Show that $a = \frac{2}{3}$.

Find ${\text{P}}\left( {X < 1} \right)$.

Given that ${\text{P}}\left( {s < X < 0.8} \right) = 2 \times {\text{P}}\left( {2s < X < 0.8} \right)$, and that 0.25 < s < 0.4 , find the value of s.

$a\left[ {\int_0^{0.5} {3x\,{\text{d}}x}  + \int_{0.5}^2 {\left( {2 - x} \right)} \,{\text{d}}x} \right] = 1$     M1

Note: Award the M1 for the total integral equalling 1, or equivalent.

$a\left( {\frac{3}{2}} \right) = 1$     (M1)A1

$a = \frac{2}{3}$     AG

[3 marks]

EITHER

$\int_0^{0.5} {2x\,{\text{d}}x}  + \frac{2}{3}\int_{0.5}^1 {\left( {2 - x} \right)} \,{\text{d}}x$     (M1)(A1)

$= \frac{2}{3}$     A1

OR

$\frac{2}{3}\int_1^2 {\left( {2 - x} \right)} \,{\text{d}}x = \frac{1}{3}$     (M1)

so ${\text{P}}\left( {X < 1} \right) = \frac{2}{3}$      (M1)A1

[3 marks]

${\text{P}}\left( {s < X < 0.8} \right) = \int_s^{0.5} {2x\,{\text{d}}x}  + \frac{2}{3}\int_{0.5}^{0.8} {\left( {2 - x} \right)} \,{\text{d}}x$     M1A1

$= \left[ {{x^2}} \right]_s^{0.5} + 0.27$

$0.25 - {s^2} + 0.27$     (A1)

${\text{P}}\left( {2s < X < 0.8} \right) = \frac{2}{3}\int_{2s}^{0.8} {\left( {2 - x} \right)} \,{\text{d}}x$     A1

$= \frac{2}{3}\left[ {2x - \frac{{{x^2}}}{2}} \right]_{2s}^{0.8}$

$\frac{2}{3}\left( {1.28 - \left( {4s - 2{s^2}} \right)} \right)$

equating

$0.25 - {s^2} + 0.27 = \frac{4}{3}\left( {1.28 - \left( {4s - 2{s^2}} \right)} \right)$     (A1)

attempt to solve for s      (M1)

s = 0.274      A1

[7 marks]