A z-score is a unit of numerical measurement used in the field of statistical studies 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-score is useful in statistics because it allows 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
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.
Conversion of z-score to raw score is done using formula shown 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 upon whether the area from the mean for a positive value or for a negative value is desired, we use the suitable values on the z score table. Values lesser 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.
Given the following data, determine 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:
Subject | Score (x) | Mean (μ) _{ } | Standard Deviation (σ) _{ } |
Statistics | 80 | 70 | 15 |
Solution:
From the given dataset 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 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.
*Response times vary by subject and question complexity. Median response time is 34 minutes and may be longer for new subjects.
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