AS June 2023 Q3

EdexcelAS paperCurrent spec16 marksIncludes hypothesis testingPoisson Distribution

3. A machine produces cloth. Faults occur randomly in the cloth at a rate of 0.4 per square metre.

The machine is used to produce tablecloths, each of area \(A\) square metres. One of these tablecloths is taken at random.

The probability that this tablecloth has no faults is 0.0907

(a) Find the value of \(A\) (3)

The tablecloths are sold in packets of 20

A randomly selected packet is taken.

(b) Find the probability that more than 1 of the tablecloths in this packet has no faults. (3)

A hotel places an order for 100 tablecloths each of area \(A\) square metres.

The random variable \(X\) represents the number of these tablecloths that have no faults.

(i) \(\mathrm{E}(X)\)

(ii) \(\mathrm{Var}(X)\) (3)

(d) Use a Poisson approximation to estimate \(\mathrm{P}(X = 10)\) (2)

It is claimed that a new machine produces cloth with a rate of faults that is less than 0.4 per square metre.

A piece of cloth produced by this new machine is taken at random.
The piece of cloth has area 30 square metres and is found to have 6 faults.

(e) Stating your hypotheses clearly, use a suitable test to assess the claim made for the new machine. Use a 5% level of significance. (4)

(f) Write down the p-value for the test used in part (e). (1)

Mark scheme (a)

Scheme	Marks	AO
[\(F\) = no of faults in \(A\) m²] \(F \sim \mathrm{Po}(A \times 0.4)\)	M1	3.3
[\(\mathrm{P}(F = 0) \Rightarrow\)] \(0.0907 = \mathrm{e}^{-0.4A}\)	M1	3.4
\(A = \underline{\mathbf{6}}\)	A1	1.1b
	(3)

Notes

1^st M1 for selecting the correct model \(\mathrm{Po}(0.4A)\)

2^nd M1 for a correct equation - may be implied by a correct answer with no incorrect working

A1 for \(A = 6\) (or awrt 6.0)

Mark scheme (b)

Scheme	Marks	AO
[\(T\) = no. of tablecloths with no faults] \(T \sim \mathrm{B}(20, 0.0907)\)	M1	3.3
\(\mathrm{P}(T \gt 1) = 1 - \mathrm{P}(T \leqslant 1)\)	M1	1.1b
\(= 0.55276\ldots\) = awrt 0.553	A1	3.4
	(3)

Notes

1^st M1 for selecting a correct model \(\mathrm{B}(20, 0.0907)\) used, or seen if only distribution in (b)

2^nd M1 for correctly interpreting “more than 1” to reach \(1 - \mathrm{P}(T \leqslant 1)\). May be implied by A1

A1 for awrt 0.553

Mark scheme (c)

Scheme	Marks	AO
[\(X \sim \mathrm{B}(100, 0.0907)\)]	M1	3.3
(i) \(\mathrm{E}(X) = 100 \times 0.0907 = \underline{\mathbf{9.07}}\)	A1	1.1b
(ii) \(\mathrm{Var}(X) = 100 \times 0.0907 \times (1 - 0.0907) = 8.247351\) awrt 8.25	A1	1.1b
	(3)

Notes

(i) M1 for \(X \sim \mathrm{B}(100, 0.0907)\) used, or seen if only distribution in (c).
May be implied by correct \(\mathrm{E}(X)\) or \(\mathrm{Var}(X)\)

A1 for 9.07

(ii) A1 for awrt 8.25 SC - award M0A1A0 for:

using \(X \sim \mathrm{B}(100, 0.4)\) leading to \(\mathrm{E}(X) = 40\), \(\mathrm{Var}(X) = 24\)
using \(X \sim \mathrm{B}(100, p)\), \(0 \lt p \lt 1\) and \(\mathrm{E}(X) = 100p\), \(\mathrm{Var}(X) = 100p(1-p)\)

Mark scheme (d)

Scheme	Marks	AO
\(X \approx\ \sim \mathrm{Po}(9.07)\);	M1	3.4
\(\mathrm{P}(X = 10) \approx 0.11947\ldots\) 0.1195 or awrt 0.119	A1	1.1b
	(2)

Notes

M1 for selecting the correct Poisson model – ft their answer to (c)(i)

A1 for 0.1195 or awrt 0.119

Mark scheme (e)

Scheme	Marks	AO
\(\mathrm{H}_0: \lambda = 0.4\) (or \(\lambda = 12\)) \(\qquad \mathrm{H}_1: \lambda \lt 0.4\) (or \(\lambda \lt 12\))	B1	2.5
[\(Y\) = no. of faults from new machine] \(Y \sim \mathrm{Po}(12)\)	M1	1.1b/3.3
\(\mathrm{P}(Y \leqslant 6) = 0.04582\ldots\)	A1	3.4
[Significant] there is evidence to support the claim	A1	2.2b
	(4)

Notes

B1 for both hypotheses correct in terms of \(\lambda\) or \(\mu\)

M1 for selecting a suitable model. Sight or use of \(\mathrm{Po}(12)\). May be implied by 1^st A1

1^st A1 for a correct probability must be 0.046 or better

2^nd A1 for a correct conclusion in context using “claim” or “rate of faults”

Scheme	Marks	AO
\(p\)-value \(= 0.04582\ldots\) awrt 0.0458	B1ft	1.2
	(1)
	(16)