Elementary Statistics Chapter

1、6.1 - 1Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Lecture Slides Elementary Statistics Eleventh Edition and the Triola Statistics Series by Mario F. Triola6.1 - 2Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Chapter 6Normal Probability Distributio

2、ns6-1 Review and Preview6-2 The Standard Normal Distribution6-3 Applications of Normal Distributions6-4 Sampling Distributions and Estimators6-5 The Central Limit Theorem6-6 Normal as Approximation to Binomial6-7 Assessing Normality6.1 - 3Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights

3、 Reserved.Section 6-1 Review and Preview6.1 - 4Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.v Chapter 2: Distribution of datav Chapter 3: Measures of data sets, including measures of center and variationv Chapter 4: Principles of probabilityv Chapter 5: Discrete probability

4、 distributionsReview6.1 - 5Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. Chapter focus is on:v Continuous random variablesv Normal distributionsPreviewFigure 6-1Formula 6-1f x e12x22Distribution determined by fixed values of mean and standard deviation6.1 - 6Copyright 2010,

5、 2007, 2004 Pearson Education, Inc. All Rights Reserved.Section 6-2 The Standard Normal Distribution6.1 - 7Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Key ConceptThis section presents the standard normal distribution which has three properties:1. Its graph is bell-shaped.2

6、. Its mean is equal to 0 ( = 0).3. Its standard deviation is equal to 1 ( = 1).Develop the skill to find areas (or probabilities or relative frequencies) corresponding to various regions under the graph of the standard normal distribution. Find z-scores that correspond to area under the graph.6.1 -

7、8Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Uniform DistributionA continuous random variable has a uniform distribution if its values are spread evenly over the range of probabilities. The graph of a uniform distribution results in a rectangular shape.6.1 - 9Copyright 201

8、0, 2007, 2004 Pearson Education, Inc. All Rights Reserved.A density curve is the graph of a continuous probability distribution. It must satisfy the following properties:Density Curve1. The total area under the curve must equal 1.2. Every point on the curve must have a vertical height that is 0 or g

9、reater. (That is, the curve cannot fall below the x-axis.)6.1 - 10Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.Area and Probability 6.1 - 11Copyright 2010, 2

10、007, 2004 Pearson Education, Inc. All Rights Reserved.Using Area to Find ProbabilityGiven the uniform distribution illustrated, find the probability that a randomly selected voltage level is greater than 124.5 volts. Shaded area represents voltage levels greater than 124.5 volts. Correspondence betw

11、een area and probability: - 12Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Standard Normal DistributionThe standard normal distribution is a normal probability distribution with = 0 and = 1. The total area under its density curve is equal to 1.6.1 - 13Copyright 201

12、0, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Finding Probabilities When Given z-scoresv Table A-2 (in Appendix A)v Formulas and Tables insert cardv Find areas for many different regions6.1 - 14Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Finding Probabilities O

13、ther Methodsv STATDISKv Minitabv Excelv TI-83/84 Plus6.1 - 15Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Methods for Finding Normal Distribution Areas6.1 - 16Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Methods for Finding Normal Distribution Area

14、s6.1 - 17Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Table A-26.1 - 18Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.1. It is designed only for the standard normal distribution, which has a mean of 0 and a standard deviation of 1.2. It is on two pag

15、es, with one page for negative z-scores and the other page for positivez-scores.3. Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score.Using Table A-26.1 - 19Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.4

16、. When working with a graph, avoid confusion between z-scores and areas.z ScoreDistance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of Table A-2.AreaRegion under the curve; refer to the values in the body of Table A-2.5. The part of the z-scor

17、e denoting hundredths is found across the top.Using Table A-26.1 - 20Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0C at the freezing point of water. Tests on a large

18、sample of these instruments reveal that at the freezing point of water, some thermometers give readings below 0 (denoted by negative numbers) and some give readings above 0 (denoted by positive numbers). Assume that the mean reading is 0C and the standard deviation of the readings is 1.00C. Also ass

19、ume that the readings are normally distributed. If one thermometer is randomly selected, find the probability that, at the freezing point of water, the reading is less than 1.27. Example - Thermometers6.1 - 21Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.P(z 1.27) = Example

20、- (Continued)6.1 - 22Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Look at Table A-26.1 - 23Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.P (z 1.27) = 0.8980Example - cont6.1 - 24Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

21、The probability of randomly selecting a thermometer with a reading less than 1.27 is 0.8980.P (z 1.27) = 0.8980Example - cont6.1 - 25Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Or 89.80% will have readings below 1.27.P (z 1.23) = 0.8907Example - Thermometers Again6.1 - 27C

