Answer
$0.0497$
Work Step by Step
Given $p=0.53,q=1−p=0.47,n=250, check np=132.5>5,nq=117.5>5$,
we can use a normal distribution approximation to the problem with
$\mu=np=132.5,\sigma=\sqrt {npq}=7.891$
For $X<120$ in a binomial distribution, we use $X<119.5$ in the normal approximation, and
$z=\frac{119.5-132.5}{7.891}=-1.6474$ thus $P(X<119.5)=P(z)=0.0497$
