Question 929314
 n=58, p= 0.3, q = .7
........
a) P(x=12) = 58C12(.3)^12(.7)^46  0r  P(x=12) = binompdf(58, .3,12) = .0355
b) normal distribution can be used to estimate this probability: both np  and nq  ≥  5
.........
Using a Casio fx-115 ES plus
mean = .3•58 = 17.4,  s = {{{sqrt(.3*.7*58)}}} = 3.49
P(x = 12) = normalpdf(12, 3.49, 17.4) = .0345 
....
c) .0355 - .0345 = .0010, the amount the probabilities differ