HP 49g+ / Probability UTPN « Next Oldest | Next Newest »

 ▼ Thomas Unregistered Posts: 21 Threads: 5 Joined: Jan 1970 07-24-2006, 02:19 PM Hi I have some problems with the normal distribution, ecpecially with the utpn-function. when i type in mean, variance and the x-variable i don't get the correct numbers. for example, if the mean is 11.5, the variance 961 and the x 0.1129 it equals 0.643311 but it should equal 0.545 according to the normal distrubution table. has anyone an idea? many thanks in advance ▼ hugh steers Unregistered Posts: 536 Threads: 56 Joined: Jul 2005 07-24-2006, 02:33 PM let p(x) be the probability from -inf to x. so p(1) = 84.1%, p(2)=97.7% and so on, so you know which way round i'm counting. i get p(.1129) = .544945, so that's where that number is coming from, its when x is already normalised. however, 1-p((.1129-11.5)/sqrt(961)) = .643312 so thats where the other number is coming from, first you normalise, then, in this case, you're calulating 1-p(x). sometimes this is called r(x) = 1-p(x) for some calculators. hope this helps, ▼ Thomas Unregistered Posts: 21 Threads: 5 Joined: Jan 1970 07-25-2006, 04:23 AM Thanks for your help, I still don't understand how to get the result 0.544945 by calculator. if i try it manually (by interpolation) it is no problem. by calclator it equals .643312. do you know what i have to type in to get the right result? thx ▼ James M. Prange (Michigan) Unregistered Posts: 1,041 Threads: 15 Joined: Jan 2005 07-25-2006, 06:16 AM Quote: Thanks for your help, I still don't understand how to get the result 0.544945 by calculator. if i try it manually (by interpolation) it is no problem. by calclator it equals .643312. do you know what i have to type in to get the right result? Well, if I type in 0 1 .1129 [+/-] UTPN [ENTER] on any RPL calculator, it returns .544945081834. 11.5 961 .1129 UTPN [ENTER] returns .643311996636. My guess is that you're misinterpreting your normal distribution table. With a variance of 961, the standard deviation is 31. The difference between your mean and x values is 11.5-.1129=11.3871, so your x value is 11.3871/31=.367325806452 standard deviations below the mean, so try .3673 in your normal distribution table. Regards,James Edited: 25 July 2006, 6:59 a.m. after one or more responses were posted ▼ Thomas Unregistered Posts: 21 Threads: 5 Joined: Jan 1970 07-25-2006, 06:58 AM hi again, it seems to be working now and i am starting to understand, i really was missinterpreting the whole thing, your post was very helpful. thank you for your assistance. regards thomas ▼ James M. Prange (Michigan) Unregistered Posts: 1,041 Threads: 15 Joined: Jan 2005 07-25-2006, 08:01 AM You're welcome. I see that you replied while I was editing my reply. The thing is that normal distribution tables assume a mean of 0 and a standard deviation (square root of the variance) of 1. Having separate tables for various different mean and standard deviation values wouldn't be practical, so to use the table, you have to compute how many standard deviations from the mean your x value happens to be; this is sometimes called a "z value". Since the normal distribution is symmetrical, tables typically cover "z" values on only one side of 0. The RPL calculators' UTPN function takes the mean, variance, and "x" values, and return the "Upper Tail Probability", that is, the probability from x to positive infinity. There's no need find the "z" value. Regards,James

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