Find the cumulative distribution function $F(t)$, for $0 \leqslant t \leqslant 2$.

Sketch the graph of $F(t)$ for $0 \leqslant t \leqslant 2$, clearly indicating the coordinates of the endpoints.

Given that $P(T < a) = 0.75$, find the value of $a$.

$F(t) = \int_0^t {\left( {x - \frac{{{x^3}}}{4}} \right){\text{d}}x{\text{ }}\left( { = \int_0^t {\frac{{x(4 - {x^2})}}{4}{\text{d}}x} } \right)}$     M1

$= \left[ {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{{16}}} \right]_0^t{\text{ }}\left( { = \left[ {\frac{{{x^2}(8 - {x^2})}}{{16}}} \right]_0^t} \right){\text{ }}\left( { = \left[ {\frac{{ - 4 - {x^2}{)^2}}}{{16}}} \right]_0^t} \right)$     A1

$= \frac{{{t^2}}}{2} - \frac{{{t^4}}}{{16}}{\text{ }}\left( { = \frac{{{t^2}(8 - {t^2})}}{{16}}} \right){\text{ }}\left( { = 1 - \frac{{{{(4 - {t^2})}^2}}}{{16}}} \right)$     A1

Note:     Condone integration involving $t$ only.

Note:     Award M1A0A0 for integration without limits eg, $\int {\frac{{t(4 - {t^2})}}{4}{\text{d}}t = \frac{{{t^2}}}{2} - \frac{{{t^4}}}{{16}}}$ or equivalent.

Note:     But allow integration $+$ $C$ then showing $C = 0$ or even integration without $C$ if $F(0) = 0$ or $F(2) = 1$ is confirmed.

[3 marks]

correct shape including correct concavity     A1

clearly indicating starts at origin and ends at $(2,{\text{ }}1)$     A1

Note:     Condone the absence of $(0,{\text{ }}0)$.

Note:     Accept 2 on the $x$-axis and 1 on the $y$-axis correctly placed.

[2 marks]

attempt to solve $\frac{{{a^2}}}{2} - \frac{{{a^4}}}{{16}} = 0.75$ (or equivalent) for $a$     (M1)

$a = 1.41{\text{ }}( = \sqrt 2 )$     A1

Note:     Accept any answer that rounds to 1.4.

[2 marks]