Problem
In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 72 and a standard deviation of 8 . Grades are assigned such that the top $10 \%$ receive A's, the next $20 \%$ received B's, the middle $40 \%$ receive C's, the next $20 \%$ receive D's, and the bottom $10 \%$ receive F's. Find the lowest score on the final exam that would qualify a student for an $A$, a $B$, a $C$, and a $D$. Click here to view Page 1 of the Standard Normal Table. Click here to view Page 2 of the Standard Normal Table. The lowest score that would qualify a student for an $\mathrm{A}$ is (Round up to the nearest integer as needed.)
Solution
Step 1 ：The problem is asking for the lowest score that would qualify a student for an A. This corresponds to the top 10% of the scores. In a normal distribution, this corresponds to a z-score of approximately 1.28 (from the standard normal table).
Step 2 ：The z-score is calculated as \((X - μ) / σ\), where X is the score, μ is the mean, and σ is the standard deviation.
Step 3 ：We can rearrange this formula to solve for X: \(X = z * σ + μ\).
Step 4 ：Substituting the given values, we get \(X = 1.28 * 8 + 72\).
Step 5 ：Calculating the above expression, we find that the lowest score that would qualify a student for an A is \(\boxed{83}\).
