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by blahblah3
3228 days ago
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If M ~ N(u,v_m) and W ~ N(u,v_w) [where N refers to the normal distribution], with v_m > v_w, then P(M > T) > P(W > T) for all T > u. I.e if the trait is approximately normally distributed with equal means among two groups, the group with the higher variance will exhibit more extreme values. Why is this a change in argument? |
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