1. © 2010 Pearson Prentice Hall. All rights reserved Chapter The Normal Probability Distribution © 2010 Pearson Prentice Hall. All rights reserved 3 7

2. Section 7.3 Applications of the Normal Distribution 7-2

3. © 2010 Pearson Prentice Hall. All rights reserved 7-3

4. © 2010 Pearson Prentice Hall. All rights reserved 7-4

5. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. * What is the probability that a randomly selected steel rod has a length less than 99.2 cm? * Based upon information obtained from Stefan Wilk. EXAMPLEFinding the Probability of a Normal Random Variable Interpretation: If we randomly selected 100 steel rods, we would expect about 4 of them to be less than 99.2 cm. 7-5

6. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. What is the probability that a randomly selected steel rod has a length between 99.8 and 100.3 cm? Interpretation: If we randomly selected 100 steel rods, we would expect about 42 of them to be between 99.8 cm and 100.3 cm. EXAMPLEFinding the Probability of a Normal Random Variable 7-6

7. © 2010 Pearson Prentice Hall. All rights reserved The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source: www.columbia.edu/cu/gsas/departments/psychology/department.html.) EXAMPLEFinding the Percentile Rank of a Normal Random Variable What is the percentile rank of a student who earns a combined GRE score of 1300? The area under the normal curve is a probability, proportion, or percentile. Here, the area under the normal curve to the left of 1300 represents the percentile rank of the student. Area left of 1300 = Area left of (z = 1.33) = 0.91 (rounded to two decimal places) Interpretation: The student scored at the 91 st percentile. This means the student scored better than 91% of the students who took the GRE. 7-7

8. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer must discard all rods less than 99.1 cm or longer than 100.9 cm. What proportion of rods must be discarded? EXAMPLEFinding the Proportion Corresponding to a Normal Random Variable The proportion is the area under the normal curve to the left of 99.1 cm plus the area under the normal curve to the right of 100.9 cm. Area left of 99.1 + area right of 100.9 = (Area left of z = -2) + (Area right of z = 2) = 0.0228 + 0.0228 = 0.0456 Interpretation: The proportion of rods that must be discarded is 0.0456. If the company manufactured 1000 rods, they would expect to discard about 46 of them. 7-8

9. © 2010 Pearson Prentice Hall. All rights reserved 7-9

10. © 2010 Pearson Prentice Hall. All rights reserved 7-10

11. © 2010 Pearson Prentice Hall. All rights reserved The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) EXAMPLEFinding the Value of a Normal Random Variable What is the score of a student whose percentile rank is at the 85 th percentile? The z-score that corresponds to the 85 th percentile is the z-score such that the area under the standard normal curve to the left is 0.85. This z-score is 1.04. x = µ + zσ = 1049 + 1.04(189) = 1246 Interpretation: The proportion of rods that must be discarded is 0.0456. If the company manufactured 1000 rods, they would expect to discard about 46 of them. 7-11

12. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer wants to accept 90% of all rods manufactured. Determine the length of rods that make up the middle 90% of all steel rods manufactured. EXAMPLEFinding the Value of a Normal Random Variable Area = 0.05 z 1 = -1.645 and z 2 = 1.645 x 1 = µ + z 1 σ = 100 + (-1.645)(0.45) = 99.26 cm x 2 = µ + z 2 σ = 100 + (1.645)(0.45) = 100.74 cm Interpretation: The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between 99.26 cm and 100.74 cm. 7-12