S2 January 2005 Q6

EdexcelOld spec16 marksIncludes hypothesis testingPoisson Distribution

6. Over a long period of time, accidents happened on a stretch of road at random at a rate of 3 per month.

Find the probability that

(a) in a randomly chosen month, more than 4 accidents occurred, (3)

(b) in a three-month period, more than 4 accidents occurred. (2)

At a later date, a speed restriction was introduced on this stretch of road. During a randomly chosen month only one accident occurred.

(c) Test, at the 5% level of significance, whether or not there is evidence to support the claim that this speed restriction reduced the mean number of road accidents occurring per month. (4)

The speed restriction was kept on this road. Over a two-year period, 55 accidents occurred.

(d) Test, at the 5% level of significance, whether or not there is now evidence that this speed restriction reduced the mean number of road accidents occurring per month. (7)

Mark scheme (a)

Scheme	Marks
Let \(X\) represent number of accidents/month \(\therefore X \sim \mathrm{Po}(3)\)	B1
\(\mathrm{P}(X \gt 4) = 1 - \mathrm{P}(X \leqslant 4);\ = 1 - 0.8153 = \underline{0.1847}\)	M1; A1
	(3)

Notes

(corrected from the printed mark scheme: the printed scheme has 1 – 0.8513; \(\mathrm{Po}(3)\) tables give \(\mathrm{P}(X \leqslant 4) = 0.8153\))

Mark scheme (b)

Scheme	Marks
Let \(Y\) represent number of accidents in 3 months \(\therefore Y \sim \mathrm{Po}(3 \times 3 = 9)\)	B1
\(\mathrm{P}(Y \gt 4) = 1 - 0.0550 = \underline{0.9450}\)	B1
	(2)

Notes

1st B1 can be implied

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0: \lambda = 3;\ \mathrm{H}_1: \lambda \lt 3\) \(\alpha = 0.05\)	B1
\(\mathrm{P}(X \leqslant 1 \mid \lambda = 3) = 0.1991;\ \gt 0.05\)	B1; M1
\(\therefore\) Insufficient evidence to support the claim that the mean number of accidents has been reduced.	A1ft
	(4)

Notes

1st B1 both

2 tailed; allow B0 B1 M1 (0.025) A0

(NB: CR: \(X \leqslant 0\); \(X = 1\) not in CR; same conclusion \(\Rightarrow\) B1, M1, A1)

Mark scheme (d)

Scheme	Marks
\(\mathrm{H}_0: \lambda = 24 \times 3 = 72;\ \mathrm{H}_1: \lambda \lt 72\)	B1 B1
\(\alpha = 0.05 \Rightarrow \text{CR}: z \lt -1.6449\)	B1
Using Normal approximation with \(\mu = \sigma^2 = 72\)	B1
\(z = \dfrac{55.5 - 72}{\sqrt{72}} = -1.94454\ldots\)	M1 A1
Since \(-1.944\ldots\) is in the CR, \(\mathrm{H}_0\) is rejected. There is evidence that the restriction has reduced the number of accidents.	A1ft
	(7)
	(16 marks)

Notes

1st B1 can be implied \(\lambda = 72\)

2nd B1 both \(\mathrm{H}_0\) & \(\mathrm{H}_1\)

3rd B1 \(-1.6449\)

4th B1 can be implied

M1 standardisation with \(\pm 0.5\), \(\mu\) & \(\sigma\)

A1 awrt \(-1.94/5\)

A1ft context & clear evidence

Alternative (d)

Scheme	Marks
\(p = 0.0262 \lt 0.05\)	B1

B1 awrt 0.026, equivalent to \(-1.6449\)