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Your 4 th homework is assigned. It is due on 12 th of Feb, 11:59 pm.

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##### Your 4 th homework is assigned. It is due on 12 th of Feb, 11:59 pm.

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**Your 4 th homework is assigned. It is due on 12th of Feb,**11:59 pm.**c) 0.35**d) 0.2522 e) 0.2616 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 work • The probability remains same for each individual (trial) • Individuals are independent**By using TI-84:**d) P(X = 3) = = P(X ≤ 3) – P(X ≤ 2) binomcdf(10,.35,3) - binomcdf(10,.35,2) e) P(X ≤ 2) binomcdf(10,.35,2)**Thinking Challenge**• The communications monitoring company Postini has reported that 91% of e-mail 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?**Thinking Challenge**• What 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)= 0.5117)**Content**• Two Types of Random Variables • Probability Distributions for Discrete Random Variables • The Binomial Distribution • Hypergeometric Distributions • Probability Distributions for Continuous Random Variables • The Normal Distribution • Uniform Distribution**4.4**Hypergeometric Distribution**Characteristics of a HypergeometricRandom 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.**Hypergeometric Probability Distribution Function**[x = Maximum [0, n – (N – r), …, Minimum (r, n)] where . . .**Hypergeometric Probability Distribution Function**N = Total number of elements r = Number of S’s in the N elements n = Number of elements drawn x = Number of S’s drawn in the n elements**a. Hypergeometric**b. Binomial**Thinking Challenge**A 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**Thinking Challenge**• 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)**4.5**Probability Distributions for Continuous Random Variables**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:**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.**Continuous Probability Density Function**P(a<x<b)=P(a≤x≤b) sinceP(x=a)=P(x=b)=0**4.6**The Normal Distribution**Importance of Normal Distribution**• Describes many random processes or continuous phenomena • Can be used to approximate discrete probability distributions • Example: binomial • Basis for classical statistical inference**Mean Median Mode**Normal Distribution • ‘Bell-shaped’ & symmetrical • Mean, median, mode are equal • f • ( • x • ) • x**Probability Density Function**where µ = Mean of the normal random variable x = Standard deviation π = 3.1415 . . . e = 2.71828 . . . P(x < a) is obtained from a table of normal probabilities**Normal Distribution Probability**• Probability is area under curve! f(x) x c d**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.**s**= 1 m z = 0 1.96 The Standard Normal Table:P(0 < z < 1.96) Standard Normal Probability Table (Portion) .06 Z .04 .05 1.8 .4671 .4678 .4686 .4750 .4750 1.9 .4738 .4744 2.0 .4793 .4798 .4803 2.1 .4838 .4842 .4846 Probabilities**s**= 1 .3962 .3962 m = 0 The Standard Normal Table:P(–1.26 z 1.26) Standard Normal Distribution P(–1.26 ≤ z ≤ 1.26) = .3962 + .3962 = .7924 z –1.26 1.26**s**= 1 .500 .500 .3962 m = 0 The Standard Normal Table:P(z > 1.26) Standard Normal Distribution P(z > 1.26) = .5000 – .3962 = .1038 z 1.26**s**= 1 .4973 .4772 m = 0 The Standard Normal Table:P(–2.78 z –2.00) Standard Normal Distribution P(–2.78 ≤ z ≤ –2.00) = .4973 – .4772 = .0201 –2.78 –2.00 z**s**= 1 .4834 .5000 m = 0 The Standard Normal Table:P(z > –2.13) Standard Normal Distribution P(z> –2.13) = .4834 + .5000 = .9834 z –2.13**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.**Each distribution would require its own table.**f(x) That’s an infinite number of tables! x Non-standard Normal Distribution Normal distributions differ by mean & standard deviation.**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 formula has a standard normal distribution. The value z describes the number of standard deviations between x and µ.**Standard Normal Distribution** = 1 m = 0 z One table! Standardize theNormal Distribution Normal Distribution s x**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 z Show 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.**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).**Normal Distribution**Standard Normal Distribution = 10 = 1 .0478 = 5 6.2 = 0 .12 z x Non-standard Normal μ = 5, σ = 10: P(5 < x < 6.2)**Normal Distribution**Standardized Normal Distribution .0478 = 5 = 0 z 3.8 x -.12 Non-standard Normal μ = 5, σ = 10: P(3.8 x 5) = 10 = 1**Normal Distribution**Standard Normal Distribution .1664 .0832 .0832 z 2.9 5 7.1 x -.21 0 .21 Non-standard Normal μ = 5, σ = 10: P(2.9 x 7.1) = 10 = 1**Standard Normal Distribution**Normal Distribution = 1 = 10 .5000 .3821 .1179 = 0 .30 z = 5 8 x Non-standard Normal μ = 5, σ = 10: P(x 8)**Normal Distribution**Standard Normal Distribution = 10 = 1 .1179 .0347 .0832 z x 7.1 8 .21 .30 Non-standard Normal μ = 5, σ = 10: P(7.1 X 8) = 5 = 0**Normal Distribution Thinking Challenge**You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000hours and = 200hours. What’s the probability that a bulb will last A. between 2000 and 2400 hours? B. less than 1470hours?**Normal Distribution**Standard Normal Distribution = 200 = 1 .4772 z x = 0 2.0 = 2000 2400 Solution* P(2000 x 2400)**Normal Distribution**Standard Normal Distribution = 200 = 1 .5000 .4960 .0040 z x –2.65 = 0 1470 = 2000 Solution* P(x 1470)**What is z, given P(z) = .1217?**Standard Normal Probability Table (Portion) .01 = 1 Z .00 0.2 .1217 0.0 • .0000 .0040 .0080 0.1 .0398 .0438 .0478 0.2 .0793 .0832 .0871 = 0 • ? z 0.3 .1217 .1179 .1255 Finding z-Values for Known Probabilities .31**Standard Normal Distribution**= 10 = 1 .1217 .1217 • ? = 0 .31 z = 5 • x Finding x Values for Known Probabilities Normal Distribution 8.1**Using a TI: invNorm(p, μ, σ) returns the 100pth percentile**of N(μ, σ).**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.**Solution**a) 3.7, 2.2, 2.95, 1.45 b) 1.9 • P(z > 0.84)=0.20 P(z > 1.96)=0.025 So, for cum laude: 0.84=(x-2.7)/0.5x= 3.12; for summa cum laude: 1.96=(x-2.7)/0.5 x= 3.68**4.7**Descriptive Methods forAssessing Normality