Completed 3_4-5 Lecture Notes Accessible

3.4 Measures of Relative Position: Relative position describes a data value’s location in the data set relative to all the other data values. Questions about relative position include – How far is the data value above the mean? – Is the data value in the top 10% of values? Measures of relative position include: z-scores, percentiles, and quartiles Consider an upper-level physics class and a student made a 25 on the mid-term exam. This might or might not be a passing grade. If the professor assigns grades relative to all the grades on the exam, then a grade of 25 might be a passing grade. To know how good or bad this exam score is when compared to all the other exam scores, you need a measure of relative position. Z-scores A z-score reports the number of standard deviations a data value is above or below the mean. We will place the data values (x) on the top row below the graph. We will place the z-scores below the data values on the bottom row below the graph. Begin by marking off all of the special data values μ + 1 σ , etc. Since a data value at μ + 1 σ is one standard deviation above the mean, the z-score at μ + 1 σ is 1. Similarly, a data value at μ + 2 σ is two standard deviations above the mean, the z-score at μ + 2 σ is 2, and a data value at μ + 3 σ is three standard deviations above the mean, the z-score at μ + 3 σ is 3. Now we can see the development of our number line. Clearly, to continue the pattern, the z-score for the mean must be zero. This should make sense because a data value right at the mean is none above and none below the mean. Next we can see that a data value at μ − 1 σ is one standard deviation below the mean, and its z-score must be − 1 to continue the pattern. Thus, we know that data values that are greater than the mean must have a positive z-score and data values below the mean must have negative z-scores.

So, the z-score for a data value at μ − 2 σ must be − 2 , and the z-score for a data value at μ − 3 σ must be − 3 . Notice that both the mean and the standard deviation are being used, therefore, z-scores are designed to be used in what shape distribution? Roughly symmetrical Example Consider a uniform distribution with mean 10 and standard deviation 2. Find the z-scores for the data values 6 and 16. First set up your graph with the mean in the center, and mark all special locations μ − 1 σ = 10 − 2 = 8 , μ − 2 σ = 6 , μ − 3 σ = 4 , μ + 1 σ = 10 + 2 = 12 , μ + 2 σ = 14 , μ − 3 σ = 16 Next, count how many standard deviations 6 is away from the mean of 10. The data value of 6 is 2 standard deviations below the mean. Thus, the z-score for a data value of 6 is − 2. The same process will work for the data value of 16. The data value of 16 is 3 standard deviations above the mean, thus the z-score for a data value of 16 is 3. Mean and Standard Deviation of z-scores Mean and SD of z-scores: The mean of the z-scores is always 0 . The z-score for the mean is always 0 because the mean is 0 std. devs. below the mean and 0 std. devs. above the mean. Since the mean is the balance point, all data sets will balance at a z-score of 0, and thus the mean of the z-scores is 0. The standard deviation of the z-scores is always 1 . The distance between μ and μ + 1 σ is 1 standard deviation (or 1 σ ). The corresponding z-scores are 0 and 1, and the distance between 0 and 1 is 1, so the standard deviation for the z-scores is exactly that distance, 1.

Calculations In General, our method of simply counting the number of standard deviations away from the mean will only work for data values that are very special (like μ + 1 σ , and μ + 2 σ , etc.). How do we find the z-score for values that are in between these special values? Quite simply we use the formula that measures the distance between the data value and the mean, then divides by the standard deviation so that we will know how many standard deviations fit within that distance. The formula for the z-score is on your formula chart. z = x − μ σ Examples Consider our uniform distribution with mean 10 and standard deviation 2. Find the z-score for the data value 9 and interpret this z-score. z = x − μ σ = 9 − 10 2 = − 1 2 =− 0.5 The z-score for the data value of 9 is − 0.5 . This means that the data value of 9 is ½ of a standard deviation below the mean. Consider the physics exam that had a mean of 20 and standard deviation of 2.3. Find the z-score for the exam grade of 25 and interpret. z = x − μ σ = 25 − 20 2.3 = 5 2.3 = 2.174 The z-score for the exam grade of 25 is 2.174 . This means that the exam grade of 25 is 2.174 standard deviations above the mean. This physics student did quite well on the exam! Rounding Recall that the mean, standard deviation and variance are all rounded to one more decimal place than the data. For proportions (probabilities) and z-scores, we will round to 3 decimal places. Other numbers we will not worry about rounding. Why are z-scores useful? Z-scores allow us to compare data values across different populations. Example Zeke competed in the 100m dash. The mean finish time was 12.7s with standard deviation 0.4s. Evelyn competed in the mile race which had a mean finish time of 391.2s with a standard deviation of 12s. If Zeke’s finish time was 13s and Evelyn’s finish time was 397.1s, who finished their race in better standing relative to their competitors? Zeke Evelyn z = x − μ σ = 13 − 12.7 0.4 = 0.3 0.4 = 0.75 z = x − μ σ = 397.1 − 391.2 12 = 5.9 12 = 0.492 In a race you want to be faster. Faster means that it took you less time to finish. Evelyn did better because her z-score is lower, which indicates that she took less time to finish relative to her competition than Zeke did relative to his competition.