2 Your 4 th homework is assigned. It is due on 12th of Feb, 11:59 pm.
3 There are only two possible outcomes: work or do not work a) Adult not working during summer vacation.b)The experiment consist of 10 identical trials. A trial for this experiment is an individual.There are only two possible outcomes: work or do not workThe probability remains same for each individual (trial)Individuals are independent
5 Thinking ChallengeThe communications monitoring company Postini has reported that 91% of messages are spam. Suppose your inbox contains 25 messages.What are the mean and standard deviation of the number of real messages you should expect to find in your inbox?What is the probability that you will find only 1 or 2 real messages?
6 Thinking ChallengeWhat are the mean and standard deviation of the number of real messages you should expect to find in your inbox? (2.25, 1.43)What is the probability that you will find only 1 or 2 real messages?(P(X=1)+P(X=2)= )
7 Content Two Types of Random Variables Probability Distributions for Discrete Random VariablesThe Binomial DistributionHypergeometric DistributionsProbability Distributions for Continuous Random VariablesThe Normal DistributionUniform Distribution
8 Hypergeometric Distribution 4.4Hypergeometric Distribution
9 Characteristics of a Hypergeometric Random Variable The experiment consists of randomly drawing n elements without replacement from a set of N elements, r of which are S’s (for success) and (N – r) of which are F’s (for failure).The hypergeometric random variable x is the number of S’s in the draw of n elements.
10 Hypergeometric Probability Distribution Function [x = Maximum [0, n – (N – r), …, Minimum (r, n)]where . . .
11 Hypergeometric Probability Distribution Function N = Total number of elementsr = Number of S’s in the N elementsn = Number of elements drawnx = Number of S’s drawn in the n elements
13 Thinking ChallengeA carton of 12 eggs has 4 rotten eggs and 8 good eggs. Three eggs are chosen at random from the carton to make a three-egg omelet.Let X = the number of rotten eggs chosen. What is the probability that the sample will consist of one rotten egg and two good eggs, that is, what is P(X = 1)?(a) 81/220 (b) 192/220 (c) 112/220 (d) 56/220
14 Thinking Challenge a) b) c) d) 5 cards are picked from a deck of 52 cards without replacement. What is the probability that 2 of the selected cards will be Ace?a) b) c) d)
15 Probability Distributions for Continuous Random Variables 4.5Probability Distributions for Continuous Random Variables
16 Continuous Probability Density Function The graphical form of the probability distribution for a continuous random variable x is a smooth curve that might appear as below:
17 Continuous Probability Density Function This curve, a function of x, is denoted by the symbol f(x) and is variously called a probability density function (pdf), a frequency function, or a probability distribution.The areas under a probability distribution correspond to probabilities for x. The area A beneath the curve between two points a and b is the probability that x assumes a value between a and b.
18 Continuous Probability Density Function P(a<x<b)=P(a≤x≤b) sinceP(x=a)=P(x=b)=0
19 The Normal Distribution 4.6The Normal Distribution
20 Importance of Normal Distribution Describes many random processes or continuous phenomenaCan be used to approximate discrete probability distributionsExample: binomialBasis for classical statistical inference
21 Normal Distribution f ( x ) x ‘Bell-shaped’ & symmetrical Mean, median, mode are equalf(x)xMean Median Mode
22 Probability Density Function where µ = Mean of the normal random variable x = Standard deviation π = e = P(x < a) is obtained from a table of normal probabilities
24 Normal Distribution Probability Probability is area under curve!f(x)xcd
25 Standard Normal Distribution The standard normal distribution is a normal distribution with µ = 0 and = 1. A random variable with a standard normal distribution, denoted by the symbol z, is called a standard normal random variable.
