Applying the Standard Normal Distribution (Z) to Real Data

In a previous post, I introduced the concept of the standard normal variable Z and how it relates to probability. Here I will demonstrate how to translate actual data to Z and make probability interpretations (under the assumption of normality).

Example 1: Assume we are interested in tire mileage, and we know that tires have a population mean mileage μ = 36,500 σ = 5000 x ~ normal.


What is the probability that a tire (x) will last more than 40,000 miles?

To solve this problem, we translate our real word data to data that is distributed standard normal, or we convert x = 40,000 to some value of z and base our probability on z.

Step 1: Calculate Z 

z =  (x-μ)/σ = (40,000-36,500)/5000 = .70

Step 2: Find the corresponding value from the z-table

For z =.70 the value from the table is .7580

Step 3: Interpret and Calculate Probability

The probability of a value > 40,000 is the same as finding a z > .70. Based on the rules we defined for positive z-values, we know that the probability of a value greater than .70 = 1-.7580 = .2420.  This implies that the probability that Z will exceed .70 and that our original data (x =mileage) will exceed 40,000 is 24.2 %.


Example 2: Assume we are interested in tire mileage, and we know that tires have a population mean mileage μ = 36,500 σ = 5000  x ~ normal.


What is the probability that a tire (x) will last less than 30,100 miles?

 To solve this problem, we translate our real word data to data that is distributed standard normal, or we convert x = 40,000 to some value of z and base our probability on z.

Step 1: Calculate Z


z =  (x-μ)/σ = (30,100-36,500)/5000 = -1.28

Step 2: Find the corresponding value from the z-table 

For z = -1.28 the value from the table is .8999 ~ .90

Step 3: Interpret and Calculate Probability 

The probability of a value of z < -1.28 and the probability that x (mileage) will be less than 30,100 is 1-.90 = .10 or 10%.




Example 3: Assume we are interested in tire mileage, and we know that tires have a population mean mileage μ = 36,500 σ = 5000 x ~ normal.


What is the probability that a tire (x) will last less than 50,000 miles?


Step 1: Calculate Z



z =  (x-μ)/σ = (50,000-36,500)/5000 =  2.7


Step 2: Find the corresponding value from the z-table 


For z = 2.7 the value from the table is .9965 


Step 3: Interpret and Calculate Probability 

The probability of z  being less than 2.7 or x (tire mileage) being less than 50,000 is 99.65%. ( remember, for a positive Z, the probability of getting a lower value of Z is obtained from taking the corresponding value directly from the standard normal table)


Example 4: Assume we are interested in tire mileage, and we know that tires have a population mean mileage μ = 36,500 σ = 5000 x ~ normal.


What is the probability that a tire (x) will last between 30,000 and 40,000 miles?

Step 1:  Calculate P( x < 30,000)  

z =  (x-μ)/σ = (30,000-36,500)/5000 = -1.3

Based on the value from the z table, and a negative z value, the probability of getting a lower x (or the probability of mileage < 30,000) = 1-.9032 = .0968 ~ 9.7%

Step 2: Calculate P( x > 40,000)

z =  (x-μ)/σ = (40,000-36,500)/5000 =.70

From example 1 we know this turns out to be 24.2%.

Step 3:  Calculate P(30,000 <= x <= 40,000)

This is the same as finding the probability of a z between -1.3 and .70. We get this by subtracting the results from step 1 and 2:

1 - .097 - .242 = .661 ~ 66.1 % or
100 - 9.7% - 24.2%= 66.1%