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Answer :
The minimum score required for admission in the university, considering only the top 20% of marks, can be determined using the given information. With a mean of 400 and a standard deviation of 25, the minimum score for admission is calculated to be 425.
To find the minimum score for admission, we need to determine the cutoff point that corresponds to the top 20% of marks. In a normal distribution, the mean marks would be at the 50th percentile. Since we are considering the top 20%, the cutoff point would be at the 80th percentile.
To calculate this cutoff point, we can use the concept of z-scores. A z-score measures the number of standard deviations a particular value is from the mean. In this case, we need to find the z-score that corresponds to the 80th percentile.
The 80th percentile corresponds to a z-score of approximately 0.84. Using the z-score formula, z = (x - μ) / σ, where x is the score, μ is the mean, and σ is the standard deviation, we can rearrange the formula to solve for x.
Plugging in the known values, we have 0.84 = (x - 400) / 25. Rearranging the equation and solving for x, we find x = 425.
Therefore, the minimum score required for admission is 425. This means that any applicant who scores 425 or above on the entrance exam will qualify for admission to the university.
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