Section 7.2 Applications of the Normal Distribution 307 Finding a normal value corresponding to a given area The following procedure is used to calculate the normal value corresponding to an area to the left. Step 1. In an empty cell, select the Insert Function icon and highlight Statistical in the category field. Step 2. Click on the NORM.INV function and press OK. Step 3. Enter the area to the left of the desired normal value in the Probability field. Step 4. Enter the value for the mean in the Mean field and the value for the standard deviation in the Standard deviation field. Step 5. Click OK. Figure H Figure H illustrates finding the normal value that has an area of 0.98 to its left, where �� = 100 and �� = 15 (Example 7.17). SECTION 7.2 Exercises Exercises 1– 8 are the Check Your Understanding exercises located within the section. Understanding the Concepts In Exercises 9–10, fill in each blank with the appropriate word or phrase. 9. The process of converting a value x from a normal distribution to a z-score is known as . standardization 10. A value that is two standard deviations below the mean will have a z-score of . −2 In Exercises 11–16, determine whether the statement is true or false. If the statement is false, rewrite it as a true statement. 11. z-scores follow a standard normal distribution. True 12. A z-score indicates how many standard deviations a value is above or below the mean. True 13. If a normal population has a mean of �� and a standard deviation of ��, then the area to the left of �� is less than 0.5. False 14. If a normal population has a mean of �� and a standard deviation of ��, then the area to the right of �� + �� is less than 0.5. True 15. If a normal population has a mean of �� and a standard deviation of ��, then P(X = ��) = 1. False 16. If a normal population has a mean of �� and a standard deviation of ��, then P(X = ��) = 0. True Practicing the Skills 17. A normal population has mean �� = 20 and standard deviation �� = 4. a. What proportion of the population is less than 18? 0.3085 b. What is the probability that a randomly chosen value will be greater than 25? 0.1056 18. A normal population has mean �� = 9 and standard deviation �� = 6. a. What proportion of the population is less than 20? b. What is the probability that a randomly chosen value will be greater than 5? 0.7486 Tech: 0.7475 19. A normal population has mean �� = 25 and standard deviation �� = 11. a. What proportion of the population is greater than 34? 0.2061 Tech: 0.2066 b. What is the probability that a randomly chosen value will be less than 10? 0.0869 Tech: 0.0863 20. A normal population has mean �� = 61 and standard deviation �� = 16. a. What proportion of the population is greater than 100? b. What is the probability that a randomly chosen value will be less than 80? 0.8830 Tech: 0.8825 21. A normal population has mean �� = 47 and standard deviation �� = 3. a. What proportion of the population is between 40 and 50? b. What is the probability that a randomly chosen value will be between 50 and 55? 0.1549 Tech: 0.1548 22. A normal population has mean �� = 35 and standard deviation �� = 8. a. What proportion of the population is between 20 and 30? b. What is the probability that a randomly chosen value will be between 30 and 40? 0.4714 Tech: 0.4680 23. A normal population has mean �� = 12 and standard deviation �� = 3. What is the 40th percentile of the population? 11.25 Tech: 11.24 24. A normal population has mean �� = 56 and standard deviation �� = 8. What is the 85th percentile of the population? 64.32 Tech: 64.29 25. A normal population has mean �� = 46 and standard deviation �� = 9. What is the 19th percentile of the population? 38.08 Tech: 38.10 26. A normal population has mean �� = 71 and standard deviation �� = 33. What is the 91st percentile of the population? 115.22 Tech: 115.24 Working with the Concepts 27. Check your blood pressure: In a recent study, the Centers for Disease Control and Prevention reported that diastolic blood pressures of adult women in the United States are approximately normally distributed with mean 80.5 and standard deviation 9.9.
navidi_monk_elementary_statistics_2e_ch7-9
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