Application of Z-score or standard score on evaluating a students performance. It includes the formula for z-value, interpretation of in relation to the mean and stdev, sample step by step calculation and interpretationof z-score, and sample real scenario application.
2. A z-score (aka, a standard score)
indicates how many Standard Deviation
an element or a score is from the mean
is from the mean. A z-score can be
calculated from the following formula
3. • It can be found using the formula z = (x – μ) / σ, : read z is
equal to the quotient between the difference of the raw
score and mean and the standard deviation.
Where:
• z - is the z - score
• x - is the raw score
• μ - is the (population) mean, but u can use x bar too
• .σ - is the standard deviation
4. • A z-score less than zero (0) represents an element
less than the mean.
• A z-score greater than 0 represents an element
greater than the mean.
• A z-score equal to 0 represents an element equal to
the mean.
5. • A z-score equal to 1 represents an element that is 1 standard
deviation greater than the mean; a z-score equal to 2, 2
standard deviations greater than the mean; etc.
• A z-score equal to -1 represents an element that is 1 standard
deviation less than the mean; a z-score equal to -2, 2 standard
deviations less than the mean; etc.
• If the number of elements in the set is large, about 68% of the
elements have a z-score between -1 and 1; about 95% have a z-
score between -2 and 2; and about 99% have a z-score between
-3 and 3.
6. In a 20 item test, the average score of the class is 16 and with 3.5
SD. Cherrie Pie scored 18.
z = (x – μ) / σ
z = (18– 16) /3.5
z = 2/3.5z = 0.57
That means that Cherrie Pie`s score is 0.57 (less than 1) standard
deviations above the mean.
7. Other than said above, we can use Cherri Pies`s z-score of .57 to
see how well she performed compared to other students or look at
her class standing.
• We may have to look through the z-table which can be found
here http://www.statisticshowto.com/tables/z-table/
• Now take 0.57
• y - is 0.5
• x - is 0.07
We would get the z-value of 0.2157
8.
9. • Multiply the z-value by 100: 0.2157*100 = 21.57
• It means that 21.57% of the class performed better than Cherrie
Pie
• Substract 21.57 from 100: 100-21.57 = 78.43
• It means that Cherrie Pie performed better than 78.43% of the
test class.
• That would help the teacher on making decisions with regards to
Cherri Pie`s performance.
