If a test has a mean score of 75, with a standard deviation of 11, what % of students scored an A (90 or higher) ?
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To determine the percentage of students who scored an A (90 or higher) on the test, we need to calculate the z-score for a score of 90 and then find the corresponding area under the standard normal distribution curve.
Step 1: Calculate the z-score using the formula: z = (x - μ) / σ
Where:
- x is the value we're interested in (90 in this case)
- μ is the mean of the distribution (75 in this case)
- σ is the standard deviation (11 in this case)
z = (90 - 75) / 11
z = 1.3636
Step 2: Look up the z-score in a standard normal distribution table. This table provides the area under the curve to the left of a given z-score.
Looking up the z-score of 1.36 in the table, we find that the area to the left of this z-score is 0.9131.
Step 3: Subtract the area from 1 to find the percentage of students who scored 90 or higher.
Percentage = (1 - 0.9131) * 100
Percentage = 8.69%
Therefore, approximately 8.69% of students scored an A (90 or higher) on the test.