Question 1185639
The original distribution is B(90, 0.90).  Use the normal distribution N(np, npq) = N(81, 8.1) to approximate the probability {{{P(X[B] < 72)}}}.  
(Both np and n(1-p) are greater than 5, so we can do the normal approximation.)


{{{P(X[B] < 72) =  P(X[N] < 71.5) = P(Z < (71.5 - 81)/sqrt(8.1) = -3.338) = 0.0004}}}.


The answer is thus (c).