Standard Normal Distribution

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Standard Normal Distribution

Since there is a different normal distribution for every pair of µ and σ (mean and standard deviation), we often transform the variable by subtracting the mean and dividing by the standard deviation to produce a standardized variable, also called a Z Score.

The formula for the Z score is as follows:

              z = x-µ/σ

If X is normally distributed N(µ, σ), the standardized variable Z has a  standard normal distribution.  Its mean is 0 and its standard deviation is 1. Denoted N(0, 1).  This table summarizes the detail:

Detail of the Characteristics of the Standard Normal Distribution

Parameters	µ = population mean; σ = population standard deviation
PDF	f(z) = 1/√2π e-z^2 where z = x-µ/σ
Domain	-∞ < z ∞
Mean	0
Standard Deviation	1

Since every transformed normal distribution will look the same, we can use a common scale, usually labeled from -3 to +3 (Empirical Rule, +/-3 σ away from the mean).  Since f(z) is a PDF, the entire area under the curve is 1.  The probability of an event P(z1 < z < z2) is a definite integral of f(z).

The Z-score table.

Tables of normal probabilities have been prepared for you so you can look up any normal desired area. One such table is the Z score, also called the Cumulative Standard Distribution.   This table is in Appendix C of the text.  (I have also scanned the chart and uploaded it to Wiggio.)

Finding z for a given area

We can also use the charts to find a corresponding z to a given area. For example, what z score defines the “top 1%” of a normal distribution? You would look to the areas for one that is approximately .99. That z-score is approx.. 2.33

Finding Areas by Using Standardized Variables

Ex. John took an economics exam and scored an 86. The class mean was 75 with a standard deviation of 7.  What percentile is John in? That is What us P(X < 86)? We need first to calculate John’s standardized Z-score.

              Z = (x - µ)/σ = (86 – 75)/7 = 11/7 = 1.57

This says that John’s score is 1.57 standard deviations above the mean. From the Z-score table, we get that P(Z < 1.57) = .9418, so John is approximately in the 94th percentile. That means that his score was better than 94 percent of the class.

Inverse Normal

How can we find the various normal percentiles (5th, 10th, 25th…95th, etc) known as the inverse normal? That is, how can we find X for a given area? We simply turn the standardizing transformation around.

              x = µ + zσ  (solving for x in z = (x-µ)/σ)

Ex. Suppose John’s economics professor has decided that any student who scores below the 10th percentile must retake the exam.  The exam scores are normal with µ = 75 and σ = 7. What is the score that would require a student to retake the exam?

(The approximate z-score for the 10th percentile is z = -1.28.)

The three ways to solve this problem are:

-        Use the table to find z = -1.28 to satisfy P(Z < -1.28) = .10

-        Substitute the values into z = x - µ/σ to get -1.28 = (x-75)/7

-        Solve for x to get x = 75 – (1.28)(7) = 66.04 (or 66 after rounding)

Students who got a 66 or less would have to retake the exam.




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