## Question 493:

1

To calculate standard scores, also called z-scores, we need to take each data-point and subtract it from the mean, then divide that result by the standard deviation. This process converts the distribution of exam scores to the standard normal form which has a mean of 0 and standard deviation of 1. By doing this we can use the properties of the normal curve to make some judgements about how unusual these scores are.

So the z-scores for each officer score are :

1. (195-80)/20 = 5.75
2. (175-80)/20 = 4.75
3. (191-80)/20 = 5.55
4. (189-80)/20 = 5.45

To interpret these z-scores, we would read them as officer 1's score was 5.75 standard deviations above the mean and officer 2's score was 4.75 standard deviations above the mean. The empirical rule of the standard normal curve tells us that we can expect around 95% of values to fall within +/- 2 standard deviations of the mean and 99% of values would fall around +/1 3 standard deviations from the mean. All these scores are all above 3, so we already know these are very rare scores indeed. That alone might be evidence enough, but lets go further.

To find out exactly how unusual they are, we'd need to look them up using a z-table or you can use the z-score to percentile calculator.  Most z-tables only go to +/-3 standard deviations. Enter each z-score in the calculator. I'd select 1-sided area  since we're interested in whether these scores are unusually high, not just unusually far from the mean (change the decimals to 13)I get percentages of 99.9999994738%, 99.9998954496%, 99.9999983874%, 99.9999972076%.

So indeed, these are VERY unusually high scores. We would interpret them as offiicer 1 scored higher than 99.9999994738% of all other test takers ever...and so on. Given that all four of these cronies scored this high, I'd be suspicious.