Standardizing Normally Distributed Random Variables

Published: 10 September 2013
on channel: jbstatistics
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I discuss standardizing normally distributed random variables (turning variables with a normal distribution into something that has a standard normal distribution). I work through an example of a probability calculation, and an example of finding a percentile of the distribution. It is assumed that you can find values from the standard normal distribution, using either a table or a computer.

The mean and variance of adult female heights in the US is estimated from data found in a National Health Statistics Report:

McDowell MA, Fryar CD, Ogden CL, Flegal KM. Anthropometric reference data for children and adults: United States, 2003-2006. National health statistics re- ports; no 10. Hyattsville, MD: National Center for Health Statistics. 2008.

For those using R, here is the R code for the examples used in this video:

American female heights example (approximately normally distributed with a mean of 162.2 and a standard deviation of 6.8).

Finding the probability that a randomly selected female is taller than 170.5 cm.
Easiest way:
1-pnorm(170.5,162.2,6.8)
[1] 0.111121

Standardizing route:
1-pnorm((170.5-162.2)/6.8,0,1)
[1] 0.111121
The default in R's pnorm is the standard normal distribution (mean=0, SD=1), so the mean and SD can be left out when dealing with the standard normal.
1-pnorm((170.5-162.2)/6.8)
[1] 0.111121

Finding the probability that a randomly selected female has a height between 150.5 and 170.5.
Easiest way:
pnorm(170.5,162.2,6.8)-pnorm(150.5,162.2,6.8)
[1] 0.8462162

Standardizing route:
pnorm((170.5-162.2)/6.8)-pnorm((150.5-162.2)/6.8)
[1] 0.8462162

10th percentile of heights of adult American females.
Easiest:
qnorm(.1,162.2,6.8)
[1] 153.4854

Alternatively, via the standard normal distribution:
qnorm(.1)
[1] -1.281552
That's the 10th percentile of the standard normal distribution. Converting to the distribution of heights,
-1.281552*6.8+162.2
[1] 153.4854


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