Z-scores are expressed in terms of standard deviations from their means. Resultantly, these z-scores have a distribution with a mean of 0 and a standard deviation of 1. The formula for calculating the standard score is given below:

As the formula shows, the standard score is basically the score, short the mean score, partitioned by the standard deviation. In this way, how about we come back to our two inquiries.

How well did Sarah perform in her English Literature coursework compared to the other 50 students?

To address this inquiry, we can re-express it as: What rate (or number) of understudies scored higher than Sarah and what rate (or number) of understudies scored lower than Sarah? To begin with, how about we repeat that Sarah scored 70 out of 100, the mean score was 60, and the standard deviation was 15 (see beneath).

In terms of z-scores, this gives us:

The z-score is 0.67 (to 2 decimal spots), yet now we have to work out the rate (or number) of understudies that scored higher and lower than Sarah. To do this, we have to allude to the standard ordinary conveyance table.

This table helps us to identify the probability that a score is greater or less than our z-score score. To utilize the table, which is simpler than it may see first sight, we start with our z-score, 0.67 (if our z-score had in excess of two decimal spots, for instance, our own was 0.6667, we would gather it together or down as needs be; thus, 0.6667 would end up 0.67). The y-hub in the table features the initial two digits of our z-score and the x-pivot the subsequent decimal spot. Consequently, we start with the y-hub, discovering 0.6, and after that move along the x-pivot until we find 0.07, before at long last perusing off the fitting number; for this situation, 0.2514. This implies the likelihood of a score being more noteworthy than 0.67 is 0.2514. On the off chance that we take a gander at this as a rate, we just occasions the score by 100; henceforth 0.2514 x 100 = 25.14%. As it were, around 25% of the class showed signs of improvement mark than Sarah (about 13 understudies since there is nothing of the sort as a component of an understudy!).

Returning to our question, “How very much did Sarah perform in her English Literature coursework contrasted with the other 50 understudies?”, plainly we can see that Sarah showed improvement over an enormous extent of understudies, with 74.86% of the class scoring lower than her (100% – 25.14% = 74.86%). We can likewise perceive how well she performed comparative with the mean score by subtracting her score from the mean (0.5 – 0.2514 = 0.2486). Subsequently, 24.86% of the scores (0.2486 x 100 = 24.86%) were lower than Sarah’s, yet over the mean score. Notwithstanding, the key finding is that Sarah’s score was not perhaps the best imprint. It wasn’t even in the top 10% of scores in the class, despite the fact that from the outset locate we may have anticipated that it should be. This leads us onto the subsequent question.