[tex]\mathbb P(142<X<174)=\mathbb P\left(\dfrac{142-142}{16}<\dfrac{X-142}{16}<\dfrac{174-142}{16}\right)=\mathbb P(0<Z<2)[/tex]
Approximately 95% of any normal distribution lies within two standard deviations of the mean, i.e. [tex]\mathbb P(-2<Z<2)\approx0.95[/tex]. Because the distribution is symmetric, you have [tex]\mathbb P(-2<Z<2)=2\mathbb P(0<Z<2)[/tex], so [tex]\mathbb P(0<Z<2)\approx\dfrac{0.95}2=0.475\approx0.48[/tex].
