# How to derive quantiles of a non-standard normal distribution [duplicate]

Let X = [11.17, 9.52, 10.69, 9.84, 10.84, 9.88, 10.28, 12.23, 8.49, 10.79] be a normally distributed data with mean = 10.37 and standard deviation = 1.02. How to determine the corresponding value to +1.02 and -1.02 standard deviations, which covers 69% area of the probability distribution.

## marked as duplicate by Glen_b, Xi'an, Andy, gung♦, Scortchi♦Nov 25 '14 at 9:49

• The question is unclear as stated: do you ask what is the mean+standard deviation value? 10.37+1.02=11.39... A random variable does not take an actual value along the x-axis, but any value when realised. Please rephrase title and text. – Xi'an Nov 25 '14 at 5:10
• @Xi'an No, I didn't asked for mean + standard deviation (i.e., 10.37 + 1.02 = 11.39). The min = 8.49 and the max = 12.23. How to determine the exact lower and upper values which covers 0.69% (derived from 1.02 from left and right side of the mean) area. – mani Nov 25 '14 at 5:59
• Please rephrase your title to How to derive quantiles of a non-standard distribution. The use of the same value 1.02 both for the empirical standard deviation and for the 84% quantile of the N(0,1) distribution is quite confusing. – Xi'an Nov 25 '14 at 6:08
• The highly detailed and informative answer provided by Glen_b to your earlier question already makes this point that you seem confused between the standard deviation of your sample and the normal quantiles. – Xi'an Nov 25 '14 at 6:11
• what do you mean by 84% quantile? – mani Nov 25 '14 at 6:17

Since for a standard normal $X\sim N(0,1)$, $-1.02\le X\le 1.02$ with probability $0.69$, for a normal $Y\sim N(10.37,1.02)$, $$9.33=-1.02\cdot1.02+10.37\le Y\sim1.02\cdot X+10.37\le 1.02\cdot 1.02+10.37=11.41$$