Intro to Probabiltiy theory notes for Elements Class.pptx

Approximately a normal distribution the probability

Info icon This preview shows pages 90–98. Sign up to view the full content.

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.
Image of page 90

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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
Image of page 91
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
Image of page 92

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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.
Image of page 93
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.
Image of page 94

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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).
Image of page 95
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).
Image of page 96

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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 .
Image of page 97
Image of page 98
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern