Sabino took the ACT in 2003 and received a composite score of 22. If the mean and standard deviation for that year were, respectively, 20.8 and 4.8, what percent of all of the students who took the exams scored better than him?
To determine the percentage of students who scored better than Sabino on the ACT in 2003, you need to find his percentile rank. This can be done by using a standard normal distribution table.
First, calculate the Z-score of Sabino's composite score using the formula:
Z = (X - μ) / σ
Where:
X is the individual score
μ is the mean
σ is the standard deviation
In this case, Sabino's score is 22, the mean is 20.8, and the standard deviation is 4.8.
Z = (22 - 20.8) / 4.8
Z = 1.2 / 4.8
Z = 0.25
Next, locate the corresponding Z-score in the standard normal distribution table, also known as the Z-table. The Z-value of 0.25 corresponds to the area between the mean and the Z-score.
The Z-table provides the area to the left of the Z-score. Since we are interested in the students who scored better than Sabino, we need to find the area to the right of the Z-score.
Subtract the area to the left from 1 to get the area to the right:
Area to the right = 1 - Area to the left
Finally, multiply the area to the right by 100 to get the percentage:
Percentage = (Area to the right) * 100
So, using the Z-table, you can find the percentage of students who scored better than Sabino on the ACT in 2003.