S2 June 2010 Q5

EdexcelOld spec15 marksIncludes hypothesis testingPoisson Distribution

5. A company has a large number of regular users logging onto its website. On average 4 users every hour fail to connect to the company’s website at their first attempt.

(a) Explain why the Poisson distribution may be a suitable model in this case. (1)

Find the probability that, in a randomly chosen 2 hour period,

(b)

(i) all users connect at their first attempt,

(ii) at least 4 users fail to connect at their first attempt. (5)

The company suffered from a virus infecting its computer system. During this infection it was found that the number of users failing to connect at their first attempt, over a 12 hour period, was 60.

(c) Using a suitable approximation, test whether or not the mean number of users per hour who failed to connect at their first attempt had increased. Use a 5% level of significance and state your hypotheses clearly. (9)

Mark scheme (a)

Scheme	Marks
Connecting occurs at random/independently, singly or at a constant rate	B1
	(1)

Notes

B1 Any one of randomly/independently/singly/constant rate. Must have context of connection/logging on/fail

Mark scheme (b)

Scheme	Marks
Po (8)	B1
(i) \(\mathrm{P}(X = 0) = 0.0003\)	M1A1
(ii) \(\mathrm{P}(X \geqslant 4) = 1 - \mathrm{P}(X \leqslant 3)\)	M1
\(= 1 - 0.0424\) \(= 0.9576\)	A1
	(5)

Notes

B1 Writing or using Po(8) in (i) or (ii)
(i) M1 for writing or finding \(\mathrm{P}(X = 0)\)
A1 awrt 0.0003
(ii) M1 for writing or finding \(1 - \mathrm{P}(X \leqslant 3)\)
A1 awrt 0.958

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \lambda = 4\ (48) \quad \mathrm{H}_1 : \lambda \gt 4\ (48)\)	B1
N(48,48)	M1 A1
Method 1 \(\mathrm{P}(X \geqslant 59.5) = \mathrm{P}\left(Z \geqslant \dfrac{59.5 - 48}{\sqrt{48}}\right)\)	M1 M1 A1
\(= \mathrm{P}(Z \geqslant 1.66)\) \(= 1 - 0.9515\) \(= 0.0485\)	A1
\(0.0485 \lt 0.05\) Reject H₀. Significant. 60 lies in the Critical region	M1
The number of failed connections at the first attempt has increased.	A1 ft
	(9)
	(15 marks)

Notes

Method 2

Scheme	Marks
\(\dfrac{x - 0.5 - 48}{\sqrt{48}} = 1.6449\)	M1 M1 A1
\(x = 59.9\)	A1

B1 both hypotheses correct. Must use \(\lambda\) or \(\mu\)
M1 identifying normal
A1 using or seeing mean and variance of 48
These first two marks may be given if the following are seen in the standardisation formula : 48 and \(\sqrt{48}\) or awrt 6.93

M1 for attempting a continuity correction (Method 1: 60 \(\pm\) 0.5 / Method 2: \(x \pm 0.5\) )
M1 for standardising using their mean and their standard deviation and using either Method 1 [59.5, 60 or 60.5. accept \(\pm\) z.] Method 2 [ (\(x \pm 0.5\)) and equal to a \(\pm z\) value)
A1 correct z value awrt \(\pm\)1.66 or \(\pm\dfrac{59.5 - 48}{\sqrt{48}}\), or \(\dfrac{x - 0.5 - 48}{\sqrt{48}} = 1.6449\)
A1 awrt 3 sig fig in range 0.0484 – 0.0485, awrt 59.9

M1 for “reject H₀” or “significant” maybe implied by “correct contextual comment”
If one tail hypotheses given follow through “their prob” and 0.05 , \(p \lt 0.5\)
If two tail hypotheses given follow through “their prob” with 0.025, \(p \lt 0.5\)
If one tail hypotheses given follow through “their prob” and 0.95 , \(p \gt 0.5\)
If two tail hypotheses given follow through “their prob” with 0.975, \(p \gt 0.5\)
If no H₁ given they get M0

A1 ft correct contextual statement followed through from their prob and H₁. need the words number of failed connections/log ons has increased o.e.
Allow “there are more failed connections”
NB A correct contextual statement alone followed through from their prob and H₁ gets M1 A1