Problem source

An experiment produces a random number $T$ uniformly distributed on $[0,1]$. Let $X$ be the larger root of the equation
\[x^{2}+2x+T=0.\]
What is the probability that $X>-1/3$? Find $\mathbb{E}(X)$ and show that $\mathrm{Var}(X)=1/18$. The experiment is repeated independently 800 times generating the larger roots $X_{1}, X_{2}, \dots, X_{800}$. If
\[Y=X_{1}+X_{2}+\dots+X_{800}.\]
find an approximate value for  $K$ such that
\[\mathrm{P}(Y\leqslant K)=0.08.\]

Solution source

$(x+1)^2+T-1 = 0$ so the larger root is $-1 + \sqrt{1-T}$

\begin{align*}
&& \mathbb{P}(X > -1/3) &= \mathbb{P}(-1 + \sqrt{1-T} > -1/3) \\
&&&=  \mathbb{P}(\sqrt{1-T} > 2/3)\\
&&&=  \mathbb{P}(1-T > 4/9)\\
&&&= \mathbb{P}\left (T < \frac59 \right) = \frac59
\end{align*}

Similarly, for $t \in [-1,0]$

\begin{align*}
&& \mathbb{P}(X \leq t) &= \mathbb{P}(-1 + \sqrt{1-T} \leq t) \\
&&&=  \mathbb{P}(\sqrt{1-T} \leq t+1)\\
&&&=  \mathbb{P}(1-T \leq (t+1)^2)\\
&&&= \mathbb{P}\left (T  \geq  1-(t+1)^2\right) = (t+1)^2 \\
\Rightarrow && f_X(t) &= 2(t+1) \\
\Rightarrow && \E[X] &= \int_{-1}^0 x \cdot f_X(x) \d x \\
&&&= \int_{-1}^0 x2(x+1) \d x \\
&&&= \left [\frac23x^3+x^2 \right]_{-1}^0 \\
&&&= -\frac13 \\
&& \E[X^2] &=  \int_{-1}^0 x^2 \cdot f_X(x) \d x \\
&&&= \int_{-1}^0 2x^2(x+1) \d x \\
&&&= \left [ \frac12 x^4 + \frac23x^3\right]_{-1}^0 \\
&&&= \frac16 \\
\Rightarrow && \var[X] &=  \E[X^2] - \left (\E[X] \right)^2 \\
&&&= \frac16 - \frac19 = \frac1{18}
\end{align*}

Notice that by the central limit theorem $\frac{Y}{800} \approx N( -\tfrac13, \frac{1}{18 \cdot 800})$. Also notice that $\Phi^{-1}(0.08) \approx -1.4 \approx -\sqrt{2}$

Therefore we are looking for roughly $800 \cdot (-\frac13 -\frac{1}{\sqrt{18 \cdot 800}} \sqrt{2})) = -267-9 = -276$