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Lecture 9

Lecture 9 (week 10)

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Department
Quantitative Methods
Course
QMS 102
Professor
Clare Chua
Semester
Fall

Description
11/9/2010 Quiz 2 covered materials in Chapter 5 Chapter 6 Normal Probability Chap pter 7 Distribution Chapter 6 Continuous Probability Distribution Normal Probability Distribution The focus is on a continuous, bell-shaped diisttrbuuttionn Discuss the properties Normal Distribution Uniform Distribution Normal Distribution is symmetricaThe uniform distribution is symmetrical shaped, implying that most valuesand therefore the mean = median. cluster around the mean, equal to median. Properties Normal Probability Distribution 1. The shape is determines by mean, 2 and standard deviation, 8 2. The highest point on the normal curve is median and the mode of the distribution. 3. The normal distribution is symmetrical: the curves shape to the left of the mean is the mirror image of its shape to the right of the mean. Thus, mean = median 4.Theaiioffhenomaalure exend onfniittyy in both directions and never touch the When a mathematical horizontal axis. Thus, it has an infinite range. expression is available to - B < X < +B represent a continuous variable, 5. However, the tails get close enough to the theprobability that various the total area under the normal curve equalst within certain ranges 1. or intervals can be calculated. 6. Since the normal curve is symmetrical, the area under the normal curve to the right of the mean (2) equals the area under the normal curve to the left of the mean, and each of these areas equals 0.5. 1 www.notesolution.com 11/9/2010 Normal Probability Distribution Normal Probability Distribution However, the EXACT probability of a particular value from a Normal Distribution is ZERO. a b e.g. P(X=140) = 0 What information do we need to calculate P(a x b)? The normal probability distribution 2 is defined by the equation - x- 2=95 140 f(x)= 1 e 2 2 or all x on the realline Normal Probability Distribution Normal Probability Distribution a b a b What information do we need to calculate P(a x b)? Tocaculaete Normal Pa x b)):: Random X= measured Variable e.g. X= a persons weight 2 and 8 Parameters 2 and 8 Suppose that the weights of adults are normally distributed Presentation of Solution with a mean of 170.0lbs and a standard deviation of 25.0 lbs. Follow the normal template page 269 What is the probability that a person weighs from 150 to 190 lbs? X = ____________________ 8 = 25.0 Note: For example: X= number of .... WRONG Calculating normal probability = Findingarea under a portion of the normal curve = Normal = 150 2=170190 X= apersons weight Refer to the calculator Lesson 5: P ( ) = P ( X symbol # ) = Ncd ( L, U, , ) = 0.xxxx Page 169 (words) ( , , ) 4 decimals 2 = 170.0 lbs 8 = 25.0 lbs normal Must draw a diagram P( 150 x 190) = Ncd (150,190, 25, 170) Lower upper 8 (Mean, , value, shading) 2 2 www.notesolution.com11/9/2010 Normal Probability Exercises Inverse Normal Page 269 to 270 Q6.1 to Q6.6 Inverse Normal Probability The weights of adults are normally distributed with a mean of 170.0 lbs Normal Probability Distribution and a standard deviation of 25.0 lbs. 1) What is the maximum weight of the lightest 30% of the population? Present your solution as follows (use the template) Given the value X, find The Inverse Normal X = persons weight the probability (or Probability means: 2 = 170.0 8 = 25.0 Normal the area under the Given the probability Area= 8 = 25.0 curve)) P(X @? ))0.30 0.3 (area), find the x Area value?
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