Let \(X\) denote the number of times Joe misses the chute. The distribution for \(X\) is binomial with \(n=500\) trials and success probability \(p=0.2\) (successful in regards to missing the chute).
You then know that the average number of times that he misses is \(np=500\times0.2=100\), with a standard deviation of \(\sqrt{np(1-p)}\approx8.944\). These parameters will come in handy in just a moment.
You're asked to compute \(\mathbb P(X<80)\). With the continuity correction, this is approximated by
\(\mathbb P(X<80)\approx\mathbb P(X<79.5)=\mathbb P\left(\dfrac{X-100}{8.944}<\dfrac{79.5-100}{8.944}\right)=\mathbb P(Z<-2.292)\approx0.011=1.1\%\)
