S2 June 2009 Q7

EdexcelOld spec15 marksContinuous Random Variables

Figure 1: graph of y = f(x), an isosceles triangle rising from O to a maximum of 1/2 at x = 2 and falling to the x-axis at x = 4 — Figure 1

Figure 1 shows a sketch of the probability density function \(\mathrm{f}(x)\) of the random variable \(X\). The part of the sketch from \(x = 0\) to \(x = 4\) consists of an isosceles triangle with maximum at (2, 0.5).

(a) Write down \(\mathrm{E}(X)\). (1)

The probability density function \(\mathrm{f}(x)\) can be written in the following form.

\[\mathrm{f}(x) = \begin{cases} ax & 0 \leqslant x \lt 2 \\ b - ax & 2 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}\]

(b) Find the values of the constants \(a\) and \(b\). (2)

(d) Find the lower quartile of \(X\). (3)

(e) State, giving a reason, whether \(\mathrm{P}(2 - \sigma \lt X \lt 2 + \sigma)\) is more or less than 0.5 (2)

Scheme	Marks
\(\mathrm{E}(X) = 2\) (by symmetry)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(0 \leqslant x \lt 2\), gradient \(= \dfrac{\frac{1}{2}}{2} = \dfrac{1}{4}\) and equation is \(y = \tfrac{1}{4}x\) so \(a = \tfrac{1}{4}\)	B1
\(b - \tfrac{1}{4}x\) passes through (4, 0) so \(b = 1\)	B1
	(2)

Notes

B1 for value of \(a\). B1 for value of \(b\)

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(X^2) = \displaystyle\int_0^2 \left(\tfrac{1}{4}x^3\right)\mathrm{d}x + \displaystyle\int_2^4 \left(x^2 - \tfrac{1}{4}x^3\right)\mathrm{d}x\)	M1M1
\(= \left[\dfrac{x^4}{16}\right]_0^2 + \left[\dfrac{x^3}{3} - \dfrac{x^4}{16}\right]_2^4\)	A1
\(= 1 + \dfrac{64 - 8}{3} - \dfrac{256 - 16}{16} \quad = \quad 4\tfrac{2}{3}\) or \(\tfrac{14}{3}\)	M1A1
\(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 = \tfrac{14}{3} - 2^2\ ,\ = \tfrac{2}{3}\) (so \(\sigma = \sqrt{\tfrac{2}{3}} = 0.816\)) (*)	M1 A1cso
	(7)

Notes

1^st M1 for attempt at \(\displaystyle\int ax^3\) using their \(a\). For attempt they need \(x^4\). Ignore limits.

2^nd M1 for attempt at \(\displaystyle\int bx^2 - ax^3\) use their \(a\) and \(b\). For attempt need to have either \(x^3\) or \(x^4\). Ignore limits

1^st A1 correct integration for both parts
3^rd M1 for use of the correct limits on each part
2^nd A1 for either getting 1 and \(3\dfrac{2}{3}\) or awrt 3.67 somewhere or \(4\dfrac{2}{3}\) or awrt 4.67

4^th M1 for use of \(\mathrm{E}(X^2) - [\mathrm{E}(X)]^2\) must add both parts for \(\mathrm{E}(X^2)\) and only have subtracted the mean² once. You must see this working

3^rd A1 \(\sigma = \sqrt{\dfrac{2}{3}}\) or \(\sqrt{0.66667}\) or better with no incorrect working seen.

Mark scheme (d)

Scheme	Marks
\(\mathrm{P}(X \leqslant q) = \displaystyle\int_0^q \tfrac{1}{4}x\,\mathrm{d}x = \tfrac{1}{4}\), \(\tfrac{q^2}{2} = 1\) so \(q = \sqrt{2} = 1.414\) awrt 1.41	M1A1,A1
	(3)

Notes

M1 for attempting to find LQ, integral of either part of \(\mathrm{f}(x)\) with their ‘\(a\)’ and ‘\(b\)’ = 0.25
Or their \(\mathrm{F}(x) = 0.25\) i.e. \(\dfrac{ax^2}{2} = 0.25\) or \(bx - \dfrac{ax^2}{2} + 4a - 2b = 0.25\) with their \(a\) and \(b\)

If they add both parts of their \(\mathrm{F}(x)\), then they will get M0.
1^st A1 for a correct equation/expression using their ‘\(a\)’
2^nd A1 for \(\sqrt{2}\) or awrt 1.41

Mark scheme (e)

Scheme	Marks
\(2 - \sigma = 1.184\) so \(2 - \sigma,\ 2 + \sigma\) is wider than IQR, therefore greater than 0.5	M1,A1
	(2)
	(15 marks)

Notes

M1 for a reason based on their quartiles

Possible reasons are \(\mathrm{P}(2 - \sigma \lt X \lt 2 + \sigma) = 0.6498\) allow awrt 0.65
1.184 < LQ(1.414)

A1 for correct answer > 0.5

NB you must check the reason and award the method mark. A correct answer without a correct reason gets M0 A0