S2 June 2013 Q3

EdexcelOld spec13 marksIncludes hypothesis testingPoisson Distribution

3. An online shop sells a computer game at an average rate of 1 per day.

(a) Find the probability that the shop sells more than 10 games in a 7 day period. (3)

Once every 7 days the shop has games delivered before it opens.

(b) Find the least number of games the shop should have in stock immediately after a delivery so that the probability of running out of the game before the next delivery is less than 0.05 (3)

In an attempt to increase sales of the computer game, the price is reduced for six months. A random sample of 28 days is taken from these six months. In the sample of 28 days, 36 computer games are sold.

(c) Using a suitable approximation and a 5% level of significance, test whether or not the average rate of sales per day has increased during these six months. State your hypotheses clearly. (7)

Mark scheme (a)

Scheme	Marks
\(X \sim \mathrm{Po}(7)\)	B1
\(\mathrm{P}(X \gt 10) = 1 - \mathrm{P}(X \leqslant 10)\)	M1
\(= 1 - 0.9015\)
\(= 0.0985\) awrt 0.0985	A1
	(3)

Notes

B1 stating or using Po(7)

M1 stating or using \(1 - \mathrm{P}(X \leqslant 10)\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(X \gt d) \lt 0.05\) Or \(\mathrm{P}(X \geqslant d) \lt 0.05\)
\(\mathrm{P}(X \leqslant d) \gt 0.95\) \(\mathrm{P}(X \lt d) \gt 0.95\)	M1
\(\mathrm{P}(X \leqslant 11) = 0.9467\) \(\mathrm{P}(X \lt 12) = 0.9467\)
\(\mathrm{P}(X \leqslant 12) = 0.9730\) \(\mathrm{P}(X \lt 13) = 0.9730\)	A1
Least number of games = 12 Least number of games 13	A1
	(3)

Notes

M1 using or writing \(\mathrm{P}(X \gt d) \lt 0.05\) or \(\mathrm{P}(X \lt d) \gt 0.95\) (condone \(\geqslant\) instead of \(\gt\) and \(\leqslant\) instead of \(\lt\)) May be implied by correct answer. Different letters may be used.

1st A1 \(\mathrm{P}(X \leqslant 12)\)/ \(\mathrm{P}(X \lt 13) =\) awrt 0.973 or \(\mathrm{P}(X \leqslant 11)\) / \(\mathrm{P}(X \lt 12) =\) awrt 0.947 May be implied by a correct answer

2nd A1 12 or 13

NB An answer of 12/13 on its own with no working gains M1A1A1

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \lambda = 1, (\mu = 28) \quad \mathrm{H}_1 : \lambda \gt 1 (\mu \gt 28)\)	B1
\(Y \sim \mathrm{Po}(28)\) approximated by N(28,28)	B1
\(\mathrm{P}(Y \geqslant 36) = \mathrm{P}\left(Z \geqslant \frac{35.5 - 28}{\sqrt{28}}\right)\) or \(1.6449 = \dfrac{x - 0.5 - 28}{\sqrt{28}}\)	M1M1
\(= \mathrm{P}(Z \geqslant 1.42)\) \(= 0.0778\) or \(1.42 \lt 1.6449\) or CR \(X \geqslant 37.2\)	A1
\(0.0778 \gt 0.05\) so do not reject \(\mathrm{H}_0\)/not significant. Not in CR	M1
There is no evidence that the average rate of sales per day has increased.	A1cso
	(7)
	(13 marks)

Notes

1st B1 both hypotheses correct using \(\lambda\) or \(\mu\), and 1 or 28

2nd B1 for writing or using a normal approximation with correct mean and Var (may be given if sd correct in standardisation formula)

1st M1 for use of a continuity correction 35.5 or 36.5 or \(x \pm 0.5\)

2nd M1 Standardising using their mean and their sd. If they have not written down a mean and sd then these need to be correct here to award the mark. They must use [35.5, 36.5, 36, \(x\) or \(x \pm 0.5\)] For CR must have = awrt 1.64 or 1.65

1st A1 awrt 0.0778 or 0.9222 or the statement \(1.42 \lt\) awrt 1.65/1.64 or CR \(X \geqslant 37.2\)/ \(X \gt 37.2\)

3rd M1 a correct conclusion for their probability. May be implied by a correct contextual conclusion. NB Non contextual contradicting statements gets M0

2nd A1 a correct contextual conclusion for their hypotheses and a fully correct solution with no errors seen. Need the words “rate/average number”, “sales” and “increased”oe

NB If found \(\mathrm{P}(X = 36)\) they can get B1B10M0A0M0A0