clearly stated hypotheses and standard procedure. The calculations are mechanical with no conceptual challenges or novel insights required, making it slightly easier than average even for Further Maths content.
Spec	5.04b Linear combinations: of normal distributions 5.05c Hypothesis test: normal distribution for population mean

5 A manufacturer uses three types of capacitor in a particular electronic device. The capacitances, measured in suitable units, are modelled by independent Normal distributions with means and standard deviations as shown in the table.

\cline { 2 - 3 } \multicolumn{1}{c|}{}

Capacitance

Type

Mean

Standard

deviation

3.9

0.32

7.8

0.41

30.2

0.64

Determine the probability that the total capacitance of a randomly chosen capacitor of Type B and two randomly chosen capacitors of Type A is at least 16 units.
Determine the probability that the capacitance of a randomly chosen capacitor of Type C is within 1 unit of the total capacitance of four randomly chosen capacitors of Type B. When the manufacturer gets a new batch of 1000 capacitors from the supplier, a random sample of 10 of them is tested to check the capacitances. For a new batch of Type C capacitors, summary statistics for the capacitances, $x$ units, of the random sample are as follows. $n = 10$ $$\sum x = 299.6 \quad \sum x ^ { 2 } = 8981.0$$ You should assume that the capacitances of the sample come from a Normally distributed population, but you should not assume that the standard deviation is 0.64 as for previous Type C capacitors.
In this question you must show detailed reasoning. Carry out a hypothesis test at the $5 \%$ significance level to check whether it is reasonable to assume that the capacitors in this batch have the specified mean capacitance for Type C of 30.2 units.

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5	(a)	Two A and one B ⁓ N(2 × 3.9 + 7.8, 2 × 0.322 + 0.412)

N(15.6, 0.3729)

Answer	Marks
P(≥ 16) = 0. 256 (0.25622…)	B1

M1

A1

Answer	Marks
[3]	3.3

1.1

Answer	Marks
3.4	For N and mean

For variance

Answer	Marks
BC	Allow if N stated

anywhere in answer

SOI

Answer	Marks	Guidance
5	(b)	Four B – one C ⁓ N(4×7.8 – 30.2, 4×0.412 + 0.642)

N(1, 1.082)

Answer	Marks
P(within 1 unit) = 0.473 (0.47274…)	B1

M1

A1

Answer	Marks
[3]	3.3

1.1

Answer	Marks
3.4	For N and mean

For variance

Answer	Marks
BC	Allow -1 for mean

Allow if N stated

anywhere in answer

SOI

Answer	Marks	Guidance
5	(c)	DR

H : μ = 30.2 H : μ ≠ 30.2

0 1

where μ is the population mean capacitance

Sample mean = 29.96

1( 299.62)

Est. population variance = 8981.0−

9 10

= 0.5538

29.96−30.2

Test statistic =

0.5538

10 = −1.020

Refer to t

9 Critical value (2-tailed) at 5% level is 2.262

−1.020 > –2.262 so not significant (do not reject H )

0 Insufficient evidence to suggest that the capacitance of

Answer	Marks
the batch is different from 30.2	B1

B1

M1

A1

M1

A1

M1

A1

M1

E1

Answer	Marks
[11]	3.3

1.2

1.1

3.3

1.1

3.4

1.1 2.2b

Answer	Marks
3.5a	Hypotheses in words only must

include “population”

For definition in context

FT their mean and/or sd

BC

No FT if not t

9

Answer	Marks
Or 1.020 < 2.262	Or sd = 0.7442

Or

P(t < ‒1.020) = 0.1672

Or 0.1672 > 0.025

Answer must be in

context

Question 5:
5 | (a) | Two A and one B ⁓ N(2 × 3.9 + 7.8, 2 × 0.322 + 0.412)
N(15.6, 0.3729)
P(≥ 16) = 0. 256 (0.25622…) | B1
M1
A1
[3] | 3.3
1.1
3.4 | For N and mean
For variance
BC | Allow if N stated
anywhere in answer
SOI
5 | (b) | Four B – one C ⁓ N(4×7.8 – 30.2, 4×0.412 + 0.642)
N(1, 1.082)
P(within 1 unit) = 0.473 (0.47274…) | B1
M1
A1
[3] | 3.3
1.1
3.4 | For N and mean
For variance
BC | Allow -1 for mean
Allow if N stated
anywhere in answer
SOI
5 | (c) | DR
H : μ = 30.2 H : μ ≠ 30.2
0 1
where μ is the population mean capacitance
Sample mean = 29.96
1( 299.62)
Est. population variance = 8981.0−
9 10
= 0.5538
29.96−30.2
Test statistic =
0.5538
10
= −1.020
Refer to t
9
Critical value (2-tailed) at 5% level is 2.262
−1.020 > –2.262 so not significant (do not reject H )
0
Insufficient evidence to suggest that the capacitance of
the batch is different from 30.2 | B1
B1
B1
M1
A1
M1
A1
M1
A1
M1
E1
[11] | 3.3
1.2
1.1
1.1
1.1
3.3
1.1
3.4
1.1
2.2b
3.5a | Hypotheses in words only must
include “population”
For definition in context
FT their mean and/or sd
BC
No FT if not t
9
Or 1.020 < 2.262 | Or sd = 0.7442
Or
P(t < ‒1.020) = 0.1672
Or 0.1672 > 0.025
Answer must be in
context

Show LaTeX source

5 A manufacturer uses three types of capacitor in a particular electronic device. The capacitances, measured in suitable units, are modelled by independent Normal distributions with means and standard deviations as shown in the table.

\begin{center}
\begin{tabular}{ | c | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & \multicolumn{2}{c|}{Capacitance} \\
\hline
Type & Mean & \begin{tabular}{ c }
Standard \\
deviation \\
\end{tabular} \\
\hline
A & 3.9 & 0.32 \\
\hline
B & 7.8 & 0.41 \\
\hline
C & 30.2 & 0.64 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that the total capacitance of a randomly chosen capacitor of Type B and two randomly chosen capacitors of Type A is at least 16 units.
\item Determine the probability that the capacitance of a randomly chosen capacitor of Type C is within 1 unit of the total capacitance of four randomly chosen capacitors of Type B.

When the manufacturer gets a new batch of 1000 capacitors from the supplier, a random sample of 10 of them is tested to check the capacitances. For a new batch of Type C capacitors, summary statistics for the capacitances, $x$ units, of the random sample are as follows.\\
$n = 10$

$$\sum x = 299.6 \quad \sum x ^ { 2 } = 8981.0$$

You should assume that the capacitances of the sample come from a Normally distributed population, but you should not assume that the standard deviation is 0.64 as for previous Type C capacitors.
\item In this question you must show detailed reasoning.

Carry out a hypothesis test at the $5 \%$ significance level to check whether it is reasonable to assume that the capacitors in this batch have the specified mean capacitance for Type C of 30.2 units.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2021 Q5 [17]}}

This paper (11 questions)

View full paper

Q1 6 Q2 13 Q3 10 Q4 8 Q5 17 Q6 14 Q7 10 Q8 16 Q9 6 Q10 9 Q11 11