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Answer :
To be within the top 20% of test-takers, a student would need to score at least 903 on the test.
Percentiles in a normal distribution can be calculated using z-scores which are a measure of how many standard deviations an element is from the mean. Here, we'll need to find the z-score that corresponds to the 80th percentile. Using a z-table or a statistical calculator, the z-score that corresponds to the 80th percentile is approximately 0.84.
Knowing this z-score, we can use the formula for a z-score in a normal distribution, which is:
z = (X - μ) / σ
Where:
- z is the z-score,
- X is the value in the dataset,
- μ is the mean of the dataset,
- σ is the standard deviation of the dataset.
Rearranging this formula to solve for X gives us:
X = z×σ + μ
Now we can substitute the values we have:
X = 0.84 × 300 + 650
= 902.
However, since the scores must be whole numbers, and we want to ensure that the student is in the top 20%, we need to round up to the next whole number, which gives us a final score of 903.
In conclusion, to be within the top 20% of test-takers, a student would need to score at least 903 on the test.
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