S1 January 2012 Q3

EdexcelOld spec11 marksDRVs

3. The discrete random variable \(X\) can take only the values 2, 3, 4 or 6. For these values the probability distribution function is given by

\(x\)	2	3	4	6
\(\mathrm{P}(X = x)\)	\(\dfrac{5}{21}\)	\(\dfrac{2k}{21}\)	\(\dfrac{7}{21}\)	\(\dfrac{k}{21}\)

where \(k\) is a positive integer.

(a) Show that \(k = 3\) (2)

Find

(b) \(\mathrm{F}(3)\) (1)

(d) \(\mathrm{E}(X^2)\) (2)

(e) \(\mathrm{Var}(7X - 5)\) (4)

Mark scheme (a)

Scheme	Marks
\(\dfrac{5}{21} + \dfrac{2k}{21} + \dfrac{7}{21} + \dfrac{k}{21} = 1\)	M1
\(\dfrac{12 + 3k}{21} = 1\) \(k = 3\) * AG required for both methods	A1
	(2)

Notes

M1 Award for verification. Sub in k=3 and show \(\sum \mathrm{P}(X = x) = 1\). Require at least three correct terms seen or line 2 of scheme. (corrected from the printed mark scheme: \(\sum x\mathrm{P}(X = x) = 1\))
A1 Correct solution only including verification.

Scheme	Marks
\(\dfrac{11}{21}\)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(X) = 2 \times \dfrac{5}{21} + 3 \times \dfrac{6}{21} + 4 \times \dfrac{7}{21} + 6 \times \dfrac{1}{7}\)	M1
\(= 3\dfrac{11}{21}\) or \(\dfrac{74}{21}\) or awrt 3.52	A1
	(2)

Notes

M1 At least two correct terms required for method, follow through ‘their \(k\)’ for method. Correct answer only, award M1 A1.

Mark scheme (d)

Scheme	Marks
\(\mathrm{E}(X^2) = 2^2 \times \dfrac{5}{21} + 3^2 \times \dfrac{6}{21} + 4^2 \times \dfrac{7}{21} + 6^2 \times \dfrac{1}{7}\)	M1
= 14	A1
	(2)

Notes

M1 At least two correct terms required for method. M0 if probability is squared.
Correct answer only, award M1 A1. Accept exact equivalent of 14 for A1.

Mark scheme (e)

Scheme	Marks
\(\mathrm{Var}(X) = 14 - \left(3\dfrac{11}{21}\right)^2\)	M1
\(= 1\dfrac{257}{441}\) or \(\dfrac{698}{441}\) or awrt 1.6	A1
\(\mathrm{Var}(7X - 5) = 7^2\,\mathrm{Var}(X)\)	M1
\(= 77\dfrac{5}{9}\) or \(\dfrac{698}{9}\) or awrt 77.6	A1
	(4)
	(11 marks)

Notes

M1 for use of correct formula in both. 1.6 can be implied by correct final answer.
Working needs to be clearly labelled to award first method mark without second stage of calculation.
If a new table for values of 7X – 5 is used, so Y = 7X – 5
\(\mathrm{E}(Y^2) = \dfrac{9751}{21}\) ; \(\mathrm{Var}(Y) = 77\dfrac{5}{9}\) or \(\dfrac{698}{9}\) or awrt 77.6 Award M1A1; M1A1
If any attempt to divide by 4 seen as part of working award M0 for that part.