Z scores (Z value) is the number of standard deviations a score or a value (x) away from the mean. In other words, Z-score measures the dispersion of data. Technically, Z-score tells a value (x) is how many standard deviations below or above the population mean (µ). If the Z value is positive, it indicates that the value or score (x) is above the mean. Similarly, if Z value is negative, it means the value (x) is below the mean.
What is Standard normal distribution?
A Normal Standard Distribution curve is a symmetric distribution where the area under the normal curve is 1 or 100%. The standard normal distribution is a type of special normal distribution with a mean (µ) of 0 and a standard deviation of 1.
A standard normal distribution always mean zero and has intervals that increase by 1. Each number on the horizontal line corresponds to z-score. Hence, use Z Scores to transform a given standard distribution into something that is easy to calculate probabilities on. As it can determine the likelihood of some event happening.
Any normal distribution with any value of mean (µ) and sigma can be transformed into the standard normal distribution, where the mean of zero and a standard deviation of 1. This is also called standardization.
A Z-scores tells how many standard deviation a value or score is from the mean (µ). For example if a Z-score negative 3 means the value (x) is 3 standard deviation left of the mean. Similarly, if the Z-score is positive 2.5 means the value (x) is 2.5 standard right of the mean (µ).
This is a common transformation, so there is a reference chart that allows us to look up values. Those values correlate to the value under the normal distribution curve – in other words, what’s the chance of an event happening. We use the Z table to find the percent chance.
How to interpret the Z-score table
Most importantly, Z-score helps to calculate how much area that specific Z-score is associated with. A Z-score table is also known as a standard normal table used to find the exact area. Z-score table tells the total quantity of area contained to the left side of any score or value (x).
There are two methods to read the Z-table:
Case 1: Use Z-table to see the area under the value (x)
In the Z-table top row and the first column corresponds to the Z-values and all the numbers in the middle corresponds to the areas.
For example a Z-score of -1.53 has an area of 0.0630 to the left of it. In other words, p(Z<-1.53) = 0.0630.
Standard normal table also used to determine the area to the right of any Z-value by subtracting the area on the left from 1. Simply, 1-AreaLeft = Arearight
For example Z-score of 0.83 has an area of 0.7967 to the left of it. So, Area of right is 1-0.7967 = 0.2033.
Case 2: Use the Z-table to see what that score is associated with a specific area.
- Pick the right Z row by reading down the right column.
- Read across the top to find the decimal space.
- Finally, find the intersection and multiply by 100.
For example: Value of Z corresponds to an area of 0.9750 to the left of it is 1.96.
How to create Z score in excel
Using Z score, find the percentage by using the formula: 1-NORMSDIST(Z), where Z is your calculated Z Score.
What if the Z Score is off the Chart?
Z score tables sometimes only go up to 3. But depending on the spread of the population, z scores could go on for a while. A Z score of 3 refers to 3 standard deviations. That would mean that more than 99% of the population was covered by the z score. There’s not a lot left, but there is some. Use Excel to find the actual value if your table doesn’t go that high.
How do you Calculate a Z Score?
The formula for transforming a score or observation x from any normal distribution to a standard normal score is :
Calculating a Z Score for a Population
Z Score for a Sample
Great example of using a Z score to determine how well you scored on a test (compared to the rest of the field) here.
How many parts in a population will be longer or greater than some number?
Z score examples using standard deviation
Example 1: Longer than
Hospital stays for admitted patients at a certain hospital are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to last for longer than 300 hours?
- x̅ = 200
Z= x- x̅/s =300-200/75= 100/75= 1.33
Z score from the table for 1.33 = 0.9082
Since, we are looking for longer, solution is P(x>300) = P(Z>1.33) = 1- P(Z<1.33)= 1-0.9082 = 0.0918 = 9.18%
Example 2: Less than
Hospital stays, for admitted patients at a certain hospital are measured in hours and were found to be normally distributed with an average of 200 hours and a standard deviation of 75 hours. How many of these stays can be expected to last less than 75 hours?
- x̅ = 200
Z= x- x̅/s =75-200/75= -125/75= -1.667
Z score from the table for -1.667 = 0.0475
Since, we are looking for less than, solution is = 4.75%
Example 3: Both less than AND greater than. (Percentage outside the range)
The mean inside diameter of a sample of 200 washers produced by a machine is 0.502 inches and the standard deviation is 0.005 inches. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 inches, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.
- x̅ = 0.502
Z= x- x̅/s =0.496-0.502/0.005= -1.2
Z score from the table for -1.2 = 0.1151
Since, we are looking for less than, solution for lower bound = 11.51%
- x̅ = 0.502
Z= x- x̅/s =0.508-0.502/0.005= 1.2
Z score from the table for 1.2 = 0.8849
Since, we are looking for More than, solution is 1-0.8849 = 0.1151 = 11.51%
So, total is 23.02%
Example 4: Both upper and lower bound. (Percentage OUTSIDE the range.)
The weights of 500 American men were taken and the sample mean was found to be 194 pounds with a standard deviation of 11.2 pounds. What percentages of men have weights between 175 and 225 pounds.
- x̅ = 194
Z= x- x̅/s =175-194/11.2= -1.6964
Z score from the table for -1.6964= 0.0455
P(X<175) = 4.55%
- x̅ =194
Z= x- x̅/s =225-194/11.2= 2.7678
Z score from the table for 2.7678 = 0.9971=99.71%
Since, we are looking for weights between 175 and 225, P(175<x<225) = 99.71%-4.55% = 95.16%
Z Transformation Example with Standard Deviation
A batch of batteries with an average of 60v and a Standard Deviation of 4v. If 9 batteries are selected at random, what is the probability that the total voltage of the batteries is greater than 530?
What is the probability that the average voltage is less than 62?
Z Test for Two Proportion
What do you do when the sample size is less than 20?
Great question! You’d apply student t-scores.
Six Sigma Green Belt Z Score Questions
Question: This formula Z = (X – μ)/σ is used to calculate a Z score that, with the appropriate table, can tell a Belt what ____________________________________.
A) Ratio the area under the curve is to the total population
B) Number of Standard Deviations are between X and μ
C) The Median of the sample population is
D) Proportion of the data is between X and μ
Question: A battery manufacturer was considering changing suppliers for a particular part. The purchasing manager required that the average cost of the part be less than or equal to $32 in order to stay within budget. A sample of the 32 initial deliveries had a Mean of the new product upgrade price of $28 with an estimated Standard Deviation of $3. Based on the data provided, the Z value for the data assuming a Normal Distribution is?
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Z Score Practice Questions and Z Charts
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