Question 427216
Let X = r.v. representing the weight of certain containers.
Then the distribution in question is Y = 36X.  This is also a normal distribution with mean E(Y) = E(36X) = 36E(X) = 36*16 = 576, and variance Var(Y) = Var(36X) = {{{36^2Var(X) = 1296Var(X) = 1296(0.6)^2}}} = 466.56.  The standard deviation is thus {{{sqrt(466.56) = 21.6}}}.  You're looking for the probability {{{P(Y > 580) = P(Z = (Y - 576)/21.6 > (580 - 576)/21.6 = 0.185)}}}, or P(Z > 0.185).  Now use any table of standard normal probabilities.