$P(X\textgreater60) = P(Z\textgreater\frac{60-55.6}{2.2}) = p(Z\textgreater2) =1-0.9772 = 0.0228$ 
The average weight of dogs that come to a certain vet\textquotesingle s office is 55.6 lbs, with a standard deviation of 2.2 lbs. If the weights are normally distributed, what percent of dogs weight more than 60 lbs? 
66.8\% 
47.2\% 
33.4\% 
15.9\% 
2.28\% 