Answer:
27.43% of the students scored 83 or more in the exam.
Explanation:
We can use the Z-score formula to find the percentage of students scoring 83 or more:
Z = (X - μ) / σ
where X is the test score, μ is the mean test score, and σ is the standard deviation.
Substituting the given values, we get:
Z = (83 - 80) / 5 = 0.6
Using a Z-score table or calculator, we can find the percentage of students scoring 83 or more:
P(Z > 0.6) = 1 - P(Z < 0.6) = 1 - 0.7257 ≈ 0.274
27.43% of the students scored 83 or more in the exam.
OR. USING PROPERTIES OF NORMAL DISTRIBUTION
convert the raw score of 83 into a standardized score, also known as a z-score.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = raw score
μ = mean
σ = standard deviation
Substituting the values given in the question, we get:
z = (83 - 80) / 5
z = 0.6
This means that a student who scored 83 on the exam has a z-score of 0.6.
Next, we need to find the area under the normal distribution curve to the right of this z-score. We can use a standard normal distribution table or calculator for this purpose.
Using a standard normal distribution table, we find that the area to the right of z = 0.6 is approximately 0.2743.
Therefore, the percentage of students scoring 83 or more in the exam is approximately:
0.2743 x 100% = 27.43%
So, about 27.43% of students scored 83 or higher on the exam.