Problem source

\begin{questionparts}
\item Three real numbers are drawn independently from the continuous 
rectangular distribution on $[ 0, 1 ]\,$. The random variable $X$ is the maximum of the 
three numbers. Show that the probability that $X \le 0.8$ is $0.512\,$, and calculate the
 expectation of $X$. 
\item $N$ real numbers are 
drawn independently from a continuous rectangular distribution on $[ 0, a ]\,$. 
The random variable $X$ is the maximum of the $N$ numbers. 
A hypothesis test  with a significance level of 5\% is carried out using  the value, $x$, of
  $X $. 
The null hypothesis is that $a=1$ and 
the alternative hypothesis is that $a<1 \,$. The form of the test is such that
$H_0$ is rejected 
if $x < c\,$, for some chosen number $c\,$.
Using the approximation 
 $2^{10} \approx 10^3\,$, determine the smallest 
integer value of $N$ such that if $x \le 0.8$ 
the null hypothesis will be rejected.
With this value of $N$,
write down the probability that the null hypothesis is rejected if $a = 0.8\,$, 
and find  the probability that the null hypothesis is rejected if $a = 0.9\,$.
\end{questionparts}

Solution source

\begin{align*}
\P(X \leq 0.8) &= \P(X_1 \leq 0.8,X_2 \leq 0.8,X_3 \leq 0.8) \\
&= 0.8^3 \\
&= 0.512
\end{align*}

\begin{align*}
&& \P(X < c) &= c^3 \\
\Rightarrow && f_X(x) &= 3x^2 \\
\Rightarrow && \E[X] &= \int_0^1 x \cdot (3x^2) \, dx \\
&& &= \left [ \frac{3}{4}x^4 \right]_0^1 \\
&&&= \frac{3}{4}
\end{align*}

$X$ is distributed the maximum of $N$ numbers on $[0,a]$.
\begin{align*}
H_0 : & x= 1 \\
H_1 : & x < 1
\end{align*}

\begin{align*}
&&\P(X < c) &= c^N \\
&&&= \frac1{20} \\
\Rightarrow && N &= -\frac{\log(20)}{\log(c)}
\end{align*}
where $c = 0.8$, we have 
\begin{align*}
N &= \frac{\log(20)}{\log(5/4)} \\
&= \frac{\log(5)+\log(4)}{\log(5)-\log(4)} \\
&= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1}
\end{align*}

\begin{align*}
&& 2^{10} &\approx 10^{3} \\
&& 10\log(2) &\approx 3 (\log(5) + \log(2)) \\
&& 7\log(2) &\approx 3 \log(5) \\
&& \frac{\log(5)}{2\log(2)} &\approx \frac{7}{6}
\end{align*}

\begin{align*}
&= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1}
&= \frac{\frac{7}{6} + 1}{\frac{7}{6} -1} \\
&= 13
\end{align*}

Since $2^{10} > 10^3$  then $N=14$ is the value we seek.

$\P(X < 0.8 | a= 0.8) = 1$
$\P(X < 0.8 | a= 0.9, N=14) = \frac{8^{14}}{9^{14}}$