AS June 2022 Q2

EdexcelAS paperCurrent spec10 marksIncludes hypothesis testingPoisson Distribution

2. Xena catches fish at random, at a constant rate of 0.6 per hour.

(a) Find the probability that Xena catches exactly 4 fish in a 5-hour period. (2)

The probability of Xena catching no fish in a period of \(t\) hours is less than 0.16

(b) Find the minimum value of \(t\), giving your answer to one decimal place. (3)

Independently of Xena, Zion catches fish at random with a mean rate of 0.8 per hour.

Xena and Zion try using new bait to catch fish. The number of fish caught in total by Xena and Zion after using the new bait, in a randomly selected 4-hour period, is 12

(c) Use a suitable test to determine, at the 5% level of significance, whether or not there is evidence that the rate at which fish are caught has increased after using the new bait. State your hypotheses clearly and the p-value used in your test. (5)

Scheme	Marks	AO
\(X \sim \mathrm{Po}(3)\)	M1	3.3
\(\mathrm{P}(X = 4) = 0.1680\ldots\)	A1	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{e}^{-0.6 \times t} \lt 0.16\) oe	M1	3.1b
\(-0.6 \times t \lt \ln 0.16\)	dM1	1.1b
[\(t \gt 3.054\ldots\)] \(t = 3.1\)	A1	1.1b
	(3)

Notes

M1: Forming a correct equation from the information given. Condone \(\mathrm{e}^{-0.6 \times t} = 0.16\) or finding \(\mathrm{P}(X = 0)\) for [\(t = 3.1\)] 0.155… and [\(t = 3\)] 0.165… or \(\mathrm{P}(X = 0)\) for [\(\lambda = 1.84\)] 0.158… and [\(\lambda = 1.83\)] 0.1604…

dM1: Dependent on the 1st method mark. A correct method to solve their inequality/equation.
Or [\(t = 3.05\)] 0.1604 or [\(\lambda = 1.835\)] 0.159…

A1: 3.1

NB An answer of 3.1 gains 3/3

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{H}_0: \lambda = 1.4 \qquad \mathrm{H}_1: \lambda \gt 1.4\)	B1	2.5
\(J \sim \mathrm{Po}(5.6)\)	B1	3.3
Method 1 \(\mathrm{P}(J \geqslant 12) = 1 - \mathrm{P}(J \leqslant 11)\) \(= 1 - 0.9875\ldots\) Method 2 \(\mathrm{P}(J \geqslant 11) =\) awrt 0.0282 and \(\mathrm{P}(J \geqslant 10) =\) awrt 0.0591…	M1	1.1b
Method 1: \(= 0.01(248\ldots)\) Method 2: \(J \geqslant 11\)	A1	1.1b
\(0.01(24) \lt 0.05\) or \(12 \gt 11\) or 12 is in the critical region or 12 is significant or Reject \(\mathrm{H}_0\). There is evidence at the 5% level of significance that the rate of fish caught may have increased.	A1	2.2b
	(5)
	(10 marks)

Notes

B1: Both hypotheses in terms of \(\lambda\) or \(\mu\). Allow 5.6 instead of 1.4

B1: Writing or using \(\mathrm{Po}(5.6)\)

M1: For writing or using \(1 - \mathrm{P}(J \leqslant 11)\) Implied by a correct probability or CR
Allow \(\mathrm{P}(J \leqslant 10) =\) awrt 0.972 and \(\mathrm{P}(J \leqslant 9) =\) awrt 0.941

A1: 0.01 or better (allow truncation eg 0.0124)

NB Allow M1 A1 if \(\mathrm{P}(J \leqslant 11) = 0.9875\ldots\) is written on its own

A1: Independent of hypotheses. A correct conclusion based on their probability with 0.05 conclusion in context (bold words) Do not accept contradicting statements.