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

Your 4 th homework is assigned. It is due on 12 th 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.

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

  2. 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

  3. 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)

  4. 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?

  5. 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)

  6. 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

  7. 4.4 Hypergeometric Distribution

  8. 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.

  9. Hypergeometric Probability Distribution Function [x = Maximum [0, n – (N – r), …, Minimum (r, n)] where . . .

  10. 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

  11. a. Hypergeometric b. Binomial

  12. 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

  13. 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)

  14. 4.5 Probability Distributions for Continuous Random Variables

  15. 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:

  16. 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.

  17. Continuous Probability Density Function P(a<x<b)=P(a≤x≤b) sinceP(x=a)=P(x=b)=0

  18. 4.6 The Normal Distribution

  19. 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

  20. Mean Median Mode Normal Distribution • ‘Bell-shaped’ & symmetrical • Mean, median, mode are equal • f • ( • x • ) • x

  21. 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

  22. Effect of Varying Parameters ( & )

  23. Normal Distribution Probability • Probability is area under curve! f(x) x c d

  24. 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.

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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.

  31. 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.

  32. 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.

  33. Standard Normal Distribution  = 1 m = 0 z One table! Standardize theNormal Distribution Normal Distribution s  x

  34. 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.

  35. 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).

  36. 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)

  37. 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

  38. 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

  39. 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)

  40. 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

  41. 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?

  42. Normal Distribution Standard Normal Distribution = 200 = 1 .4772 z x = 0 2.0 = 2000 2400 Solution* P(2000  x 2400)

  43. Normal Distribution Standard Normal Distribution = 200 = 1 .5000 .4960 .0040 z x –2.65 = 0 1470 = 2000 Solution* P(x  1470)

  44. 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

  45. Standard Normal Distribution = 10 = 1 .1217 .1217 • ? = 0 .31 z = 5 • x Finding x Values for Known Probabilities Normal Distribution 8.1

  46. Using a TI: invNorm(p, μ, σ) returns the 100pth percentile of N(μ, σ).

  47. 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.

  48. 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)/0.5x= 3.12; for summa cum laude: 1.96=(x-2.7)/0.5 x= 3.68

  49. 4.7 Descriptive Methods forAssessing Normality

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