Stats notes >> continuous distributions >> z-scores explained with examples
## Z-score explainedThe z-score is associated with the normal distribution and it is a number that may be used to: - tell you where a score lies compared with the rest of the data, above/below mean.
- compare scores from different normal distributions
Let's take a closer look at each of these uses. ## Z-score and distance from the mean, and Z-tableThe Z-score is a number that may be calculated for each data point in a set of data.The number is continuous and may be negative or positive, and there s no max/min value. The z-score tells us how "far" that data point is from the mean. This distance from the mean is measured in terms of standard deviation. We may make statements such as "the data point(score) is 1 standard deviation above the mean" and "the score is 3 standard deviations above the mean", which means the latter score is three times further from the mean.
- 50? The z-score is (50-50)/10 = 0. Interpretation: student score is 0 distance (in units of standrd deviations) from the mean, so the student has scored average.
- 60? The z-score is (60-50)/10 = 1. Interpretation: student has scored above average - a distance of 1 standard deviation above the mean.
- 69.6? The z-score is (69.6-50)/10 = 1.96. Interpretation: student has scored above average - a distance of 1.96 above the average score.
Looking back the final example. The z-score was 1.96. Now if we look in the z-table we find that 2.5 per cent of the scores is above 1.96. So we can say that in the population this student did better than 97.5 per cent of students, or that only 2.5 per cent of students scored higher. What about for the student with z-score of zero - what proportion of students did better? ## Comparing scores from different normal distributions using the z-scoreSuppose a student sits 2 exams, getting 55 in a verbal test and 60 in a numerical reasoning test. The class scores for each exam are normally distributed. For the verbal test, the mean is 50 and standard deviation 5; for the numerical test, the mean is 50 and standard deviation is 12. Now it is plain to see that the student did above average for each test, and did better at numerical reasoning. How did this student perform relative to everyone else? We can answer this by calculating the z-score. The z-score for the verbal test is (55-50)/5 = 1. The z-score for the numerical test is (60-50)/12 = 0.83 Since the z-score for the verbal test is larger than for the numerical test, the student did better in the verbal than in the numerical test compared to everyone else. Another way to see this is that z-score of 1 for verbal implies about 16 per cent did better at verbal; a z-score of 0.83 for numerical implies about 20 per cent did better at numerical.
Something to ponder. Say you see the histogram for your data is positively skewed. Can you then proceed to use the Z-score as we have done on this page? If you really want to know the answer, or indeed if you know the answer and would like to share it then proceed to the thread on this topic (link below)
Discuss the z-score on the forum |