AS June 2024 Q4

EdexcelAS paperCurrent spec8 marksContinuous Random Variables

4. The continuous random variable \(X\) is uniformly distributed over the interval \([2, 7]\)

(a) Write down the value of \(\mathrm{E}(X)\) (1)

(b) Find \(\mathrm{P}(1 \lt X \lt 4)\) (1)

(d) Find \(\mathrm{E}\left(\dfrac{3}{X^2}\right)\) (3)

Scheme	Marks	AO
\(\mathrm{E}(X) =\) 4.5	B1	1.1b
	(1)

Scheme	Marks	AO
\(\left[\mathrm{P}(1 \lt X \lt 4) = \mathrm{P}(2 \lt X \lt 4) =\right] \dfrac{2}{5}\)	B1	1.1b
	(1)

Mark scheme (c)

Scheme	Marks	AO
\(2X^2 - 15X + 27 \gt 0 \rightarrow \big[\{X \lt 3\} \cup \{X \gt 4.5\}\big]\)	M1	3.1a
\(\mathrm{P}(\{X \lt 3\} \cup \{X \gt 4.5\}) = \dfrac{3-2}{7-2} + \dfrac{7-4.5}{7-2}\left[= 1 - \dfrac{4.5-3}{7-2}\right]\)	M1	1.1b
\(= \frac{7}{10}\) or 0.7	A1	1.1b
	(3)

Notes

1^st M1: Attempting to find roots. Can be implied by sight or use of 3 and 4.5

2^nd M1: Attempt to find probability for their “outside” region from U[2, 7]
Must have both correct ft probability statements and at least one correct ft probability
e.g. \(\mathrm{P}(X \lt 3) = 0.2\) and \(\mathrm{P}(X \gt 4.5) = 0.5\) followed by \(0.2 \times 0.5\) is M0M1A0

A1: \(\frac{7}{10}\) oe

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{E}\left(\dfrac{3}{X^2}\right) = \displaystyle\int_2^7 \dfrac{3}{x^2} \times \dfrac{1}{(7-2)}\,\mathrm{d}x\)	M1	3.1a
\(= \left[-0.6x^{-1}\right]_2^7\)	M1	1.1b
\(= \frac{3}{14}\)	A1	1.1b
	(3)
	(8 marks)

Notes

1^st M1: use of \(\mathrm{E}(\mathrm{g}(x))\) by setting up correct integral (ignore limits)

2^nd M1: integration with limits

A1: \(\frac{3}{14}\) oe from correct working