### Normal Distribution Problem**Question:**The scores on a test approximate a normal distribution with a mean score of 72 and a standard deviation of 9. Approximately what percent of the students taking the test received a score greater than 90?[Graph/Diagram Explanation: Not provided in the image]**Solution:**To determine the percentage of students who received a score greater than 90, we can use the properties of the normal distribution.Given:- Mean (μ) = 72- Standard deviation (σ) = 9- Score (X) to evaluate = 90First, we need to calculate the z-score for X = 90 using the z-score formula:\[ z = \frac{X - μ}{σ} \]Substituting the given values:\[ z = \frac{90 - 72}{9} \]\[ z = \frac{18}{9} \]\[ z = 2 \]Next, we consult a standard normal (z) table or use a statistical calculator to find the probability corresponding to a z-score of 2. The z-table gives us the area to the left of z.The cumulative probability for z = 2 is approximately 0.9772. This represents the probability that a student scored less than 90.Finally, to find the percentage of students who scored greater than 90, we subtract this cumulative probability from 1:\[ P(X > 90) = 1 - P(X < 90) \]\[ P(X > 90) = 1 - 0.9772 \]\[ P(X > 90) = 0.0228 \]Thus, approximately 2.28% of the students received a score greater than 90.

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3. The scores on a test approximate a normal distribution with a mean score of 72 and a standard deviation of 9. Approximately what percent of the student taking the test received a score greater than 90?

3. The scores on a test approximate a normal distribution with a mean score of 72 and a standard deviation of 9. Approximately what percent of the student taking the test received a score greater than 90?

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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