- #1
war485
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Homework Statement
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
Homework Equations
standard deviation = square root of variance
z = (value - mean) / standard deviation
z (normal) table
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!