S2 January 2013 Q4

EdexcelOld spec14 marksContinuous Random Variables

4. The continuous random variable \(X\) is uniformly distributed over the interval \([-4, 6]\).

(a) Write down the mean of \(X\). (1)

(b) Find \(\mathrm{P}(X \leqslant 2.4)\) (2)

The continuous random variable \(Y\) is uniformly distributed over the interval \([a, 4a]\).

(d) Use integration to show that \(\mathrm{E}(Y^2) = 7a^2\) (4)

(e) Find \(\mathrm{Var}(Y)\). (2)

(f) Given that \(\mathrm{P}\left(X \lt \frac{8}{3}\right) = \mathrm{P}\left(Y \lt \frac{8}{3}\right)\), find the value of \(a\). (3)

Scheme	Marks
Mean = 1	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(X \leqslant 2.4) = (2.4 - -4) \times \dfrac{1}{10}\)	M1
\(= 0.64\) or \(\dfrac{16}{25}\)	A1
	(2)

Notes

M1 \((2.4 - -4) \times \dfrac{1}{10}\) or \(1 - (6 - 2.4) \times \dfrac{1}{10}\) o.e

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(-3 \lt X - 5 \lt 3) = \mathrm{P}(2 \lt X \lt 6)\)	M1
\(= 0.4\)	A1
	(2)

Notes

M1 finding \(\mathrm{P}(2 \lt X \lt 6)\) or \(\mathrm{P}(X \gt 2)\) or \(1 - \mathrm{P}(X \lt 2)\). May be implied by a correct answer if there is no incorrect working. Do not ignore subsequent incorrect working.

NB if they change the distribution to U[-9,1] then M1 is for finding \(\mathrm{P}(-3 \lt X \lt 1)\) or \(\mathrm{P}(X \gt -3)\) or \(1 - \mathrm{P}(X \lt -3)\). May be implied by a correct answer if there is no incorrect working. Do not ignore subsequent incorrect working.

Mark scheme (d)

Scheme	Marks
\(\displaystyle\int_a^{4a} \frac{y^2}{4a - a}\,\mathrm{d}y = \left[\frac{y^3}{9a}\right]_a^{4a}\)	M1 M1 dep
\(= \dfrac{64a^3 - a^3}{9a}\)	A1
\(= 7a^2\) *AG	A1cso
	(4)

Notes

NB remember the answer is given (AG) so they must show their working

1st M1 writing or using \(\displaystyle\int_a^{4a} y^2\mathrm{f}(y)\,\mathrm{d}y\) with correct limits used at some point. Condone omission of d\(y\). \(\mathrm{f}(y)\) does not need to be correct.

2nd M1 dependent on previous M being awarded. Attempting to integrate at \(y^n \to \dfrac{y^{n+1}}{n+1}\)

1st A1 correct expression - the correct limits must be substituted.

2nd A1 cso

Mark scheme (e)

Scheme	Marks
\(\mathrm{Var}(Y) = \dfrac{1}{12}(4a - a)^2\) or \(\mathrm{Var}(Y) = 7a^2 - \left(\dfrac{5}{2}a\right)^2\)	M1
\(= \dfrac{3}{4}a^2\)	A1cso
	(2)

Notes

M1 either use of \(\dfrac{(b - a)^2}{12}\) or \(\mathrm{E}(Y^2) - [\mathrm{E}(Y)]^2\):- they may use their part (d) for \(\mathrm{E}(Y^2)\)

Mark scheme (f)

Scheme	Marks
\(\dfrac{2}{3} = \dfrac{1}{3a}\left(\dfrac{8}{3} - a\right)\)	M1 A1
\(a = \dfrac{8}{9}\)	A1
	(3)
	(14 marks)

Notes

M1 using \(\dfrac{1}{3a}\left(\dfrac{8}{3} - a\right) =\) a probability or \(\dfrac{1}{3a}\left(4a - \dfrac{8}{3}\right) =\) a probability

An answer of \(\frac{8}{9}\) with no incorrect working gains M1A1A1