Distribution Families

## The distribution and its two parameters A normal (Gaussian) distribution is fully specified by two numbers: the mean `mu`, which locates the peak, and the standard deviation `sigma`, which sets the width. Here `mu = 100` and `sigma = 15`. Everything else about the curve — its symmetry, its bell shape, the share of mass in any region — follows from those two parameters alone. Note the notation trap: `N(100, 15)` is written by some authors with 15 as the standard deviation and by others with the second slot holding the variance. Always say which you mean; a candidate who silently treats 15 as a variance gets `sigma = 3.87` and every subsequent number wrong. ## Standardising: the z-score A z-score re-expresses a raw value as a number of standard deviations from the mean: ``` z = (x - mu) / sigma ``` For `x = 130`: `z = (130 - 100) / 15 = 30 / 15 = 2`. Standardising maps any normal distribution onto the standard normal, the one with mean 0 and standard deviation 1. That is why a single table or a single function suffices for all normal distributions: the shape never changes, only the location and the scale, so shifting and rescaling reduces every case to one reference curve. A z-score is also unit-free, which is what makes a score from one test comparable to a score from another with a different mean and spread. ## What the 68-95-99.7 rule states For a normal distribution: ``` P(|z| < 1) ≈ 0.68 -> about 68% within one SD P(|z| < 2) ≈ 0.95 -> about 95% within two SDs P(|z| < 3) ≈ 0.997 -> about 99.7% within three SDs ``` Translated to this score scale, roughly 68% of scores lie between 85 and 115, roughly 95% between 70 and 130, and roughly 99.7% between 55 and 145. ## Getting to the answer The question asks for one tail. About 95% of the mass lies inside two SDs, so about 5% lies outside — but that 5% is the two tails together. Symmetry splits it evenly: ``` P(X > 130) = P(z > 2) ≈ (1 - 0.95) / 2 = 0.025 ``` About 2.5%, or one score in forty. Forgetting the halving step and answering 5% is the single most common error on this question. ## Approximate versus exact The three percentages are deliberately rounded for mental arithmetic. The precise values are: ``` P(|z| < 1) = 0.6827 -> one tail 0.1587 P(|z| < 2) = 0.9545 -> one tail 0.0228 P(|z| < 3) = 0.9973 -> one tail 0.00135 ``` So the exact answer above 130 is 2.28%, not 2.50%. The gap matters in one direction especially: the 95% figure people quote for confidence intervals corresponds to `z = 1.96`, not `z = 2`. Using 2 as shorthand is fine for a whiteboard estimate and wrong for a published number. Being explicit about which one you are using is part of a good answer. ## Reading the tails A few landmarks are worth carrying in your head, because interviewers ask them in either direction: - Above one SD: about 16% (half of the 32% outside one SD). Here that is scores above 115. - Above two SDs: about 2.5%. Here, above 130. - Above three SDs: about 0.15%. Here, above 145. The reverse direction is equally fair game: "what score marks the top 16%?" is answered by `mu + 1*sigma = 115` without any table at all. For a z that is not a whole number — say `z = 1.47` for a score of 122 — the rule gives no exact answer; you interpolate roughly (about 7%) or look up the normal CDF. ## The assumption behind the arithmetic Every number above depends on the variable actually being normal. The rule is not a general property of symmetric or bell-ish distributions: a heavy-tailed symmetric distribution can put far more than 5% beyond two SDs, and a skewed distribution has no symmetry to split a tail with. Applying the rule to a bounded or strongly skewed quantity — a count, a duration, a proportion — produces confident nonsense, sometimes including probability mass on impossible negative values. Stating that caveat unprompted is what separates a mechanical answer from a statistical one.