[tex]\mathbb P(20<X<25)=\mathbb P\left(\dfrac{20-20}5<\dfrac{X-20}5<\dfrac{25-20}5\right)=\mathbb P(0<Z<1)[/tex]
Since approximately 68% of any normal distribution lies within one standard deviation of the mean, and the distribution is symmetric, it follows that
[tex]\mathbb P(0<Z<1)=\dfrac12\mathbb P(|Z|<1)=\dfrac{0.68}2=0.34[/tex]
