Examples of Z-score Calculations

Examples of Z-score Calculations 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 - ÃŽ ¼)/ ÏÆ' The description of each part of the formula is: x is the value of our variableÃŽ ¼ is the value of our population mean.ÏÆ'Â  is the value of the population standard deviation.z is the z-score. Â   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? Â   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.