What is the mean of a lopsided normal distribution?

I've become interested lately in using the breeder's equation from genetics (used for selective breeding involving additive traits). In order to use it, though, you have to know the mean of your parental group. For example, if I have a normally distributed population and I eliminate the bottom 20% of the population from the breeding pool, then the remaining 80% is my parental group I use in breeding the next generation. I'll use the mean of that group along with an estimate of the narrow heritability of the trait and the mean of my original population to figure out where the mean of my next generation stands. There's just one problem: not being very statistically literate, the one thing I'm not clear on is how exactly I would compute the mean for the parental group given that I'm faced with a lopsided distribution by making a cut from one end of curve. To throw out a random problem or two, lets say that I have fixed the original mean as 100 points and defined a standard deviation as being 15 points on my scale. How might I figure out the parental mean if I were to retain only the top 91.5% or 78.1% of my normally distributed population? How do I know what mean the top x % of a normal distribution actually would possess? Can someone please show me the general formula for working problems of this class? Thanks for your help. I actually found a Java applet that will allow you to plug in certain numbers but I'd still like to learn the math behind it and how to start from the desired percentiles themselves rather than retrieving percentiles by randomly plugging in certain intervals: http://onlinestatbook.com/2/java/normal.html

Vensten said:

I've become interested lately in using the breeder's equation from genetics (used for selective breeding involving additive traits). In order to use it, though, you have to know the mean of your parental group. For example, if I have a normally distributed population and I eliminate the bottom 20% of the population from the breeding pool, then the remaining 80% is my parental group I use in breeding the next generation. I'll use the mean of that group along with an estimate of the narrow heritability of the trait and the mean of my original population to figure out where the mean of my next generation stands. There's just one problem: not being very statistically literate, the one thing I'm not clear on is how exactly I would compute the mean for the parental group given that I'm faced with a lopsided distribution by making a cut from one end of curve. To throw out a random problem or two, lets say that I have fixed the original mean as 100 points and defined a standard deviation as being 15 points on my scale. How might I figure out the parental mean if I were to retain only the top 91.5% or 78.1% of my normally distributed population? How do I know what mean the top x % of a normal distribution actually would possess? Can someone please show me the general formula for working problems of this class? Thanks for your help.

Click to expand...

Edit: My unedited post was wrong, I believe. I assumed something that wasn't relevant.

Are you fluent in integral calculus? There is an "easy" way to find your new mean: the bottom 20% has a z-score of approximately -0.85. After standardizing your data, the expected value (given that your data lie in the top 80%) should be:
\(\displaystyle \displaystyle \dfrac{1}{\sqrt{2\pi}}\int_{-0.85}^{\infty} xe^{-x^2/2} dx\approx 0.278\)

That is, 0.278 standard deviations from your mean, in the positive direction.

Vensten said:

For example, if I have a normally distributed population and I eliminate the bottom 20% of the population from the breeding pool, then the remaining 80% is my parental group I use in breeding the next generation. I'll use the mean of that group . . .

Click to expand...

If you cut off 20%, the remaining area under the distribution is 0.8. The mean will be where half of that amount lies in the normal distribution. Look where the value of F (cumulative distribution) lies in the table (or using your applet). Counting downward from 1, look for F = 1-(0.8/2) = 0.6. You can also think of that as counting upward from 0.2 till you reach half of 0.8 -- also gives F=0.6 in the normal distribution to be the point where half of the lopped-off distribution occurs. From the table, I find that to be x=0.25, meaning 0.25 standard deviations above the original mean.

If you cut a fraction A from the area, then the relevant value of F is 1 - (1-A)/2