S1 January 2011 Q6

EdexcelOld spec14 marksDRVs

6. The discrete random variable \(X\) has the probability distribution

\(x\)	1	2	3	4
\(\mathrm{P}(X = x)\)	\(k\)	\(2k\)	\(3k\)	\(4k\)

(a) Show that \(k = 0.1\) (1)

Find

(b) \(\mathrm{E}(X)\) (2)

(d) \(\mathrm{Var}(2 - 5X)\) (3)

Two independent observations \(X_1\) and \(X_2\) are made of \(X\).

(e) Show that \(\mathrm{P}(X_1 + X_2 = 4) = 0.1\) (2)

(f) Complete the probability distribution table for \(X_1 + X_2\) (2)

\(y\)	2	3	4	5	6	7	8
\(\mathrm{P}(X_1 + X_2 = y)\)	0.01	0.04	0.10		0.25	0.24

(g) Find \(\mathrm{P}(1.5 \lt X_1 + X_2 \leqslant 3.5)\) (2)

Mark scheme (a)

Scheme	Marks
\(k + 2k + 3k + 4k = 1\) or \(10k = 1\) \(k = 0.1\) (*) [allow verification with a comment e.g. “so \(k\) = 0.1”]	B1cso
	(1)

Notes

B1 for a clear attempt to use sum of probabilities = 1. Must see previous line as well as \(k\) = 0.1

A correct expression for \(\mathrm{E}(X)\) or \(\mathrm{E}(X^2)\) that is later divided by 4 scores M0

Mark scheme (b)

Scheme	Marks
\(\mathrm{E}(X) = 1 \times 0.1 + 2 \times 0.2 + 3 \times 0.3 + 4 \times 0.4 = 3\)	M1 A1
	(2)

Notes

M1 for a completely correct expression. May be implied by correct answer of 3 or \(30k\)
A1 for 3 only.

Mark scheme (c)

Scheme	Marks
\(\mathrm{E}(X^2) = 1 \times 0.1 + 4 \times 0.2 + 9 \times 0.3 + 16 \times 0.4 = 10\)	M1 A1
	(2)

Notes

M1 for a completely correct expression. May be implied by correct answer of 10 or \(100k\)
A1 for 10 only.
[ For \(\mathrm{E}(X^2) = 0.1 + 0.8 + 2.7 + 6.4 - 9 = 1\) scores M0A0 but accept this as Var(\(X\)) in (d)]

Mark scheme (d)

Scheme	Marks
\(\mathrm{Var}(X) = 10 - 9\ (= 1)\)	M1
\(\mathrm{Var}(2 - 5X) = 5^2\,\mathrm{Var}(X) = 25\)	M1 A1
	(3)

Notes

1st M1 for using Var(\(X\)) = \(\mathrm{E}(X^2) - \mathrm{E}(X)^2\), f.t their values from (b) and (c)
Allow this mark for Var(\(X\)) = 10−9 or better. May be implied if this is seen in (c).
2nd M1 for \(5^2\,\mathrm{Var}(X)\) or 25Var(\(X\)) can f.t. their Var(\(X\)). Allow \(-5^2\) if it later becomes +25
A1 for 25 only. Dependent upon both Ms
Forming distribution for \(Y\) =2-5\(X\) gets M1 for \(\mathrm{E}(Y^2)\)=194 then M1A1 for 194-169=25

Mark scheme (e)

Scheme	Marks
\(\mathrm{P}(1, 3) + \mathrm{P}(2, 2) = 2 \times 0.1 \times 0.3 + 0.2 \times 0.2 = 0.1\) (*)	M1 A1cso
	(2)

Notes

M1 for correctly identifying (1, 3) or (3, 1) and (2, 2) as required cases (\(3k^2 + 4k^2\) or better)
A1 cso for 0.1 only but must see evidence for M1

Mark scheme (g)

Scheme	Marks
\(\mathrm{P}(2) + \mathrm{P}(3) = 0.05\)	M1A1
	(2)
	(14 marks)

Notes

M1 for P(2) + P(3). May be implied by correct answer of 0.05
A1 for 0.05 only.

Correct answer only can score full marks in parts (b), (c), (f) and (g)