S4 June 2016 Q6

EdexcelOld spec15 marksConfidence Intervals Continuous Random Variables

6. A random sample of size \(n\) is taken from the random variable \(X\), which has a continuous uniform distribution over the interval \([0, a]\), \(a \gt 0\)

The sample mean is denoted by \(\bar{X}\)

(a) Show that \(Y = 2\bar{X}\) is an unbiased estimator of \(a\) (2)

The maximum value, \(M\), in the sample has probability density function

\[\mathrm{f}(m) = \begin{cases} \dfrac{nm^{n-1}}{a^n} & 0 \leqslant m \leqslant a \\ 0 & \text{otherwise} \end{cases}\]

(b) Find \(\mathrm{E}(M)\) (2)

The estimator \(S\) is defined by \(S = \dfrac{n + 1}{n}M\)

Given that \(n \gt 1\)

(d) state which of \(Y\) or \(S\) is the better estimator for \(a\). Give a reason for your answer. (7)

Mark scheme (a)

Scheme	Marks
\(\mathrm{E}(Y) = 2\mathrm{E}\left(\bar{X}\right)\) \(= 2 \times \dfrac{a}{2}\)	M1
\(= a\)	A1cso
	(2)

Notes

M1 for \(2\mathrm{E}\left(\bar{X}\right)\)

A1 For \(2 \times \dfrac{a}{2}\) leading to \(a\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}(M) = \displaystyle\int_0^a \dfrac{nm^n}{a^n}\,\mathrm{d}m\)	M1
\(= \left[\dfrac{nm^{n+1}}{a^n(n + 1)}\right]_0^a\)
\(= \dfrac{na}{n + 1}\)	A1
	(2)

Notes

M1 attempting to integrate correct expression

Mark scheme (c)

Scheme	Marks
\(\mathrm{Var}(M) = \displaystyle\int_0^a \dfrac{nm^{n+1}}{a^n}\,\mathrm{d}m - \left(\dfrac{na}{n + 1}\right)^2\)	M1A1
\(= \left[\dfrac{nm^{n+2}}{a^n(n + 2)}\right]_0^a - \dfrac{n^2a^2}{(n + 1)^2}\)	M1d
\(= na^2\left(\dfrac{(n + 1)^2 - n(n + 2)}{(n + 1)^2(n + 2)}\right)\)
\(= \dfrac{na^2}{(n + 2)(n + 1)^2}\)	A1cso
	(4)

Notes

M1 for attempting to integrate a correct expression for \(\mathrm{E}(X^2)\)
A1 correct \(\mathrm{E}(X^2)\)
M1d dependent on previous M mark, using correct formula for Var(\(M\))

Mark scheme (d)

Scheme	Marks
\(\mathrm{E}(S) = \dfrac{n + 1}{n}\mathrm{E}(M) = \dfrac{n + 1}{n} \times \dfrac{na}{n + 1} = a\)	B1
\(\mathrm{Var}(S) = \left(\dfrac{n + 1}{n}\right)^2\dfrac{na^2}{(n + 2)(n + 1)^2} = \dfrac{a^2}{n(n + 2)}\)	B1
\(\mathrm{Var}(Y) = 4\,\mathrm{Var}\left(\bar{X}\right)\)	M1
\(= 4 \times \dfrac{a^2}{12n}\) \(= \dfrac{a^2}{3n}\)	A1
As \(n \gt 1\) \(n(n + 2) \gt 3n\); therefore \(\mathrm{Var}(S) \lt \mathrm{Var}(Y)\)	M1;M1
\(\therefore\ S\) is the better estimator	A1cso
	(7)
	(15 marks)

Notes

B1 for \(\dfrac{n + 1}{n}\mathrm{E}(M) = a\) or \(\dfrac{n + 1}{n} \times \dfrac{na}{n + 1} = a\)

M1 using \(4\,\mathrm{Var}\left(\bar{X}\right)\)

NB Failure to show \(S\) is unbiased gains a maximum of 5/7 lose first B1 and final A1