Science, Tech, Math › Math Examples of Z-score Calculations Share Flipboard Email Print Natee Meepian / EyeEm / Getty Images Math Statistics Statistics Tutorials Formulas Probability & Games Descriptive Statistics Inferential Statistics Applications Of Statistics Math Tutorials Geometry Arithmetic Pre Algebra & Algebra Exponential Decay Functions Worksheets By Grade Resources View More By Courtney Taylor Professor of Mathematics Ph.D., Mathematics, Purdue University M.S., Mathematics, Purdue University B.A., Mathematics, Physics, and Chemistry, Anderson University Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra." our editorial process Courtney Taylor Updated September 21, 2018 One type of problem that is typical in an introductory statistics course is to find the z-score for some value of a normally distributed variable. After providing the rationale for this, we will see several examples of performing this type of calculation. Reason for Z-scores There are an infinite number of normal distributions. There is a single standard normal distribution. The goal of calculating a z - score is to relate a particular normal distribution to the standard normal distribution. The standard normal distribution has been well-studied, and there are tables that provide areas underneath the curve, which we can then use for applications. Due to this universal use of the standard normal distribution, it becomes a worthwhile endeavor to standardize a normal variable. All that this z-score means is the number of standard deviations that we are away from the mean of our distribution. Formula The formula that we will use is as follows: z = (x - μ)/ σ The description of each part of the formula is: x is the value of our variableμ is the value of our population mean.σ is the value of the population standard deviation.z is the z-score. Examples Now we will consider several examples that illustrate the use of the z-score formula. Suppose that we know about a population of a particular breed of cats having weights that are normally distributed. Furthermore, suppose we know that the mean of the distribution is 10 pounds and the standard deviation is 2 pounds. Consider the following questions: What is the z-score for 13 pounds?What is the z-score for 6 pounds?How many pounds corresponds to a z-score of 1.25? For the first question, we simply plug x = 13 into our z-score formula. The result is: (13 – 10)/2 = 1.5 This means that 13 is one and a half standard deviations above the mean. The second question is similar. Simply plug x = 6 into our formula. The result for this is: (6 – 10)/2 = -2 The interpretation of this is that 6 is two standard deviations below the mean. For the last question, we now know our z -score. For this problem we plug z = 1.25 into the formula and use algebra to solve for x: 1.25 = (x – 10)/2 Multiply both sides by 2: 2.5 = (x – 10) Add 10 to both sides: 12.5 = x And so we see that 12.5 pounds corresponds to a z-score of 1.25.