Module 6: Normal Distribution
The Standard Normal Distribution
Barbara Illowsky & OpenStax et al.
The standard normal distribution is a normal distribution of standardized values called z-scores. A z-score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
x = μ + (z)(σ) = 5 + (3)(2) = 11
The z-score is three.
The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation [latex]displaystyle{z}=frac{{x - mu}}{{sigma}}[/latex] produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean μ and standard deviation σ.
The following two videos give a description of what it means to have a data set that is “normally” distributed.
Z-Scores
If X is a normally distributed random variable and X ~ N(μ, σ), then the z-score is:
[latex]displaystyle{z}=frac{{x - mu}}{{sigma}}[/latex]The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. Values of x that are larger than the mean have positive z-scores, and values of x that are smaller than the mean have negative z-scores. If x equals the mean, then x has a z-score of zero.
Example
Suppose X ~ N(5, 6). This says that x is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:
[latex]displaystyle{z}=frac{{x - mu}}{{sigma}}[/latex]= [latex]displaystyle{z}=frac{{17-5}}{{6}}={2}[/latex]
This means that x = 17 is two standard deviations (2σ) above or to the right of the mean μ = 5. The standard deviation is σ = 6.
Notice that: 5 + (2)(6) = 17 (The pattern is μ + zσ = x)
Now suppose x = 1. Then: [latex]displaystyle{z}=frac{{x - mu}}{{sigma}}[/latex] = [latex]displaystyle {z}=frac{{1-5}}{{6}} = -{0.67}[/latex]
(rounded to two decimal places)
This means that x = 1 is 0.67 standard deviations (–0.67σ) below or to the left of the mean μ = 5. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)
Summarizing, when z is positive, x is above or to the right of μ and when zis negative, x is to the left of or below μ. Or, when z is positive, x is greater than μ, and when z is negative x is less than μ.
try it
What is the z-score of x, when x = 1 andX ~ N(12,3)?
[latex]displaystyle {z}=frac{{1-12}}{{3}} = -{3.67}[/latex]
Example
Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N(5, 2). Fill in the blanks.
- Suppose a person lost ten pounds in a month. The z-score when x = 10 pounds is z = 2.5 (verify). This z-score tells you thatx =