A2 October 2020 Q8

EdexcelCurrent spec11 marksContinuous Random Variables

8. A circle, centre \(O\), has radius \(x\) cm, where \(x\) is an observation from the random variable \(X\) which has a rectangular distribution on \([0, \pi]\)

(a) Find the probability that the area of the circle is greater than 10 cm² (3)

(b) State, giving a reason, whether the median area of the circle is greater or less than 10 cm² (1)

The triangle \(OAB\) is drawn inside the circle with \(OA\) and \(OB\) as radii of length \(x\) cm and angle \(AOB\) \(x\) radians.

(c) Use algebraic integration to find the expected value of the area of triangle \(OAB\).
Give your answer as an exact value. (7)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{P}(\pi X^2 \gt 10) \quad \Rightarrow \quad \mathrm{P}\left(X \gt \sqrt{\dfrac{10}{\pi}}\right)\)	M1	3.1a
\(= \dfrac{\pi - \sqrt{\dfrac{10}{\pi}}}{\pi}\)	M1	2.1
\(= 0.43209\ldots\) = awrt 0.432	A1	1.1b
	(3)

Notes

1^st M1 reduce the problem to a probability about \(X\)

2^nd M1 for use of the uniform distribution (a correct expression ft their value 1.784..)

A1 for awrt 0.432

Mark scheme (b)

Scheme	Marks	AO
P(area > median) = 0.5; since (a) < 0.5 therefore median < 10	B1	2.2a
	(1)

Notes

B1 for statement that median < 10 supported by argument about answer to (a) being < 0.5

Alternative

Median area is given by \(\pi \times \left(\dfrac{\pi}{2}\right)^2 = 7.751\ldots \lt 10\) so median < 10

Mark scheme (c)

Scheme	Marks	AO
Area of triangle \(= 0.5x^2\sin x\)	M1	3.1a
\(\mathrm{E}(\text{area}) = \displaystyle\int_{[0]}^{[\pi]} \tfrac{1}{\pi}\tfrac{1}{2}x^2\sin x\,\mathrm{d}x\)	M1	1.1b
\(= \tfrac{1}{2\pi}\displaystyle\int_{[0]}^{[\pi]} x^2\,\mathrm{d}(-\cos x) = \tfrac{1}{2\pi}\left\{\left[-x^2\cos x\right]_{[0]}^{[\pi]} - \int_{[0]}^{[\pi]} -2x\cos x\,\mathrm{d}x\right\}\)	M1	2.1
\(= \left\{\left[\dfrac{-x^2\cos x}{2\pi}\right]_{[0]}^{[\pi]}\right\} + \left[\dfrac{x\sin x}{\pi}\right]_{[0]}^{[\pi]} - \tfrac{1}{\pi}\displaystyle\int_{[0]}^{[\pi]} \sin x\,\mathrm{d}x\)	M1 A1	1.1b 1.1b
\(= \dfrac{\pi^2}{2\pi} - 0 + 0 - 0 + \left(-\dfrac{1}{\pi}\right) - \left(\dfrac{1}{\pi}\right) =,\ \underline{\dfrac{\pi}{2} - \dfrac{2}{\pi}}\)	M1 A1	1.1b 1.1b
	(7)
	(11 marks)

Notes

1^st M1 for a correct expression for area in terms of \(x\)

2^nd M1 for realisation that need to use \(\mathrm{E}(\mathrm{g}(X))\) formula and a correct expression (ignore limits)

3^rd M1 for attempt to use integration by parts

4^th M1 for a 2^nd use of integration by parts

1^st A1 for correct integration (ignore limits)

5^th M1 for clear use of the correct limits

2^nd A1 for \(\tfrac{\pi}{2} - \tfrac{2}{\pi}\)

[(c) is an extended problem and also involves work from pure for the integration]