The grades for Economics students at a large university are found to be normally distributed with a mean of 76 and a standard deviation of 4. What proportion of the students are expected to have a grade between 84 and 88?
Use same method as described in your previous post.
To find the proportion of students expected to have a grade between 84 and 88, we need to calculate the z-scores corresponding to those grades and then use the standard normal distribution table.
First, we calculate the z-scores using the formula:
z = (x - μ) / σ
Where:
x is the value we want to convert to a z-score (in this case, 84 and 88)
μ is the mean of the distribution (76)
σ is the standard deviation of the distribution (4)
For 84:
z = (84 - 76) / 4 = 2
For 88:
z = (88 - 76) / 4 = 3
Next, we refer to the standard normal distribution table (also known as the z-table) to find the proportions corresponding to these z-scores.
Looking up the z-score of 2 in the table, we find that the proportion to the left of 2 is approximately 0.9772.
Looking up the z-score of 3 in the table, we find that the proportion to the left of 3 is approximately 0.9987.
To find the proportion between 84 and 88, we subtract the proportion corresponding to the lower z-score from the proportion corresponding to the higher z-score.
Proportion = 0.9987 - 0.9772 = 0.0215
So, approximately 0.0215 or 2.15% of the students are expected to have a grade between 84 and 88.