# Probability problem gone horribly wrong

1. Jun 11, 2009

### war485

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!

2. Jun 12, 2009

### CompuChip

Let <W> be the average weight of 40 women, and <M> the average weight of 30 men. Both are again normally distributed. Then consider X = <W> - <M> like in the first problem.

3. Jun 12, 2009

### war485

I did that. Still getting same answer. Does the expected value change when you take a random sample from a population? Maybe I did my variance formula wrong? Is it n-1 instead of n?

4. Jun 12, 2009

### CompuChip

Ah, I see that is precisely what you did, I just couldn't see that right away because you all compiled it into one line.
I don't really see anything wrong, then.
It's not really weird that the probability is small (it is non-zero, but very small). I mean, for the average weight of the men to be less than that of the women, you would have to have the statistical oddity of picking significantly light (as in: a few standard deviations below average) men and/or significantly heavy (a few sigma above average) women. And not just 1 of them, but several.

5. Jun 12, 2009

### war485

never had such a small answer before (pretty much close to zero) so I thought something was wrong. Thanks for checking. :D