1. Slide Slide 1 Chapter 6 Normal Probability Distributions 6-1 Overview 6-2 The Standard Normal Distribution 6-3 Applications of Normal Distributions 6-4 Sampling Distributions and Estimators 6-5 The Central Limit Theorem 6-6 Normal as Approximation to Binomial 6-7 Assessing Normality

2. Slide Slide 2 Section 6-1 Overview

3. Slide Slide 3 Chapter focus is on:  Continuous random variables  Normal distributions Overview Figure 6-1 Formula 6-1 f(x) =  2  x-x-  )2)2 ( e2e2

4. Slide Slide 4 Section 6-2 The Standard Normal Distribution

5. Slide Slide 5 Key Concept This section presents the standard normal distribution which has three properties: 1. It is bell-shaped. 2. It has a mean equal to 0. 3. It has a standard deviation equal to 1. It is extremely important to develop the skill to find areas (or probabilities or relative frequencies) corresponding to various regions under the graph of the standard normal distribution.

6. Slide Slide 6 Definition  A continuous random variable has a uniform distribution if its values spread evenly over the range of probabilities. The graph of a uniform distribution results in a rectangular shape.

7. Slide Slide 7  A density curve is the graph of a continuous probability distribution. It must satisfy the following properties: Definition 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. (That is, the curve cannot fall below the x-axis.)

8. Slide Slide 8 Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. Area and Probability

9. Slide Slide 9 Example A statistics professor plans classes so carefully that the lengths of her classes are uniformly distributed between 50.0 min and 52.0 min. That is, any time between 50.0 min and 52.0 min is possible, and all of the possible values are equally likely. If we randomly select one of her classes and let x be the random variable representing the length of that class, then x has a distribution that can be graphed as:

10. Slide Slide 10 Example Kim has scheduled a job interview immediately following her statistics class. If the class runs longer than 51.5 minutes, she will be late for her job interview. Find the probability that a randomly selected class will last longer than 51.5 minutes given the uniform distribution illustrated below. Figure 6-3

11. Slide Slide 11 There are many different kinds of normal distributions, but they are both dependent upon two parameters… Population Mean Population Standard Deviation

12. Slide Slide 12 Definition  The standard normal distribution is a probability distribution with mean equal to 0 and standard deviation equal to 1, and the total area under its density curve is equal to 1.

13. Slide Slide 13 Finding Probabilities - Table A-2  Table A-2 (Inside back cover of textbook, Formulas and Tables card, Appendix)  STATDISK  Minitab  Excel  TI-83/84

14. Slide Slide 14 Table A-2 - Example

15. Slide Slide 15 z Score Distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2. Area Region under the curve; refer to the values in the body of Table A-2. Using Table A-2

16. Slide Slide 16 If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.58 degrees. Example - Thermometers

17. Slide Slide 17 P(z < 1.58) = Figure 6-6 Example - Cont

18. Slide Slide 18 Look at Table A-2

19. Slide Slide 19 P (z < 1.58) = 0.9429 Figure 6-6 Example - cont

20. Slide Slide 20 The probability that the chosen thermometer will measure freezing water less than 1.58 degrees is 0.9429. P (z < 1.58) = 0.9429 Example - cont

21. Slide Slide 21 P (z < 1.58) = 0.9429 94.29% of the thermometers have readings less than 1.58 degrees. Example - cont

22. Slide Slide 22 If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads (at the freezing point of water) above –1.23 degrees. The probability that the chosen thermometer with a reading above -1.23 degrees is 0.8907. P (z > – 1.23) = 0.8907 Example - cont

23. Slide Slide 23 P (z > – 1.23) = 0.8907 89.07% of the thermometers have readings above – 1.23 degrees. Example - cont

24. Slide Slide 24 A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. P (z < – 2.00) = 0.0228 P (z < 1.50) = 0.9332 P ( – 2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104 The probability that the chosen thermometer has a reading between – 2.00 and 1.50 degrees is 0.9104. Example - cont

25. Slide Slide 25 If many thermometers are selected and tested at the freezing point of water, then 91.04% of them will read between –2.00 and 1.50 degrees. P (z < – 2.00) = 0.0228 P (z < 1.50) = 0.9332 P ( – 2.00 < z < 1.50) = 0.9332 – 0.0228 = 0.9104 A thermometer is randomly selected. Find the probability that it reads (at the freezing point of water) between –2.00 and 1.50 degrees. Example - Modified

26. Slide Slide 26 P(a < z < b) denotes the probability that the z score is between a and b. P(z > a) denotes the probability that the z score is greater than a. P(z < a) denotes the probability that the z score is less than a. Notation

27. Slide Slide 27 Finding a z Score When Given a Probability Using Table A-2 1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not a cumulative region from the left, work instead with a known region that is a cumulative region from the left. 2.Using the cumulative area from the left, locate the closest probability in the body of Table A-2 and identify the corresponding z score.

28. Slide Slide 28 Finding z Scores When Given Probabilities Using the same thermometers as earlier, find the temperature corresponding to the 95 th percentile. That is, find the temperature separating the bottom 95% from the top 5%. 5% or 0.05 (z score will be positive) Figure 6-10 Finding the 95 th Percentile

29. Slide Slide 29 Finding z Scores When Given Probabilities - cont Figure 6-10 Finding the 95 th Percentile 1.645 5% or 0.05 (z score will be positive)

30. Slide Slide 30 Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% (One z score will be negative and the other positive) Using the same thermometers, find the temperatures separating the bottom 2.5% from the top 2.5%.

31. Slide Slide 31 Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% (One z score will be negative and the other positive) Finding z Scores When Given Probabilities - cont

32. Slide Slide 32 Figure 6-11 Finding the Bottom 2.5% and Upper 2.5% (One z score will be negative and the other positive) Finding z Scores When Given Probabilities - cont

33. Slide Slide 33 Recap In this section we have discussed:  Density curves.  Relationship between area and probability  Standard normal distribution.  Using Table A-2.