The grades on the final examination given in a large organic chemistry class are normally distributed with a mean of 72 and a standard deviation of 8. The instructor of this class wants to assign an “A” grade to the top 10% of the scores, a “B” grade to the next 10% of the scores, a “C” grade to the next 10% of the scores, a “D” grade to the next 10% of the scores, and an “F” grade to all scores below the 60th percentile of this distribution. For a letter grade of B, find the lowest acceptable score within the established range
76.8
To find the lowest acceptable score for a letter grade of B, we need to determine the score that corresponds to the 10th percentile of the distribution.
First, we need to find the z-score corresponding to the 10th percentile. The z-score formula is given by:
z = (x - mean) / standard deviation
To find the z-score for the 10th percentile, we can use a table or a statistical calculator like Excel's NORM.INV function. For the 10th percentile, we use a cumulative probability value of 0.10.
Using a statistical calculator, the z-score corresponding to the 10th percentile is approximately -1.2816.
Next, we can use the z-score formula to find the lowest acceptable score (x) within the established range:
-1.2816 = (x - 72) / 8
Solving for x:
x - 72 = -1.2816 * 8
x - 72 = -10.2528
x = -10.2528 + 72
x ≈ 61.7472
Therefore, the lowest acceptable score for a letter grade of B is approximately 61.7472.
To find the lowest acceptable score for a letter grade of B, we need to determine the cutoff point for the top 20% of the scores (combining the top 10% for grade A and the next 10% for grade B).
First, let's calculate the z-score corresponding to the top 20% of the scores. The z-score formula is:
z = (x - μ) / σ
where x is the score, μ is the mean, and σ is the standard deviation.
To find the z-score, we'll use the z-table (also known as the standard normal distribution table). This table provides the cumulative probability (area) under the standard normal curve up to a certain z-score.
Finding the z-score:
We want to find the z-score that corresponds to the 80th percentile (subtracting from 100% because we want the upper tail). The z-value that corresponds to the 80th percentile is approximately 0.8416.
Now, using the formula:
0.8416 = (x - 72) / 8
Solving for x, we get:
x - 72 = 0.8416 * 8
x - 72 = 6.7328
x ≈ 72 + 6.7328
x ≈ 78.7328
Therefore, the lowest acceptable score within the established range for a letter grade of B is approximately 78.73.