91Ó°ÊÓ

X follows the normal distribution with a mean of 25 and a variance of 1.

Let f₁( x) denote the p.d.f of x that the person has glaucoma and f₁( x) the p.d.f of x that the person does not have glaucoma.

Let A₁denote the event that the person has glaucoma and A₂ the event that the person does not have glaucoma.

Now,

\[\begin{array}{l}P\left( {{A_1}} \right) = 0.1\\P\left( {{A_2}} \right) = 1 - 0.1 = 0.9\end{array}\]

Therefore, the conditional probability is

\[P\left( {{A_1}\left| {X = x} \right.} \right) = \frac{{P\left( {{A_1}} \right){f_1}\left( x \right)}}{{P\left( {{A_1}} \right){f_1}\left( x \right) + P\left( {{A_2}} \right){f_2}\left( x \right)}}\]

\[P\left( {{A_1}\left| {X = x} \right.} \right) \ge \frac{1}{2}\]holds if

\[P\left( {{A_1}} \right){f_1}\left( x \right) > P\left( {{A_2}} \right){f_2}\left( x \right)\]

i.\[{e^{ - \frac{1}{2}{{\left( {x - 25} \right)}^2}}} > 9{e^{ - \frac{1}{2}{{\left( {x - 20} \right)}^2}}}\]

ii.\[ - \frac{1}{2}{\left( {x - 25} \right)^2} > \log 9 - \frac{1}{2}{\left( {x - 20} \right)^2}\]

iii.\[{\left( {x - 20} \right)^2} - {\left( {x - 25} \right)^2} > 2\log 9\]

iv.\[10x - 225 > 2\log 9\]

v.\[x > 22.5 + \frac{{\log \left( 9 \right)}}{5}\]