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From |
"Buzz Burhans" <buzzb3@earthlink.net> |

To |
<statalist@hsphsun2.harvard.edu> |

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

Date |
Thu, 6 May 2010 08:58:34 -0600 |

Thanks to all who responded to my question. Your help is most appreciated. Many of the responses essentially involved using drawnorm to simulate the quantity desired, which is useful, and it was interesting to see a couple of minor variations on doing this. My particular interest was in calculating (as opposed to simulating) this quantity based on knowledge of the information I had at hand about the mean, standard deviation, and expected normality of the response. In regards to this, Brian Poi's response was what I was specifically looking for. Because he laid it out so nicely in steps, following through his response not only gave me the answer, but allowed me to think through what it represented. I appreciate it. I am most appreciative of all the responses, thanks again to all who took the time to respond and put up with my initially inarticulate description of my question. Buzz Buzz Burhans, Ph.D. Dairy-Tech Group So. Albany, VT / Twin Falls ID Phone: 802-755-6842 Cell: 208-320-0829 Fax VT: 802-755-6842 Fax ID: 208-735-1289 Email: buzzb3@earthlink.net -----Original Message----- From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Brian P. Poi Sent: Tuesday, May 04, 2010 12:44 PM To: statalist@hsphsun2.harvard.edu Subject: RE: st: how to find the integral for a portion of a normal distribution. 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. > Assuming I understand the question correctly, I think this does what you want. 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/ * * 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/

**References**:**RE: st: how to find the integral for a portion of a normal distribution.***From:*"Brian P. Poi" <bpoi@stata.com>

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