22、opyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.P (z 1.23) = 0.890789.07% of the thermometers have readings above 1.23 degrees.Example - cont6.1 - 28Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.A thermometer is randomly selected. Find the probability t

23、hat it reads (at the freezing point of water) between 2.00 and 1.50 degrees. P (z 2.00) = 0.0228P (z 1.50) = 0.9332P (2.00 z 1.50) = 0.9332 0.0228 = 0.9104The probability that the chosen thermometer has a reading between 2.00 and 1.50 degrees is 0.9104.Example - Thermometers III6.1 - 29Copyright 201

24、0, 2007, 2004 Pearson Education, Inc. All Rights Reserved.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) = 0.0228P (z 1.50) = 0.9332P (2.00 z 1.50) = 0.9332 0.0228 = 0.9104A thermometer is randomly s

25、elected. Find the probability that it reads (at the freezing point of water) between 2.00 and 1.50 degrees. Example - cont6.1 - 30Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved. P(a z a) denotes the probability that the z score is greater than a.P(z a) denotes the probability

26、 that the z score is less than a.Notation6.1 - 31Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Finding a z Score When Given a Probability Using Table A-21. Draw a bell-shaped curve and identify the region under the curve that corresponds to the given probability. If that reg

27、ion 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.6.1 - 32Copyright 2010, 2007, 2004 Pear

28、son Education, Inc. All Rights Reserved.Finding z Scores When Given Probabilities5% or 0.05(z score will be positive)Finding the 95th Percentile6.1 - 33Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Finding z Scores When Given Probabilities - contFinding the 95th Percentile1.

29、6455% or 0.05(z score will be positive)6.1 - 34Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.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 - cont6.1 - 35Copyright 2010, 2007, 2004 Pearson

30、Education, Inc. All Rights Reserved.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 - cont6.1 - 36Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Finding the Bottom 2.5% and Upper 2.5%(One z s

31、core will be negative and the other positive)Finding z Scores When Given Probabilities - cont6.1 - 37Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.RecapIn this section we have discussed:v Density curves.v Relationship between area and probability.v Standard normal distributi

32、on.v Using Table A-2.6.1 - 38Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Section 6-3 Applications of Normal Distributions6.1 - 39Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Key ConceptThis section presents methods for working with normal distribu

33、tions that are not standard. That is, the mean is not 0 or the standard deviation is not 1, or both.The key concept is that we can use a simple conversion that allows us to standardize any normal distribution so that the same methods of the previous section can be used.6.1 - 40Copyright 2010, 2007,

34、2004 Pearson Education, Inc. All Rights Reserved.Conversion Formulax z =Round z scores to 2 decimal places6.1 - 41Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Converting to a Standard Normal Distribution x z =6.1 - 42Copyright 2010, 2007, 2004 Pearson Education, Inc. All Ri

35、ghts Reserved. In the Chapter Problem, we noted that the safe load for a water taxi was found to be 3500 pounds. We also noted that the mean weight of a passenger was assumed to be 140 pounds. Assume the worst case that all passengers are men. Assume also that the weights of the men are normally dis

36、tributed 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 6.1 - 43Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Example - cont z = 1

37、74 17229 = 0.07 =29 = 1726.1 - 44Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Example - contP ( x 174 lb.) = P(z 175) = 0.4602P(x 175) = 0.32286.1 - 87Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Interpretation of ResultsGiven that the safe capacit

38、y of the water taxi is 3500 pounds, there is a fairly good chance (with probability 0.3228) that it will be overloaded with 20 randomly selected men.6.1 - 88Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Correction for a Finite PopulationN n x= nN 1finite populationcorrection

39、 factorWhen sampling without replacement and the sample size n is greater than 5% of the finite population of size N (that is, n 0.05N ), adjust the standard deviation of sample means by multiplying it by the finite population correction factor:6.1 - 89Copyright 2010, 2007, 2004 Pearson Education, I

40、nc. All Rights Reserved.RecapIn this section we have discussed:v Central limit theorem.v Practical rules.v Effects of sample sizes.v Correction for a finite population.6.1 - 90Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Section 6-6Normal as Approximation to Binomial6.1 - 9

41、1Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Key ConceptThis section presents a method for using a normal distribution as an approximation to the binomial probability distribution.If the conditions of np 5 and nq 5 are both satisfied, then probabilities from a binomial pro

