tag:blogger.com,1999:blog-241203637384931854.post7695504121888390897..comments2009-06-19T06:18:38.220-05:00Comments on Applied Math 40S (Winter 2009): Darren Kuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-241203637384931854.post-37356610348380283452009-04-16T00:32:00.000-05:002009-04-16T00:32:00.000-05:00Oh, sorry, jumped to conclusions. I see what you d...Oh, sorry, jumped to conclusions. I see what you did about 99 ≠ 100 now. Hmm, I dunno, I guess I'll wait and see what Mr. K says, because I think that 99.99999999999999 would still technically be less than 100, so to jump down to 99 excludes quite a bit of data.... I dunno.Danielnoreply@blogger.comtag:blogger.com,1999:blog-241203637384931854.post-63711833184018146942009-04-16T00:29:00.000-05:002009-04-16T00:29:00.000-05:00Good effort, but since 100 IS the mean, it has a z...Good effort, but since 100 IS the mean, it has a z-score of zero. Therefor, you would use normalcdf(-5,0) which is .50 or 50%.<br /><br />However, these steps are unnecessary, because a normal curve is symmetrical about the mean. If the mean is 100, 50% of the data lies below 100. Furthermore, 50% of the data lies above 100.Danielnoreply@blogger.com