Question

1. Conduct a search and post an example of how Z-scores are used in the real world.

2.Do a search about the probability of any event you like and post it. For example, what is the probability of winning the lottery? or a plane crash? or any other event you would find interesting?

Answer 1

The standard score (more commonly referred to as a z-score) is a very useful statistic because it (a) allows us to calculate the probability of a score occurring within our normal distribution and (b) enables us to compare two scores that are from different normal distributions. The standard score does this by converting (in other words, standardizing) scores in a normal distribution to z-scores in what becomes a standard normal distribution. To explain what this means in simple terms, let's use an example (if needed, see our statistical guide, Normal Distribution Calculations, for background information on normal distribution calculations).

Setting the scene: Part 1

A teacher sets a piece of English Literature coursework for the 50 students in his class. We make the assumption that when the scores are presented on a histogram, the data is found to be normally distributed. The mean score is 60 out of 100 and the standard deviation (in other words, the variation in the scores) is 15 marks (see our statistical guides, Measures of Central Tendency and Standard Deviation, for more information about the mean and standard deviation).

Having looked at the performance of the teacher's class, one student, Sarah, has asked the teacher if, by scoring 70 out of 100, she has done well. Bearing in mind that the mean score was 60 out of 100 and that Sarah scored 70, then at first sight it may appear that since Sarah has scored 10 marks above the 'average' mark, she has achieved one of the best marks. However, this does not take into consideration the variation in scores amongst the 50 students (in other words, the standard deviation). After all, if the standard deviation is 15, then there is a reasonable amount of variation amongst the scores when compared with the mean.

Whilst Sarah has still scored much higher than the mean score, she has not necessarily achieved one of the best marks in her class. The question arises: How well did Sarah perform in her English Literature coursework compared to the other 50 students? Before answering this question, let us look at another problem.

The teacher has a dilemma. In the next academic year, he must choose which of his students have performed well enough to be entered into an advanced English Literature class. He decides to use the coursework scores as an indicator of the performance of his students. As such, he feels that only those students that are in the top 10% of the class should be entered into the advanced English Literature class. The question arises: Which students came in the top 10% of the class?

Therefore, we are left with two questions to answer. First, how well did Sarah perform in her English Literature coursework compared to the other 50 students? Second, which students came in the top 10% of the class?

Whilst it is possible to calculate the answer to both of these questions using the existing mean score and standard deviation, this is very complex. Therefore, statisticians have come up with probability distributions, which are ways of calculating the probability of a score occurring for a number of common distributions, such as the normal distribution. In our case, we make the assumption that the students' scores are normally distributed. As such, we can use something called the standard normal distribution and its related z-scores to answer these questions much more easily.

Standard Normal Distribution and Standard Score (z-score)

When a frequency distribution is normally distributed, we can find out the probability of a score occurring by standardising the scores, known as standard scores (or z scores). The standard normal distribution simply converts the group of data in our frequency distribution such that the mean is 0 and the standard deviation is 1 (see below).

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 simply the score, minus the mean score, divided by the standard deviation. Therefore, let's return to our two questions.

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
English Literature	70	60	15

In terms of z-scores, this gives us:

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 discussed in the opening paragraphs of this guide. 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. Note the changes to the labelling of the x-axis.

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.

Setting the Scene: Part II

Clearly, the z-score statistic is helpful in highlighting how Sarah performed in her English Literature coursework and what mark a student would have to achieve to be in the top 10% of the class and qualify for the advanced English Literature class. However, we have only been talking about one distribution here, namely the distribution of scores amongst 50 students that completed a piece of English Literature coursework. What if Sarah wanted to compare how well she performed in her Maths coursework compared with her English Literature coursework?

In this case, Sarah achieved a higher mark in her Maths coursework, 72 out of 100. However, as we have already learnt, just because her Maths score (72) is higher than her English Literature score (70), we shouldn't assume that she performed better in her Maths coursework compared to her English Literature coursework. The question therefore arises: How well did Sarah perform in her Maths coursework compared to her English Literature coursework?

Clearly, the two scores (her English Literature and Maths coursework marks) come from different distributions. The distribution of 50 students that completed the English Literature coursework has a mean of 60 and standard deviation of 15. The distribution of 50 students that completed the Maths coursework, on the other hand, has a mean of 68 and a standard deviation of 6. This gives us the following:

	Score	Mean	Standard Deviation
	(X)	µ	s
English Literature	70	60	15
Maths	72	68	6

Since these scores are from two different distributions, we need to standardise them into z-scores so that they can be directly compared. This gives us:

The z-scores highlight that the student is two thirds (z = 0.67) of a standard deviation above the mean in English Literature, but also two thirds (z = 0.67) of a standard deviation above the mean in Maths. Using the standard normal distribution table, we can see that Sarah clearly performed above 'average' in both subjects although again, around 25% of the class got a better mark than her. However, the key point her is that the standard score showed that Sarah performed equally well in her English Literature and Maths coursework, even though her marks were different in both pieces. This shows the usefulness of the standard score statistic.