42、bability distribution can be approximated well by using a normal distribution with mean = np and standard deviation npq.6.1 - 92Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Review Binomial Probability Distribution1. The procedure must have a fixed number of trials.2. The tr

43、ials must be independent.3. Each trial must have all outcomes classified into two categories (commonly, success and failure).4. The probability of success remains the same in all trials.Solve by binomial probability formula, Table A-1, or technology.6.1 - 93Copyright 2010, 2007, 2004 Pearson Educati

44、on, Inc. All Rights Reserved.Approximation of a Binomial Distributionwith a Normal Distributionnp 5nq 5then = np and = npq and the random variable hasdistribution.(normal)a6.1 - 94Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Procedure for Using a Normal Distribution to Appr

45、oximate a Binomial Distribution1. Verify that both np 5 and nq 5. If not, you must use software, a calculator, a table or calculations using the binomial probability formula.2. Find the values of the parameters and by calculating = np and = npq.3. Identify the discrete whole number x that is relevan

46、t to the binomial probability problem. Focus on this value temporarily.6.1 - 95Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.4. Draw a normal distribution centered about , then draw a vertical strip area centered over x. Mark the left side of the strip with the number equal

47、to x 0.5, and mark the right side with the number equal to x + 0.5. Consider the entire area of the entire strip to represent the probability of the discrete whole number itself. 5. Determine whether the value of x itself is included in the probability. Determine whether you want the probability of

48、at least x, at most x, more than x, fewer than x, or exactly x. Shade the area to the right or left of the strip; also shade the interior of the strip if and only if x itself is to be included. This total shaded region corresponds to the probability being sought.Procedure for Using a Normal Distribu

49、tion to Approximate a Binomial Distribution6.1 - 96Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.6.Using x 0.5 or x + 0.5 in place of x, find the area of the shaded region: find the z score; use that z score to find the area to the left of the adjusted value of x; use that c

50、umulative area to identify the shaded area corresponding to the desired probability.Procedure for Using a Normal Distribution to Approximate a Binomial Distribution6.1 - 97Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Figure 6-21Finding the Probability of “At Least 122 Men”

51、Among 213 PassengersExample Number of Men Among Passengers6.1 - 98Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.DefinitionWhen we use the normal distribution (which is a continuous probability distribution) as an approximation to the binomial distribution (which is discrete)

52、, a continuity correction is made to a discrete whole number x in the binomial distribution by representing the discrete whole number x by the interval from x 0.5 to x + 0.5(that is, adding and subtracting 0.5).6.1 - 99Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.x = at lea

53、st 8 (includes 8 and above)x = more than 8 (doesnt include 8)x = at most 8 (includes 8 and below)x = fewer than 8 (doesnt include 8)x = exactly 8 6.1 - 100Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.RecapIn this section we have discussed:v Approximating a binomial distribu

54、tion with a normal distribution.v Procedures for using a normal distribution to approximate a binomial distribution.v Continuity corrections.6.1 - 101Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Section 6-7Assessing Normality6.1 - 102Copyright 2010, 2007, 2004 Pearson Educa

55、tion, Inc. All Rights Reserved.Key ConceptThis section presents criteria for determining whether the requirement of a normal distribution is satisfied. The criteria involve visual inspection of a histogram to see if it is roughly bell shaped, identifying any outliers, and constructing a graph called

56、 a normal quantile plot.6.1 - 103Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Definition A normal quantile plot (or normal probability plot) is a graph of points (x,y), where each x value is from the original set of sample data, and each y value is the corresponding z score

57、 that is a quantile value expected from the standard normal distribution. 6.1 - 104Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Procedure for Determining Whether It Is Reasonable to Assume that Sample Data are From a Normally Distributed Population1. Histogram: Construct a

58、histogram. Reject normality if the histogram departs dramatically from a bell shape. 2. Outliers: Identify outliers. Reject normality if there is more than one outlier present.3. Normal Quantile Plot: If the histogram is basically symmetric and there is at most one outlier, use technology to generat

59、e a normal quantile plot. 6.1 - 105Copyright 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.Procedure for Determining Whether It Is Reasonable to Assume that Sample Data are From a Normally Distributed Population3. ContinuedUse the following criteria to determine whether or not the dis

60、tribution is normal.Normal Distribution: The population distribution is normal if the pattern of the points is reasonably close to a straight line and the points do not show some systematic pattern that is not a straight-line pattern. 6.1 - 106Copyright 2010, 2007, 2004 Pearson Education, Inc. All R