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------------------------------------------------ The mean score in a university admission test is 85 with a standard deviation of 3. What is the 65th percentile of the scores?

To find the 65th percentile of the scores, we can use the standard normal distribution table or a calculator. The 65th percentile of the scores is approximately 85.16.

Explanation:

To find the 65th percentile of the scores, we need to find the value that separates the top 65% of the scores from the bottom 35%. Since the scores are normally distributed with a mean of 85 and a standard deviation of 3, we can use the standard normal distribution table or a calculator to find the corresponding z-score.

Using the z-score formula, z = (X - μ) / σ, where X is the desired score, μ is the mean, and σ is the standard deviation, we can rearrange the formula to solve for X:

X = z * σ + μ

For the 65th percentile, the z-score is approximately 0.3853. Plugging in the values, we get:

X = 0.3853 * 3 + 85 = 85.1559

Therefore, the 65th percentile of the scores is approximately 85.16.

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