Defining Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.
The z-score formula is given as:
$z=\frac{x-\mu}{\sigma}$
Where
$x$ : raw score
$\mu $ : mean
$\sigma $ : standard deviation
Calculating Area Percentages
Area percentages can be calculated at any given location along a standard normal distribution curve using a z-score. In order to find a specific area under a normal curve, we evaluate the z-score of the given data value, and subsequently use a z-score table to calculate the area.
- A positive z-score shows that the raw score lies above the mean. In other words, it is higher than the mean average. For example, if a z-score is equal to +1, it is 1 standard deviation above the mean.
- A negative z-score designates that the raw score lies below the mean. In other words, it is lower than the mean average. For example, if a z-score is equal to -2, it is 2 standard deviations below the mean.
- A z-score of zero means the raw score is on the mean.
A z-score can be converted to raw score using formula below:
$\chi =\mu +z\sigma $
Where:
The information on the area under the standard normal curve for any value between the mean (zero) and any z-score can be obtained with the help of a z-table. The left-most column of a z-table is for the standard deviations above or below the mean (up to one decimal place). The top row of the z-table denotes the part of the z-score in hundredths.
Depending on whether you want to find the area from the mean for a positive value or a negative value, you will use the suitable values on the z-score table. Values below than the mean are marked with a negative score in the z-table. They represent the area under the bell curve to the left of z. Values greater than the mean are marked with a positive score in the z-table. They represent the area under the bell curve to the right of z.
Practice Problem
Given the following data, how well did Mary perform in her coursework compared to the other 50 students? The maximum marks for the subject is 100. Evaluate based on:
- What percentage (or number) of students scored lower than Mary?
- What percentage (or number) of students scored higher than Mary?
Subject | Score (x) | Mean (μ) _{ } | Standard Deviation (σ) _{ } |
Statistics | 80 | 70 | 15 |
Solution:
From the given dataset, we know Mary scored 80 out of 100, the mean score was 70, and the standard deviation was 15. We need to standardize the score in order to analyze the data.
Using the formula for z-score:
$z=\frac{x-\mu}{\sigma}=\frac{80-70}{15}=\frac{10}{15}=0.67$
The z-score is 0.67 and it is positive.
To calculate the percentage (or number) of students that scored higher and lower than Mary, we need to refer to the z-score table.
To identify 0.67, read it as 0.6 + 0.07
Look in the y-axis for 0.6, and look for 0.07 on the x-axis. The resulting value is 0.7486. This means that the probability of Mary scoring more marks than the rest of the class is 0.7486.
To interpret this z-score in terms of percentage, multiply this value by 100. Hence
$0.7486\times 100=74.86\%$
It can be concluded that Mary’s score was better than almost 75% of the class.
To calculate the number of students whom Mary outperformed, calculate 75% of 50.
$75\%\times 50=37.5$
Here we can see that Mary was better than almost 38 other students in the class.
To calculate the probability of the number of students who scored higher than Mary, subtract 0.7486 from 1.
$1-0.7486=0.2514$
Hence, the percentage of students who scored higher than Mary will be
$0.2514\times 100=25.14\%$
To calculate the number of students who performed better than Mary, subtract 38 from 50
$50-38=12$
Hence, 12 students performed better than Mary.
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