A psychology professor assigns letter grades on a test according to the following scheme. A: Top 11% of scores Scores on the test are normally distributed with a mean of 71.9 and a standard deviation of 8.4 . Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.
To find the minimum score required for an A grade, we need to find the z-score that corresponds to the top 11% of scores.
Using the normal distribution table, we find that the z-score for the top 11% of scores is approximately 1.23.
Next, we can use the formula for z-score to find the corresponding raw score:
z = (X - μ) / σ
Rearranging the formula to solve for X, we have:
X = z * σ + μ
Plugging in the values, we get:
X = 1.23 * 8.4 + 71.9 ≈ 81.3
Therefore, the minimum score required for an A grade is approximately 81.3. Rounded to the nearest whole number, the minimum score required for an A grade is 81.
To find the minimum score required for an A grade, we need to determine the cutoff point for the top 11% of scores in a normally distributed distribution.
Step 1: Find the z-score corresponding to the top 11% of scores.
Using a standard normal distribution table or calculator, we can find the z-score corresponding to the top 11% of scores (percentile). The remaining 89% will be below this score.
Step 2: Find the raw score corresponding to the z-score.
Using the z-score and the formula for converting z-scores to raw scores:
Score = (z-score * standard deviation) + mean
Given:
Mean (μ) = 71.9
Standard Deviation (σ) = 8.4
Percentile (P) = 89% (11% in the top)
Step 1: Find the z-score:
The z-score can be found by subtracting the percentile from 100 and then converting it into a decimal fraction.
Z = (100 - P) / 100 = (100 - 89) / 100 = 0.11
Step 2: Calculate the raw score:
Score = (z-score * standard deviation) + mean
Score = (0.11 * 8.4) + 71.9
Score ≈ 1 + 71.9 = 72.9
Therefore, the minimum score required for an A grade is approximately 73 (rounded to the nearest whole number).
To find the minimum score required for an A grade, we need to determine the score that corresponds to the top 11% of scores.
First, we need to convert the given information into a standard normal distribution, also known as the Z-score.
The formula to calculate the Z-score is given by:
Z = (X - μ) / σ
Where:
Z is the Z-score
X is the raw score
μ is the mean of the distribution
σ is the standard deviation of the distribution
In this case, the mean (μ) is 71.9 and the standard deviation (σ) is 8.4.
To find the Z-score corresponding to the top 11% of scores, we need to find the Z-score that has a cumulative percentage of 89% (100% - 11%).
Using a standard normal distribution table or a Z-score calculator, we can find that the Z-score corresponding to a cumulative percentage of 89% is approximately 1.23.
Now, we can rearrange the Z-score formula to solve for X (the minimum score required for an A grade):
Z = (X - μ) / σ
1.23 = (X - 71.9) / 8.4
Next, we can solve for X:
1.23 * 8.4 = X - 71.9
10.332 = X - 71.9
X = 10.332 + 71.9
X = 82.232
Therefore, the minimum score required for an A grade is approximately 82.232.
Since the question asks us to round our answer to the nearest whole number, the minimum score required for an A grade is 82.