Shivanker kumar for the class of May 9th

z-score and its application 

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:

Standard Score Calculation

Z = (X-µ)/s

As the formula shows, the standard score is simply the score, minus the mean score, divided by the standard deviation. Let’s see the application of z-score.

Application –

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

To answer this question, we can re-phrase it as: What percentage (or number) of students scored higher than Sarah and what percentage (or number) of students scored lower than Sarah? First, let's reiterate that Sarah scored 70 out of 100, the mean score was 60, and the standard deviation was 15 (see below).

                Score      Mean      Standard Deviation

                (X)           µ              s

                70           60           15

In terms of z-scores, this gives us:

Z = (X-µ)/s = (70-60)/15 = .6667

Standard Score Calculation

The z-score is 0.67 (to 2 decimal places), but now we need to work out the percentage (or number) of students that scored higher and lower than Sarah. To do this, we need to refer to the standard normal distribution table.

This table helps us to identify the probability that a score is greater or less than our z-score score. To use the table, which is easier than it might look at first sight, we start with our z-score, 0.67 (if our z-score had more than two decimal places, for example, ours was 0.6667, we would round it up or down accordingly; hence, 0.6667 would become 0.67). The y-axis in the table highlights the first two digits of our z-score and the x-axis the second decimal place. Therefore, we start with the y-axis, finding 0.6, and then move along the x-axis until we find 0.07, before finally reading off the appropriate number; in this case, 0.2514. This means that the probability of a score being greater than 0.67 is 0.2514. If we look at this as a percentage, we simply times the score by 100; hence 0.2514 x 100 = 25.14%. In other words, around 25% of the class got a better mark than Sarah (roughly 13 students since there is no such thing as part of a student!).

Going back to our question, "How well did Sarah perform in her English Literature coursework compared to the other 50 students?", clearly we can see that Sarah did better than a large proportion of students, with 74.86% of the class scoring lower than her (100% - 25.14% = 74.86%). We can also see how well she performed relative to the mean score by subtracting her score from the mean (0.5 - 0.2514 = 0.2486). Hence, 24.86% of the scores (0.2486 x 100 = 24.86%) were lower than Sarah's, but above the mean score. However, the key finding is that Sarah's score was not one of the best marks. It wasn't even in the top 10% of scores in the class, even though at first sight we may have expected it to be. This leads us onto the second question.

2. Which students came in the top 10% of the class?

A better way of phrasing this would be to ask: What mark would a student have to achieve to be in the top 10% of the class and qualify for the advanced English Literature class?

To answer this question, we need to find the mark (which we call "X") on our frequency distribution that reflects the top 10% of marks. Since the mean score was 60 out of 100, we immediately know that the mark will be greater than 60. After all, if we refer to our frequency distribution below, we are interested in the area to the right of the mean score of 60 that reflects the top 10% of marks (shaded in red). As a decimal, the top 10% of marks would be those marks above 0.9 (i.e., 100% - 90% = 10% or 1 - 0.9 = 0.1).

First, we should convert our frequency distribution into a standard normal distribution. As such, our mean score of 60 becomes 0 and the score (X) we are looking for, 0.9, becomes our z-score, which is currently unknown.

The next step involves finding out the value for our z-score. To do this, we refer back to the standard normal distribution table.

In answering the first question in this guide, we already knew the z-score, 0.67, which we used to find the appropriate percentage (or number) of students that scored higher than Sarah, 0.2514 (i.e., 25.14% or roughly 25 students achieve a higher mark than Sarah). Using the z-score, 0.67, and the y-axis and x-axis of the standard normal distribution table, this guided us to the appropriate value, 0.2514. In this case, we need to do the exact reverse to find our z-score.

We know the percentage we are trying to find, the top 10% of students, corresponds to 0.9. As such, we first need to find the value 0.9 in standard normal distribution table. When looking at the table, you may notice that the closest value to 0.9 is 0.8997. If we take the 0.8997 value as our starting point and then follow this row across to the left, we are presented with the first part of the z-score. You will notice that the value on the y-axis for 0.8997 is 1.2. We now need to do the same for the x-axis, using the 0.8997 value as our starting point and following the column up. This time, the value on the x-axis for 0.8997 is 0.08. This forms the second part of the z-score. Putting these two values together, the z-score for 0.8997 is 1.28 (i.e., 1.2 + 0.08 = 1.28).

There is only one problem with this z-score; that is, it is based on a value of 0.8997 rather than the 0.9 value we are interested in. This is one of the difficulties of refer to the standard normal distribution table because it cannot give every possible z-score value (that we require a quite enormous table!). Therefore, you can either take the closest two values, 0.8997 and 0.9015, to your desired value, 0.9, which reflect the z-scores of 1.28 and 1.29, and then calculate the exact value of "z" for 0.9, or you can use a z-score calculator. If we use a z-score calculator, our value of 0.9 corresponds with a z-score of 1.282. In other words, P ( z > 1.282 ) = 0.1.

Now that we have the key information (that is, the mean score, µ, the standard deviation, s , and z-score, z), we can answer our question directly, namely: What mark would a student have to achieve to be in the top 10% of the class and qualify for the advanced English Literature class? First, let us reiterate the facts:

Score	Mean	Standard Deviation	z-score
(X)	µ	s	z
?	60	15	1.282

To find out the relevant score, we apply the following formula:

Therefore, students that scored above 79.23 marks out of 100 came in the top 10% of the English Literature class, qualifying for the advanced English Literature class as a result.

Hope you find it important and relevant!!