# Mean of a truncated normal distribution [duplicate]

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):

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

## 1 Answer

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 $$a = 0.98$$.