S4 June 2010 Q6

EdexcelOld spec14 marksPoisson Distribution

6. Faults occur in a roll of material at a rate of \(\lambda\) per m². To estimate \(\lambda\), three pieces of material of sizes 3 m², 7 m² and 10 m² are selected and the number of faults \(X_1\), \(X_2\) and \(X_3\) respectively are recorded.

The estimator \(\hat{\lambda}\), where

\[\hat{\lambda} = k(X_1 + X_2 + X_3)\]

is an unbiased estimator of \(\lambda\).

(a) Write down the distributions of \(X_1\), \(X_2\) and \(X_3\) and find the value of \(k\). (4)

(b) Find \(\mathrm{Var}(\hat{\lambda})\). (3)

A random sample of \(n\) pieces of this material, each of size 4 m², was taken. The number of faults on each piece, \(Y\), was recorded.

(d) Find \(\mathrm{Var}\left(\dfrac{1}{4}\bar{Y}\right)\). (3)

(e) Find the minimum value of \(n\) for which \(\dfrac{1}{4}\bar{Y}\) becomes a better estimator of \(\lambda\) than \(\hat{\lambda}\). (2)

Mark scheme (a)

Scheme	Marks
\(X_1 \sim \mathrm{Po}(3\lambda)\) \(X_2 \sim \mathrm{Po}(7\lambda)\) \(X_3 \sim \mathrm{Po}(10\lambda)\)	M1
\(\mathrm{E}(\hat{\lambda}) = k\,[\mathrm{E}(X_1) + \mathrm{E}(X_2) + \mathrm{E}(X_3)]\) \(= 20\lambda k\)	M1
\(\hat{\lambda}\) unbiased therefore \(20\lambda k = \lambda\)	M1
\(k = \dfrac{1}{20}\)	A1
	(4)

Notes

M1 all 3 needed. Poisson and mean
M1 adding their means
M1 putting their \(\mathrm{E}(\hat{\lambda}) = \lambda\)
A1 cao

Mark scheme (b)

Scheme	Marks
\(\mathrm{Var}(\hat{\lambda}) = \dfrac{1}{20^2}\mathrm{Var}(X_1 + X_2 + X_3)\)	M1
\(= \dfrac{1}{20^2}(3\lambda + 7\lambda + 10\lambda)\)	M1
\(= \dfrac{\lambda}{20}\)	A1ft
	(3)

Notes

M1 use of \(k^2\mathrm{Var}(X_1 + X_2 + X_3)\)
M1 using their means from part(a) as Variances and adding together
A1 cao

Mark scheme (c)

Scheme	Marks
\(Y \sim \mathrm{Po}(4\lambda)\) \(\mathrm{E}\left(\dfrac{1}{4}\bar{Y}\right) = \dfrac{1}{4} \times 4\lambda = \lambda\) therefore unbiased	M1 A1
	(2)

Notes

M1 use of \(4\lambda\)
A1 cso plus conclusion. Accept working out bias to = 0

Mark scheme (d)

Scheme	Marks
\(\mathrm{Var}\left(\dfrac{1}{4}\bar{Y}\right) = \dfrac{1}{16} \times \dfrac{4\lambda}{n}\)	M1 B1
\(= \dfrac{\lambda}{4n}\)	A1
	(3)

Notes

M1 \(\dfrac{1}{16} \times \mathrm{Var}\,\bar{Y}\)
B1 for \(\mathrm{Var}\,\bar{Y} = \dfrac{4\lambda}{n}\)
A1 cao

Mark scheme (e)

Scheme	Marks
\(\dfrac{\lambda}{4n} \lt \dfrac{\lambda}{20}\)	M1
\(n \gt 5\) therefore \(n = 6\)	A1
	(2)
	(14 marks)

Notes

M1 for \(\mathrm{Var}\left(\dfrac{1}{4}\bar{Y}\right) \lt \mathrm{Var}(\hat{\lambda})\)
A1 \(n = 6\)