5 ms·

So, can you say what you found? What is the top 20% and top 1% based on your analysis?

• senkora 3y ago
  Yeah, sure. Here's the quantiles for the median male: > for (q in c(0.01, 0.05, 0.5, 0.80, 0.95, 0.99)) { print(qnorm(q, 6, 1.63)) } [1] 2.208053 [1] 3.318889 [1] 6 [1] 7.371843 [1] 8.681111 [1] 9.791947 And, since it's a little sloppy to just take the medians of the means and standard deviations and assume that that's meaningful, here's the direct answer to the question: "What percentage of males would be thought to be above an X by the top 20% and top 1% female?" > personal_distributions = function(df) { df %>% group_by(iid) %>% summarize(mean_attr_o = mean(attr_o, na.rm = TRUE), max_attr_o = max(attr_o, na.rm = TRUE), sd_attr_o = sd(attr_o, na.rm = TRUE)) %>% ungroup()} > pdists = personal_distributions(male.df) > prob_q_thinks_person_above_score = function(df, quantile, score) { (nrow(pdists %>% filter(qnorm(quantile, mean_attr_o, sd_attr_o) >= score))) / nrow(pdists) } > for (quantile in c(0.50, 0.80, 0.99)) { for (score in c(7, 8, 9)) { print(paste(quantile, score, prob_q_thinks_person_above_score(pdists, quantile, score))) } } [1] "0.5 7 0.220216606498195" [1] "0.5 8 0.0397111913357401" [1] "0.5 9 0" [1] "0.8 7 0.620938628158845" [1] "0.8 8 0.296028880866426" [1] "0.8 9 0.0649819494584837" [1] "0.99 7 0.985559566787004" [1] "0.99 8 0.895306859205776" [1] "0.99 9 0.711191335740072" So, the data suggests that, for 71% of males, at least 1 in every 100 women will consider them at least a 9, given that the ratings of the male's attractiveness are uniformly distributed. (but, other comments seem to suggest that those ratings are probably not a uniform distribution)