Question

Joe is a fish thrower who throws fish into a chute at a processing plant. Joe misses the chute 20% of the time. Using the normal approximation with the continuity correction, which of these correctly represents the probability Joe will miss fewer than 80 times if he throws 500 fish?

Answer 1

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*0.2=100`, with a standard deviation of
`√(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((X-100)/(8.944)<(79.5-100)/(8.944)\right)=\mathbb P(Z<-2.292)\approx0.011=1.1\%`