PicoBlog

Another perfect SAT score. Not nearly as impressive given he took it in 2001 - so this equates to a z-score of 3.1 or above using the traditional 980/200 method for this time period. He is very wealthy - consistently in the top 10, giving him a wealth z-score of 6. Because it’s impossible to actually find somebody with that z-score in a simulation, I’ll just assume he is above a z-score of 4 in wealth.

Estimation ended with a mean of 146 and standard error of 8.7.

set.seed(6) g <- rnorm(60000000) c2 <- 0.84*g + rnorm(60000000)*sqrt(1-0.84^2) c1 <- 0.35*g + rnorm(60000000)*sqrt(1-0.35^2) subby1 <- data.frame(c2, c1) subby1$g = g subby2 <- subset(subby1, (subby1$c1 > 4) & (subby1$c2 > 3)) mean(subby2$g)*15 sd(subby2$g)*15 #app <- append(c(), subby2$g) app <- append(app, subby2$g) mean(app)*15 sd(app)*15

ncG1vNJzZmirlZe3prrSnqVnq6WXwLWtwqRlnKedZL1wsdKtoKaZpJ67qHnToZxmoaFivKd5zJqppGWqqrCssdGbnKuf