Intro to Probabiltiy theory notes for Elements Class.pptx

# Approximately a normal distribution the probability

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• pauloffei201440
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approximately a normal distribution the probability that a random selected observation falls within the intervals ( - , + ), ( - 2 , +2 ), and ( - 3 , + 3 ), is approximately 0.6826, 0.9544 and 0.9973, respectively.

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91 Normal Distribution Application Example:1 As a part of a study of Alzeheimer’s disease, reported data that are compatible with the hypothesis that brain weights of victims of the disease are normally distributed. From the reported data, we may compute a mean of 1076.80 grams and a standard deviation of 105.76 grams. If we assume that these results are applicable to all victims of Alzeheimer’s disease, find the probability that a randomly selected victim of the disease will have a brain that weighs less than 800 grams. 800 µ = 1076.80 σ = 105.76
92 Solution: R.V. X =‘Brain weights’, follows a Normal distribution with µ=1076.80 and σ = 105.76): z = (X – μ)/ ; z < (800 – 1076.8)/ 105.76= -2.62 = We have to find out P (x < 800) i.e P (z < -2.62). This is the area bounded by the curve, x axis and to the left of the perpendicular drawn at z = -2.62. Thus from the standard normal table this prob., p= .0044. The probability is .0044 that a randomly selected patient will have a brain weight of less than 800 grams. - 2.62 0 σ = 1

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93 Example: 2 Suppose it is known that the heights of a certain population of individuals are approximately normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What is the probability that a person picked at random from this group will be between 65 and 74 inches tall. Solution: X (height) follows a Normal distribution with mean (µ)= 70 inches and σ 3 inches.
94 z = 65-70 3 = -1.76 - 1.67 1.33 0 σ = 1 65 74 70 σ = 3 For x= 65, Z we have For x= 74, Z we have z = 74-70 3 = 1.33 P ( 65 ≤ x ≤ 74) = P ( 65-70 3 ≤ z ≤ 74-70 3 ) = P (- 1.76 ≤ z ≤ 1.33) = P ( -∞ ≤ z ≤ 1.33) – P (-∞ ≤ z ≤ -1.67) = .9082 - .0475 = .8607 The required probability asked for in our original question, is 0.8607.

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95 Example: 3 In a population of 10,000 of the people described in previous example how many would you expect to be 6 feet 5 inches tall or taller? Solution: we first find the probability that one person selected at random from the population would be 6 feet 5 inches tall or taller. That is, P ( x ≥ 77 ) = P ( z ≥ 77-70 3 ) = P ( z ≥ 2.33 ) = 1 - .9901= .0099 Out of 10,000 people we would expect 10,000 (.0099) = 99 to be 6 feet 5 inches (77 inches tall or taller).
96 Exercise: 1.Given the standard normal distribution, find the area under the curve, above the z-axis between z=-∞ and z = 2. 2. What is the probability that a z picked at random from the population of z’s will have a value between -2.55 and + 2.55? 3. What proportion of z values are between -2.74 and 1.53? 4. Given the standard normal distribution, find P ( z ≥ 2.71) 5.Given the standard normal distribution, find P(.84 ≤ z ≤ 2.45).

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97 The Normal Distribution This was first introduced in 1750 by the mathematician De Moivre, through probability theory. Let us consider a simple example of his work. Suppose we toss 10 coins simultaneously, then the expected probability of obtaining 0, 1, 2, 3, … 10 heads is given by the Binomial Expansion of (½ + ½) 10 .
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