AS June 2019 Q4

EdexcelAS paperCurrent spec14 marksDRVs

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

\(x\)	\(-3\)	\(-1\)	1	2	4
\(\mathrm{P}(X = x)\)	\(q\)	\(\dfrac{7}{30}\)	\(\dfrac{7}{30}\)	\(q\)	\(r\)

where \(q\) and \(r\) are probabilities.

(a) Write down, in terms of \(q\), \(\mathrm{P}(X \leqslant 0)\) (1)

(b) Show that \(\mathrm{E}(X^2) = \dfrac{7}{15} + 13q + 16r\) (2)

Given that \(\mathrm{E}(X^3) = \mathrm{E}(X^2) + \mathrm{E}(6X)\)

(d) Hence find \(\mathrm{P}(X^3 \gt X^2 + 6X)\) (4)

Scheme	Marks	AO
\(q + \tfrac{7}{30}\)	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{E}(X^2) = (-3)^2 \times q + (-1)^2 \times \tfrac{7}{30} + 1^2 \times \tfrac{7}{30} + 2^2 \times q + 4^2 \times r\)	M1	1.1b
\(= \tfrac{7}{15} + 13q + 16r\) (*)	A1*cso	1.1b
	(2)

Notes

M1 for at least 3 correct terms of the expression for \(\mathrm{E}(X^2)\)

A1*cso evidence of M1 scored with no incorrect working seen leading to correct answer (*)

Allow \(-3^2 \times q + -1^2 \times \tfrac{7}{30}\) etc if followed by \(9q + \ldots\) but not if simply followed by given answer

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{E}(X) = -3q + -\tfrac{7}{30} + \tfrac{7}{30} + 2q + 4r \quad \{= 4r - q\}\)	M1	3.1a
\(\mathrm{E}(X^2 + 6X) = \tfrac{7}{15} + 7q + 40r\)	A1	1.1b
\(\mathrm{E}(X^3) = (-3)^3 \times q + (-1)^3 \times \tfrac{7}{30} + 1^3 \times \tfrac{7}{30} + 2^3 \times q + 4^3 \times r\)	M1	3.4
\(= 64r - 19q\)	A1	1.1b
Sum of probabilities \(= 1\) gives: \(2q + r = \tfrac{16}{30}\) (o.e.)	M1	1.1b
Solve: \(24r - 26q = \tfrac{7}{15}\) and \(r + 2q = \tfrac{8}{15}\) e.g. \(37r = \tfrac{111}{15}\)	dM1	1.1b
So \(r = \tfrac{1}{5}\) and \(q = \tfrac{1}{6}\)	A1	1.1b
	(7)

Notes

1^st M1 for realising the need to find \(\mathrm{E}(X)\) – a correct attempt with at least 3 correct terms

1^st A1 for the correct expression (needn’t be simplified at this stage)

2^nd M1 for a correct attempt at \(\mathrm{E}(X^3)\) with at least 3 correct terms seen
Treat no \(\tfrac{7}{30}\) terms as one correct term

2^nd A1 for \(64r - 19q\) (must be simplified) or for \(24r - 26q = \tfrac{7}{15}\)

3^rd M1 for using sum of probabilities = 1 to form an equation in \(q\) and \(r\) (needn’t be simplified)
Must be correct or clearly state that \(\Sigma\)probs = 1 being attempted with only one slip

4^th dM1 for solving their 2 linear equations in \(q\) and \(r\) (dep on 3^rd M1 and 1^st or 2^nd M1)
Must see correct method to reduce to a linear equation in one variable

3^rd A1 for \(r = \tfrac{1}{5}\) and \(q = \tfrac{1}{6}\) or any exact equivalents (dep on 2 correct equations seen)

Mark scheme (d)

Scheme	Marks	AO
\(X^3 \gt X^2 + 6X \Rightarrow X(X - 3)(X + 2) \gt 0\)	M1	2.1
Use of sketch or table to see: \(-2 \lt X \lt 0\) or \(X \gt 3\)	A1	1.1b
So \(\mathrm{P}(X^3 \gt X^2 + 6X) = \mathrm{P}(X = -1 \text{ or } 4)\)	M1	2.2a
\(= \tfrac{7}{30} + \text{“}r\text{”} = \tfrac{13}{30}\)	A1ft	1.1b
	(4)
	(14 marks)

Notes

1^st M1 for 1^st stage towards solving the inequality (factorising the cubic)

1^st A1 for solving the inequality

2^nd M1 for identifying the values of \(X\) required i.e. \(-1\) and 4

2^nd A1ft for \(\tfrac{13}{30}\) or exact equivalent e.g. \(0.4\dot{3}\) (Allow ft of “their \(r\)” + \(\tfrac{7}{30}\)) (corrected from the printed mark scheme: this note prints \(\tfrac{13}{20}\), but the answer in the scheme is \(\tfrac{13}{30} = 0.4\dot{3}\))

Alternative

\(X\)	\(-3\)	\(-1\)	1	2	4
\(X^3\)	\(-27\)	\(-1\)	1	8	64
\(X^2 + 6X\)	\(-9\)	\(-5\)	7	16	40

Table 1^st M1 for at least 4 correct values for \(X^3\) and \(X^2 + 6X\) (must be labelled)

1^st A1 for all 10 correct values. [NB Can score M1A0M1A1ft in (d)]