The probability that a randomly selected variable is positive is the same as P(X>0).
The z-score for X=0 is:
[tex]z=\frac{X-\mu}{\sigma}=\frac{0-20}{25}=\frac{-20}{25}=-\frac{4}{5}=-0.8.[/tex]
From the standard normal distribution table, P(z<-0.8) has the probability:
[tex]P(z<-0.8)=0.21186[/tex]
It follows that:
[tex]\begin{gathered} P(z>-0.8)=1-P(z<-0.8)=1-0.21186 \\ \Rightarrow P(z>-0.8)=0.78814\approx0.788 \end{gathered}[/tex]
Hence, the required probability is about 0.788.
The answer is 0.788.
