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From |
"Brian P. Poi" <bpoi@stata.com> |

To |
statalist@hsphsun2.harvard.edu |

Subject |
RE: st: how to find the integral for a portion of a normal distribution. |

Date |
Tue, 4 May 2010 13:44:09 -0500 (CDT) |

On Tue, 4 May 2010, Buzz Burhans wrote:

Suppose I have an intervention applied to cows with a demonstrated mean milk yield response of +2.05 liters, sd 1.74. Suppose I am interested a 1 liter cut point. I know how to find the proportion of responses at or below the 1 liter response; and the proportion of responses at or above the one liter response. This is what you are doing here, or it can be done with a z score. My question is, if the response is normally distributed, given n cows, how many total liters were included or accumulated in the only responses below 1 liter, and how many total liters were accumulated in only the responses at or above 1 liter. (Negative responses are possible). My interest is in the total liters, not the fraction of the total population either above or below the cutpoint.

For a random variable distributed N(mu, sigma), E[X | X < 1] = mu + sigma*lambda(alpha) where alpha = (1 - mu) / sigma lambda(alpha) = -phi(alpha) / Phi(alpha) Given a sample of size n, the number of observations less than 1 is numltone = n*Phi(alpha) so the total amount of milk given that X < 1 is milk = numltone*E[X | X < 1] In Stata, . clear all . set mem 200m (204800k) . set seed 1 . drawnorm x, means(2.05) sds(1.74) n(10000000) clear (obs 10000000) . summ x if x < 1, mean . di r(sum) -188292.37 . scalar alpha = (1 - 2.05) / 1.74 . scalar lambda = -1*normalden(alpha) / normal(alpha) . scalar condmean = 2.05 + 1.74*lambda . scalar numltone = 10000000*normal(alpha) . scalar condtotal = condmean*numltone . di condtotal -187432.23 Similarly, E[X | X > 1] = mu + sigma*lambda'(alpha) where lambda' = phi(alpha) / (1 - Phi(alpha)) In Stata, . scalar lambda2 = normalden(alpha) / (1 - normal(alpha)) . scalar condmean2 = 2.05 + 1.74*lambda2 . scalar condtotal2 = condmean2*(10000000 - numltone) . di condtotal2 20687432 . sum x if x > 1, mean . di r(sum) 20685854 -- Brian Poi -- bpoi@stata.com * * For searches and help try: * http://www.stata.com/help.cgi?search * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**Follow-Ups**:**RE: st: how to find the integral for a portion of a normal distribution.***From:*"Buzz Burhans" <buzzb3@earthlink.net>

**Re: st: how to find the integral for a portion of a normal distribution.***From:*Steve Samuels <sjsamuels@gmail.com>

**References**:**RE: st: how to find the integral for a portion of a normal distribution.***From:*"Buzz Burhans" <buzzb3@earthlink.net>

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