17M.2.AHL.TZ2.H_10

pestleMathematicsAAHLPaper 217M· sl-4-8-binomial-distributionsource ↗

A continuous random variable $X$ has probability density function $f$ given by

$f(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{{x^2}}}{a} + b,}&{0 \leqslant x \leqslant 4} \\ 0&{{\text{otherwise}}} \end{array}} \right.{\text{where }}a{\text{ and }}b{\text{ are positive constants.}}$

It is given that ${\text{P}}(X \geqslant 2) = 0.75$ .

Eight independent observations of $X$ are now taken and the random variable $Y$ is the number of observations such that $X \geqslant 2$ .

Show that $a = 32$ and $b = \frac{1}{{12}}$ .

[5]

Find ${\text{E}}(X)$ .

[2]

Find ${\text{Var}}(X)$ .

[2]

Find the median of $X$ .

[3]

Find ${\text{E}}(Y)$ .

[2]

Find ${\text{P}}(Y \geqslant 3)$ .

[1]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\int\limits_0^4 {\left( {\frac{{{x^2}}}{a} + b} \right){\text{d}}x = 1 \Rightarrow \left[ {\frac{{{x^3}}}{{3a}} + bx} \right]_0^4 = 1 \Rightarrow \frac{{64}}{{3a}} + 4b = 1}$ M1A1

$\int\limits_2^4 {\left( {\frac{{{x^2}}}{a} + b} \right){\text{d}}x = 0.75 \Rightarrow \frac{{56}}{{3a}} + 2b = 0.75}$ M1A1

Note: $\int\limits_0^2 {\left( {\frac{{{x^2}}}{a} + b} \right)} \,dx = 0.25 \Rightarrow \frac{8}{{3a}} + 2b = 0.25$ could be seen/used in place of either of the above equations.

evidence of an attempt to solve simultaneously (or check given a,b values are consistent) M1

$a = 32,{\text{ }}b = \frac{1}{{12}}$ AG

[5 marks]

${\text{E}}\left( X \right) = \int\limits_0^4 {x\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x}$ (M1)

${\text{E}}(X) = \frac{8}{3}\,\,\,( = 2.67)$ A1

[2 marks]

${\text{E}}\left( {{X^2}} \right) = \int\limits_0^4 {{x^2}\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x}$ (M1)

${\text{Var}}(X) = {\text{E}}({X^2}) - {[{\text{E}}(X)]^2} = \frac{{16}}{{15}}\,\,\,( = 1.07)$ A1

[2 marks]

$\int\limits_0^m {\left( {\frac{{{x^2}}}{{32}} + \frac{1}{{12}}} \right){\text{d}}x = 0.5}$ (M1)

$\frac{{{m^3}}}{{96}} + \frac{m}{{12}} = 0.5\,\,\,( \Rightarrow {m^3} + 8m - 48 = 0)$ (A1)

$m = 2.91$ A1

[3 marks]

$Y \sim B(8,{\text{ }}0.75)$ (M1)

${\text{E}}(Y) = 8 \times 0.75 = 6$ A1

[2 marks]

${\text{P}}(Y \geqslant 3) = 0.996$ A1

[1 mark]

Examiners’ report

[N/A]