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]