26 The Standard Normal Table: P(0 < z < 1.96) Standard Normal Probability Table (Portion).06Z.04.05zm= 0s= 11.961.8.4671.4678.4686.47501.9.4750.4738.47442.0.4793.4798.48032.1.4838.4842.4846Probabilities
27 The Standard Normal Table: P(–1.26 z 1.26) Standard Normal Distributions= 1P(–1.26 ≤ z ≤ 1.26)== .7924.3962.3962–1.261.26zm= 0
28 The Standard Normal Table: P(z > 1.26) Standard Normal Distributions= 1.500.500P(z > 1.26)= – .3962= .1038.39621.26zm= 0
29 The Standard Normal Table: P(–2.78 z –2.00) Standard Normal Distributions= 1P(–2.78 ≤ z ≤ –2.00)= – .4772= .0201.4973.4772–2.78–2.00zm= 0
30 The Standard Normal Table: P(z > –2.13) Standard Normal Distributions= 1P(z > –2.13)== .9834.4834.5000–2.13zm= 0
31 The rest of the quizzes will be held in the lecture on Wednesdays. The updated dates for quizzes are as below.I will take attendance and will give extra credit to the students who have attended at least 8 of the lectures till the end of the semester.
32 Non-standard Normal Distribution Normal distributions differ by mean & standard deviation.Each distribution would require its own table.That’s an infinite number of tables!xf(x)
33 Property of Normal Distribution If x is a normal random variable with mean μ and standard deviation , then the random variable z, defined by the formulahas a standard normal distribution. The value z describes the number of standard deviations between x and µ.
34 Standardize the Normal Distribution One table! = 1zStandard Normal Distributionsm= 0x
35 Finding a Probability Corresponding to a Normal Random Variable 1. Sketch normal distribution, indicate mean, and shade the area corresponding to the probability you want.2. Convert the boundaries of the shaded area from x values to standard normal random variable zShow the z values under corresponding x values.3. Use Table in Appendix D to find the areas corresponding to the z values. Use symmetry when necessary.
36 Of course, the TI does it all: normalcdf(a,b,μ,σ) returns the probability P(a < x < b) with x distributed N(μ, σ).
37 Non-standard Normal μ = 5, σ = 10: P(5 < x < 6.2) Normal Distributionx = 5 = 106.2 = 0z = 1.12Standard Normal Distribution.0478
38 Non-standard Normal μ = 5, σ = 10: P(3.8 x 5) Normal Distributionx = 53.8z = 0-.12Standardized Normal Distribution = 10 = 1.0478
39 Non-standard Normal μ = 5, σ = 10: P(2.9 x 7.1) Normal Distribution-.21z.21Standard Normal Distribution = 10 = 1.1664.0832
41 Non-standard Normal μ = 5, σ = 10: P(7.1 X 8) = 1087.1xNormal Distribution = 1.30z.21Standard Normal Distribution.1179.0347.0832= 5= 0
42 Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours and = 200 hours. What’s the probability that a bulb will lastA. between 2000 and hours?B. less than 1470 hours?
43 Solution* P(2000 x 2400) x z .4772 = 2000 = 200 2400 = 1 Normal Distributionx= 2000 = 2002400 = 1Standard Normal Distributionz= 02.0.4772
44 Solution* P(x 1470) x z .0040 = 2000 = 200 1470 = 0 = 1 –2.65 Normal Distributionz= 0 = 1–2.65Standard Normal Distribution.5000.0040.4960
45 Finding z-Values for Known Probabilities What is z, given P(z) = .1217?z= 0= 1?.1217Standard Normal Probability Table (Portion)Z.000.20.0.0000.0040.00800.1.0398.0438.0478.0793.0832.0871.1179.1255.010.3.1217.31
46 Finding x Values for Known Probabilities Normal DistributionStandard Normal Distributionz= 0= 1.31.1217x= 5= 10?.12178.1
47 Using a TI: invNorm(p, μ, σ) returns the 100pth percentile of N(μ, σ).
48 Normal Distribution Thinking Challenge At one university, the students are given z-scores at the end of each semester rather than the traditional GPAs. The mean and the standard deviation of all students’ cumulative GPAs, on which the z-scores are based, are 2.7 and .5, respectively.a) Translate each of the following z-scores to the corresponding GPA: z=2.0, z=-1, z=0.5, z=-2.5.b) Students with z-scores below -1.6 are put on probation. What is the corresponding probationary GPA?c) The president of the university wishes to graduate the top 20% of the students with cum laude honors and the top 2.5% with summa cum laude honors. Under the assumption that the distribution is exactly normal, by using the Table Z in the appendix, determine the limits be set in terms of original GPAs.
