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Start with a normal distribution with mean M and standard deviation S. Now exclude all values below the Kth percentile. What is the mean of the remaining 100%-K% of the values as a function of M and S?

I'd just like a simple answer and don't need a proof or generalization which are provided in lengthy answers to related questions which don't actually provide a simple equation from which to compute the answer, and assume the reader knows the difference between the functions 𝜙 and Φ (I don't):

Expected value of x in a normal distribution, GIVEN that it is below a certain value (2 answers) Expectation of truncated normal (3 answers)

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  • $\begingroup$ It's just the expected value for truncated normal, you are looking for how it was derived? $\endgroup$
    – Tim ♦
    Mar 14 at 16:08

1 Answer 1

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This is the one-sided truncated normal distribution with mean and variance given by:

  • $\mathbb{E}(X | X>a) = \mu + \sigma \phi(\alpha)/Z$
  • $Var( X|X>a) = \sigma^2[1+\alpha\phi(\alpha)/Z - (\phi (\alpha)/Z)^2 ]$

where $\alpha=(a-\mu)/\sigma$ and $Z = 1-\Phi(\alpha)$.

In your case, you'd plug in $a = 0.98$.

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