1. The problem statement, all variables and given/known data Weight: men --> mean = 75, standard deviation = 15 Weight: women ---> mean = 55, standard deviation = 11 both normally distributed one woman and one man are randomly picked. 1) probability that the man weighs less than the woman 2) if there was a random sample of 30 men and 40 women from this population, find the probability that the average weight of the men is less than the average weight of the women 2. Relevant equations standard deviation = square root of variance z = (value - mean) / standard deviation z (normal) table 3. The attempt at a solution I'll try to make it as clear as possible here: 1) P(man < woman) so P(W-M > 0) then I let a dummy variable X = W - M so expected(X) = 55 - 75 = -20 variance(X) = 11^2 + 15^2 = 346 so P(X > 0) = 1 - P(X < 0) = 1 - P(Z < 1.0752) = 0.140 2) I thought this was pretty much the same as above but the variance calculation was different: Expected(X) = -20 Var(X) = 11^2 / 40 + 15^2 / 30 = 10.525 so P(X > 0) = 1 - P(X<0) = 1 - P( Z < 20 / square root of (10.525) ) = 1 - P(z < 6.1648) and clearly, I can't get a z < 6.1648 for the normal distribution! It's not in the table, and it wouldn't make sense to have a probability of 0. Help!