Now we need to determine the percentage of peers whose score goes higher and lower then that of the scores of Sarah. Find out how well Sarah performed compared to her peers.įrom the above-given data, we can deduce thatĢ) Using the Z-score table we can find out how well she performed relative to her peers. The average score was 500 (µ) and the standard deviation was 120 (σ). Imagine a group of 300 applicants who took a math test. A Z-Score Table is a table which shows the percentage of values (or area percentage) to the left of a given z-score on a standard normal distribution. To find a specific area under a normal curve, first, find the z-score of the data value and then use a Z-Score Table to find the area. Z-score Tables Area Under a Normal Curve: This z-score will tell you how many standard errors are there between the sample mean and the population means. When you have multiple samples and want to describe the standard deviation of those sample means (the standard error), use this z score formula: Z Score Formula: Standard Error of the Mean Putting the values in the equation mentioned above, Assuming it is a normal distribution, your z score would be.įrom the question above we can deduce that,Īnd the value of standard deviation (σ) is 30 The test has a mean (μ) of 140 and a standard deviation (σ) of 30. 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.Let's take an example and understand this better. In this case, Sarah achieved a higher mark in her Maths coursework, 72 out of 100. What if Sarah wanted to compare how well she performed in her Maths coursework compared with her English Literature coursework? 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. Setting the Scene: Part IIĬlearly, 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. 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. If we use a z-score calculator, our value of 0.9 corresponds with a z-score of 1.282.
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. 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!). 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 forms the second part of the z-score. This time, the value on the x-axis for 0.8997 is 0.08. 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. You will notice that the value on the y-axis for 0.8997 is 1.2.
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. When looking at the table, you may notice that the closest value to 0.9 is 0.8997.
As such, we first need to find the value 0.9 in standard normal distribution table.
We know the percentage we are trying to find, the top 10% of students, corresponds to 0.9.