S2 June 2014 (R) Q4

EdexcelOld spec14 marksContinuous Random Variables

4. The random variable \(X\) has probability density function \(\mathrm{f}(x)\) given by

\[\mathrm{f}(x) = \begin{cases} 3k & 0 \leqslant x \lt 1 \\ kx(4 - x) & 1 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]

where \(k\) is a constant.

(a) Sketch \(\mathrm{f}(x)\). (3)

(b) Write down the mode of \(X\). (1)

Given that \(\mathrm{E}(X) = \dfrac{29}{16}\)

(d) Use integration to find the value of \(k\). (5)

(e) Write down the lower quartile of \(X\). (1)

Given also that \(\mathrm{P}(2 \lt X \lt 3) = \dfrac{11}{36}\)

(f) find the exact value of \(\mathrm{P}(X \gt 3)\). (2)

Sketch: horizontal line at height 3k from x = 0 to 1, then a curve rising to a maximum and falling to meet the x-axis at 4 — Mark scheme (a)

Scheme	Marks
	B1B1B1
	(3)

Scheme	Marks
Mode = 2	B1
	(1)

Mark scheme (c)

Scheme	Marks
Mean < mode, so negative skew	B1, dB1
	(2)

Notes

1st B1 for a suitable reason which matches their mode. The mode must be a number. Must use mean.

2nd dB1 not ft, dependent on 1st B1. Correct answer from correct value of Mode.

Mark scheme (d)

Scheme	Marks
\(3k \times 1 + \displaystyle\int_1^4 \left(4kx - kx^2\right)\,\mathrm{d}x = 1\)	M1, B1
\(3k + \left[2kx^2 - \dfrac{kx^3}{3}\right]_1^4 \{= 1\}\)	M1
\(3k + \left(32k - \dfrac{64k}{3}\right) - \left(2k - \dfrac{k}{3}\right) = 1\)	M1d
\(12k = 1 \qquad \text{so } k = \dfrac{1}{12}\)	A1
	(5)

Notes

1st M1 for attempting the sum of both areas = 1, ignore limits

B1 for \(3k\) seen added to integral

2nd M1 For some correct integration, at least one \(kx^n \to kx^{n+1}\)

3rd M1d Dependent on 1st M1 being awarded. For use of correct limits.

A1 for \(k = \frac{1}{12}\)

Scheme	Marks
Lower Quartile = 1	B1
	(1)

Mark scheme (f)

Scheme	Marks
\(\mathrm{P}(1 \lt X \lt 2) = \mathrm{P}(2 \lt X \lt 3)\) by symmetry	M1
So \(\mathrm{P}(X \gt 3) = 1 - 3k - \dfrac{22}{36} = \dfrac{5}{36}\)	A1
	(2)
	(14 marks)

Notes

M1 for identifying the symmetry. May be implied by \(\mathrm{P}(1 \lt x \lt 2) = \dfrac{11}{36}\) found by any method or writing down a correct equation (ft their \(k\)). e.g. \(0.75 - 2\times\dfrac{11}{36}\) or \(\displaystyle\int_3^4 kx(4 - x)\,\mathrm{d}x\) or \(1 - 3k - \dfrac{11}{36} - \displaystyle\int_1^2 4kx - kx^2\) with their \(k\) subst in

A1 for \(\frac{5}{36}\) or exact equivalent