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

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

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

Find the median of $X$.

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

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

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