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Answer :
Final answer:
The question relates to the concepts of normal distribution and percentiles in statistics. The score obtained by the top 2.5% of students is likely to be 99.8, calculated using the mean, standard deviation, and z-score corresponding to the top 2.5%. This represents the value of test scores that is higher than 97.5% of the scores. The correct answer is option(a).
Explanation:
The subject of the question is based on statistical concepts, specifically normal distribution and percentiles. You have been given that scores on a certain exam are normally distributed with a mean of 90 and a variance of 25. That variance gives a standard deviation of 5, as the variance is the square of the standard deviation.
To find the score obtained by the top 2.5% of the students, you need to refer to the concept of z-score, which tells us how many standard deviations a data point is away from the mean. In this case, you're searching for z corresponding to the top 2.5%, which from standard z-tables is roughly 1.96.
Substituting these values, which are mean (μ = 90), standard deviation (σ = 5), and z-score (1.96) to calculate score (X): X = μ + zσ gives you : X = 90 + (1.96)(5) which approximately equals 99.8. Hence, option a. 99.8 is the likely answer.
Learn more about Normal Distribution here:
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