AS June 2018 Q4

EdexcelAS paperCurrent spec9 marksContinuous Random Variables

4. The continuous random variable \(X\) has cumulative distribution function

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 3 \\ c - 4.5x^n & 3 \leqslant x \leqslant 9 \\ 1 & x \gt 9 \end{cases}\]

where \(c\) is a positive constant and \(n\) is an integer.

(a) Showing all stages of your working, find the value of \(c\) and the value of \(n\) (7)

(b) Find the lower quartile of \(X\) (2)

Mark scheme (a)

Scheme

Marks

\(\mathrm{F}(3) = 0\) or \(\mathrm{F}(9) = 1\)

3.1a

\(c - 4.5(3^n) = 0\) and \(c - 4.5(9^n) = 1\)

1.1b

Eliminating \(c\)
\(1 + 4.5(9^n) = 4.5(3^n)\)

Eliminating \(n\)
\(\dfrac{\log\left(\frac{c-1}{4.5}\right)}{\log(9)} = \dfrac{\log\left(\frac{c}{4.5}\right)}{\log(3)}\)

1.1b

Forming a 3 term quadratic
\(4.5(3^{2n}) - 4.5(3^n) + 1 = 0\)

Forming a 3 term quadratic
\(4.5c^2 - 20.25c + 20.25 = 0\)

3.1a

Solving 3TQ leading to a value for \(n\)
\(3^n = \dfrac{1}{3} \rightarrow n = \ldots\)
or \(3^n = \dfrac{2}{3} \rightarrow n = \ldots\)

Solving 3TQ leading to a value for \(c\)
\(c = 1.5\)
or \(c = 3\)

1.1b

\(n = -1\) only (reject other solution as \(n\) is an integer)

2.3

\(c = 1.5\) only

A1ft

1.1b

(7)

1^st M1 for use of either \(\mathrm{F}(3) = 0\) or \(\mathrm{F}(9) = 1\)

1^st A1 for two correct equations in \(c\) and \(n\)

2^nd M1 for eliminating one variable

3^rd M1 for realising that a quadratic formula is required to solve

4^th M1 for solving the quadratic formula leading to at least one value of \(c\) or \(n\)

2^nd A1 for \(n = -1\) only

3^rd A1ft for \(c = 1.5\) only (ft their \(n\) in a correct equation and dep on 1^st M1)

Mark scheme (b)

Scheme	Marks	AO
[Let \(q\) = lower quartile] \(\text{“}1.5\text{”} - 4.5\left(q^{\text{“}-1\text{”}}\right) = 0.25\)	M1	1.1b
\(q = 3.6\)	A1ft	1.1b
	(2)
	(9 marks)

M1 for use of \(\mathrm{F}(q) = 0.25\)

A1ft for \(q = 3.6\) (allow follow through on their integer value of \(n\))