49 Solutiona) 3.7, 2.2, 2.95, 1.45b) 1.9P(z > 0.84)= P(z > 1.96)=0.025So,for cum laude: 0.84=(x-2.7)/0.5x= 3.12;for summa cum laude: 1.96=(x-2.7)/0.5x= 3.68
50 Descriptive Methods for Assessing Normality 4.7Descriptive Methods for Assessing Normality
51 Determining Whether the Data Are from an Approximately Normal Distribution Construct either a histogram or stem-and-leaf display for the data and note the shape of the graph. If the data are approximately normal, the shape of the histogram or stem-and-leaf display will be similar to the normal curve.
52 Determining Whether the Data Are from an Approximately Normal Distribution Compute the intervals and determine the percentage of measurements falling in each. If the data are approximately normal, the percentages will be approximately equal to 68%, 95%, and 100%, respectively; from the Empirical Rule (68%, 95%, 99.7%).
53 Determining Whether the Data Are from an Approximately Normal Distribution Find the interquartile range, IQR, and standard deviation, s, for the sample, then calculate the ratio IQR/s. If the data are approximately normal, then IQR/s ≈ 1.3.
54 Determining Whether the Data Are from an Approximately Normal Distribution Examine a normal probability plot for the data. If the data are approximately normal, the points will fall (approximately) on a straight line.Observed valueExpected z–score
55 Normal Probability Plot A normal probability plot for a data set is a scatterplot with the ranked data values on one axis and their corresponding expected z-scores from a standard normal distribution on the other axis. [Note: Computation of the expected standard normal z-scores are beyond the scope of this lecture. Therefore, we will rely on available statistical software packages to generate a normal probability plot.]
59 Other Continuous Distributions: Uniform 4.8Other Continuous Distributions: Uniform
60 Uniform DistributionContinuous random variables that appear to have equally likely outcomes over their range of possible values possess a uniform probability distribution.Suppose the random variable x can assume values only in an interval c ≤ x ≤ d. Then the uniform frequency function has a rectangular shape.
61 Probability Distribution for a Uniform Random Variable x Probability density function:Mean:Standard Deviation:
62 Uniform Distribution Example You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses between 11.5 and 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?SODA
63 Uniform Distribution Solution f(x)1.0x11.511.812.5P(11.5 x 11.8) = (Base)/(Height)= (11.8 – 11.5)/(1) = .30
64 Review Q1Consider a powder specimen that has exactly 10 anthrax spores. Suppose that the number of anthrax spores in the sample detected by the new method follows an approximate uniform distribution between 0 and 10.a. Find the probability that 8 or fewer anthrax spores are detected in the powder specimen.b. Find the probability that between 2 and 5 anthrax spores are detected in the powder specimen.c. Find the expected number of anthrax spores.d. Find the standard deviation of anthrax spores.
65 Review Q2The output from a statistical computer program indicates that the mean and standard deviation of a data set consisting of 200 measurements are $1500 and $300, respectively. Suppose the frequency distribution of the data set has normal distribution. What of percentage of data will be between $750 and $1200?
66 Review Q3Almost all companies utilize some type of year-end performance review for their employees. Human Resources(HR) at the university of Texas health Science Center provides guidelines for supervisors rating their subordinates. For example raters are advised to examine their ratings for a tendency to be too lenient or too harsh. According to HR, “if you have this tendency, consider using a normal distribution-10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable”.Suppose you are rating an employee’s performance on a scale of 1(lowest) to 100(highest). Also assume the ratings follow normal distribution with a mean of 50 and a standard deviation of 15.What is the lowest rating you should give to an “exemplary” employee if you follow the Univ. of Texas HR guidelines?What is the lowest rating you should give to an “competent” employee if you follow the Univ. of Texas HR guidelines?
67 Review Q4Suppose that 5 out of 13 liver transplants done at a hospital will fail within a year. Consider a random sample of 4 of these 13 patients. What is the probability that 1 of these patients will result in failed transplants?
68 Key Ideas Properties of Probability Distributions Discrete Distributions1. p(x) ≥ 02.Continuous Distributions1. P(x = a) = 02. P(a < x < b) = area under curve between a and b
69 Key IdeasMethods for Assessing Normality1. Histogram