S1 June 2013 (R) Q2

EdexcelOld spec4 marksDRVs

2. The discrete random variable \(X\) takes the values 1, 2 and 3 and has cumulative distribution function \(\mathrm{F}(x)\) given by

\(x\)	1	2	3
\(\mathrm{F}(x)\)	0.4	0.65	1

(a) Find the probability distribution of \(X\). (3)

(b) Write down the value of \(\mathrm{F}(1.8)\). (1)

Mark scheme (a)

Scheme

Marks

\(x\)	1	2	3
\(\mathrm{P}(X = x)\)	0.4	0.25	0.35

\(\mathrm{P}(X = 2) = \mathrm{F}(2) - \mathrm{F}(1)\) (o.e.)

\(\mathrm{P}(X = 2) = \underline{\mathbf{0.25}}\)

\(\mathrm{P}(X = 3) = \underline{\mathbf{0.35}}\)

(3)

Notes

M1 for \(\mathrm{P}(X = 1) = 0.4\) and evidence of a correct method for finding \(\mathrm{P}(X = 2)\) or \(\mathrm{P}(X = 3)\). Implied by correct ans.

1st A1 for \(\mathrm{P}(X = 2) = 0.25\)

2nd A1 for \(\mathrm{P}(X = 3) = 0.35\)

Scheme	Marks
\([\mathrm{F}(1.8) = \mathrm{P}(X \leqslant 1.8) = \mathrm{P}(X \leqslant 1) =]\ \underline{\mathbf{0.4}}\)	B1
	(1)
	(4 marks)