1. Slide Slide 1 Suppose we are interested in the probability that z is less than 1.42. P(z < 1.42) = z*z*.00.01.02.03 1.3.9032.9049.9066.9082 1.4.9192.9207.9222.9236 1.5.9332.9345.9357.9370 0.9222 P(z < 1.42) 1.42 Find the intersection of the row 1.4 and column.02. … … … …… … …… … …

2. Slide Slide 2 Suppose we are interested in the probability that z* is less than 0.58. P(z < 0.58) = z*z*.07.08.09 0.4.6808.6844.6879 0.5.7157.7190.7224 0.6.7486.7517.7549 0.7190 …………… … … … … P(z < 0.58)

3. Slide Slide 3 Find the following probability: P(-1.76 < z < 0.58) = P(z < 0.58).7190 - P(z < -1.76) P(z < -1.76) -.0392 = 0.679 8

4. Slide Slide 4 Section 6-3 Applications of Normal Distributions

5. Slide Slide 5 Z Score Formula Formula 6-2 x – µx – µ  z =z = Round z scores to 2 decimal places

6. Slide Slide 6 Figure 6-12 Converting to a Standard Normal Distribution x –   z =

7. Slide Slide 7 Strategies for finding probabilities or proportions in normal distributions 1.State the probability statement 2.Draw a picture 3.Calculate the z-score 4.Look up the probability (proportion) in the table

8. Slide Slide 8 It is safe to load a water taxi up to 3500 pounds. We will assume the mean weight of a passenger is 140 pounds. Assume the worst case that all passengers are men. Assume also that the weights of the men are normally distributed with a mean of 172 pounds and standard deviation of 29 pounds. If one man is randomly selected, what is the probability he weighs less than 174 pounds? Example – Weights of Water Taxi Passengers

9. Slide Slide 9 Example - cont z = 174 – 172 29 = 0.07 Figure 6-13  =  29   = 172

10. Slide Slide 10 Example - cont Figure 6-13 P ( x < 174 lb.) = P(z < 0.07) = 0.5279  =  29   = 172

11. Slide Slide 11 1. Don’t confuse z scores and areas. z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located in the left half of the normal distribution. 4. Areas (or probabilities) are positive or zero values, but they are never negative. Cautions to Keep in Mind

12. Slide Slide 12 Procedure for Finding Values Using Table A-2 and Formula 6-2 1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought. 2. Use Table A-2 to find the z score corresponding to the cumulative left area bounded by x. Refer to the body of Table A-2 to find the closest area, then identify the corresponding z score. 3. Using Formula 6-2, enter the values for µ, , and the z score found in step 2, then solve for x. x = µ + (z  ) (Another form of Formula 6-2) (If z is located to the left of the mean, be sure that it is a negative number.) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem.

13. Slide Slide 13 Example – Lightest and Heaviest Use the data from the previous example to determine what weight separates the lightest 99.5% from the heaviest 0.5%?

14. Slide Slide 14 x =  + (z ●  ) x = 172 + (2.575  29) x = 246.675 (247 rounded) Example – Lightest and Heaviest - cont

15. Slide Slide 15 The weight of 247 pounds separates the lightest 99.5% from the heaviest 0.5% Example – Lightest and Heaviest - cont

16. Slide Slide 16 Data on the length of time to complete registration for classes using an on-line registration system suggest that the distribution of the variable x = time to register for students at a particular university can well be approximated by a normal distribution with mean  = 12 minutes and standard deviation  = 2 minutes. What is the probability that it will take a randomly selected student less than 9 minutes to complete registration? 9 P(x < 9) = 0.0668 Look this value up in the table. Standardizes 9.

17. Slide Slide 17 Registration Problem Continued... x = time to register  = 12 minutes and  = 2 minutes What is the probability that it will take a randomly selected student more than 13 minutes to complete registration? P(x > 13) = 1 -.6915 = 0.3085 13

18. Slide Slide 18 Registration Problem Continued... x = time to register  = 12 minutes and  = 2 minutes What is the probability that it will take a randomly selected student between 7 and 15 minutes to complete registration? P(7 < x < 15) =.9332 -.0062 = 0.9270 715

19. Slide Slide 19 Will my calculator do this for me???? YES!!!!