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Answer :
Top 20% of students taking the standardized exam. The exam scores follow a normal distribution with a mean of 550 and a standard deviation of 220.
To find the minimum score required to be in the top 20% of students, we need to determine the score that corresponds to the 80th percentile of the normal distribution.
The 80th percentile represents the score below which 80% of the scores fall. Since the scores follow a normal distribution, we can use the z-score formula to convert the percentile to a z-score.
The z-score formula is given by:
z = (X - μ) / σ
where X is the score, μ is the mean, and σ is the standard deviation.
To find the minimum score, we need to solve for X using the z-score formula.
For the 80th percentile, the z-score can be found using a standard normal distribution table or a statistical calculator. In this case, it is approximately 0.8416.
0.8416 = (X - 550) / 220
Solving for X, we get:
X = 0.8416 * 220 + 550 ≈ 729.15
Therefore, the minimum score required for a student to qualify for admission at a college is approximately 729.15.
Learn more about standard deviation :
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