Probability problem gone horribly wrong

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Homework Help Overview

The problem involves calculating probabilities related to the weights of randomly selected men and women, both of which are normally distributed. The specific questions focus on the probability that a randomly selected man weighs less than a randomly selected woman, and the probability that the average weight of a sample of men is less than that of a sample of women.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of expected values and variances for the differences in weights. There is an exploration of the implications of sample sizes on these calculations, particularly regarding the variance formula.

Discussion Status

Some participants are examining the calculations and questioning whether the expected value changes with sampling. Others are considering the implications of obtaining a very small probability and discussing the statistical significance of the results. There is no clear consensus, but the discussion is productive in exploring the nuances of the problem.

Contextual Notes

Participants are navigating the complexities of probability calculations involving normal distributions, and there is mention of potential confusion regarding the use of sample sizes in variance calculations.

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

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