OPIM 303, Managerial Statistics H Guy Williams, 2006

Transcription

1 OPIM 303 Lecture 6 Page 1

2 The height of the uniform distribution is given by 1 b a Being a Continuous distribution the probability of an exact event is zero: 2 0 There is an infinite number of points in any continuous interval. For this reason we examine probabilities in a range including equal to or less than. Interval 2 P(160 lbs) = 0 P(159 weight 160) = 1.1% Whole Range 180 Keep in mind we are not counting, we are looking at continuous spans. The probability of a given interval is defined as the area under the curve. OPIM 303 Lecture 6 Page 2

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4 This tells us it is very unlikely to find an observation more than 3 standard deviations from the mean. Many things in nature, measured and a histogram created from the data, will exhibit a Normal or bell shape. OPIM 303 Lecture 6 Page 4

5 Sections 6.1, 6.2, 6.3 Bell shape implies that most values will occur around the mean. Mean = Median due to symmetrical shape. The shape of the distribution makes it extremely unlikely that very large or very small values will occur. Continuous distributions are measured whereas discrete distributions are counted. The probability that various values occur within certain ranges or intervals can be calculated. The EXACT probability of a particular value from a continuous distribution is zero. The probabilities, or portion of area under the entire curve, must add to 1. The normal distribution provides the basis for classical statistical inference because of its relationship to the central limit theorem. OPIM 303 Lecture 6 Page 5

6 Probability of the region a,b would be the integral of the f(x) eq above, but there is no known direct solution. Use the standardized table or Excel. Also, the area under the curve, in total, from positive to negative infinity, must equal 1. OPIM 303 Lecture 6 Page 6

7 The area under the curve, from positive to negative infinity, must equal 1. OPIM 303 Lecture 6 Page 7

8 Remember: the Standardized Normal Tables give the CUMULATIVE distribution. Second decimal digit irst decimal igit By standardizing the data using the below transformation formula probabilities for any data can be extracted from only one table. Any normal random variable X can be transformed to a standardized normal random variable Z. Ex. Use the standardized distribution table to find the probability that a value is greater than Z = gives cumulative distribution value of or 3.22%. Area under the curve is 1 so the blue area is given by = 96.78%. The Z value converted from X is known as the Z-Value. Note that the table only gives the area to the LEFT of the X value. OPIM 303 Lecture 6 Page 8

9 The Standardized Distribution always has mean of 0 and standard deviation of 1. We find the probability density function for the standardized normal distribution by substituting mean of 0 and standard deviation of 1 into the normal probability density function: Look for interval probability questions which involve a multiple of the standard deviation. P(-5 <= X <= 15) where the standard deviation is 10. The question is asking What is the probability of +/- 2? We know from the rules of normal data that the answer is 95.44%. OPIM 303 Lecture 6 Page 9

10 Always draw the bell curve of the problem for insight on how to solve. P(X)=54.78% Z =.12 In this example mean of 5 is transferred to mean of 0 and sigma of 10 to sigma of 1. Will have interval probabilities. Area =.4168 Area =.5832 In this case the sought shaded region is symmetrical about the mean. The mean is at.5 of the area. So could solve by 2*( ) = 16.64% OPIM 303 Lecture 6 Page 10

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12 The Excel function to solve these problems would be: = normdist(x, mean, std dev, cumulative) We will always use cumulative = True. (cumulative = False returns the f(x) value, density function) OPIM 303 Lecture 6 Page 12

13 In this example we have a distribution of peoples weights with mean 180 and standard deviation 20. Now we are asked what is the weight such that 99.9% of males will weigh less? (green) We can solve this by looking up in the body of the standardized table the percentage we want (99.9%), extract the Z value, and convert the Z to the X value we are seeking (?). From the standardized table 99.9% gives Z = convert to x 3.09* The x conversion procedure is called unstandardizing. OPIM 303 Lecture 6 Page 13

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15 Ex 6.7; OPIM 303 Lecture 6 Page 15

16 Therefore, 95% of the download times will be between 3.08 and seconds. OPIM 303 Lecture 6 Page 16

17 This returns the Standardized value. Mean is 0, std dev is 1. We will always use True = 1 P(weight < 185) = 1 normdist(185,180,20,t) (the above <= 165 should be >= 165) P(X <= 185) P(X <= 165) = =.38 OPIM 303 Lecture 6 Page 17

18 Normal / Standardized Examples: (These solutions come from the standard normal curve cumulative table.) Ex 6.1; P(X > 9) = 1 P(X < 9) = = Mu = 7, sigma = 2 (On Campus website) Ex. 6.2; Mu = 7, sigma = 2 9) = P(X < 9) P(7 < X) = = P(7 < X< Ex. 6.3; P(X < 7 OR X> 9) = = Mu = 7, sigma = 2 (This is the complement of Ex. 6.2 above.)using P(X < 9) = = Ex 6.4; P(X < 9) P(X < 5) = = Mu = 7, sigma = 2 P(5 < X < 9) = Ex 6.4a; P(3 < X 11) = P(X < 11) P(X < OPIM 303 Lecture 6 Page 18

19 3) = = Mu = 7, sigma = 2 Ex 6.4b; P(1 < X < 13) = P(X < 13) P(X < 1) = = Mu = 7, sigma = 2 Ex 6.5; P(X < 3.5) = P(X < 3.5) = Mu = 7, sigma = 2 OPIM 303 Lecture 6 Page 19

20 Ex 6.6; complete? How much time, in seconds, will elapse before 10% of the downloads are [In this case the area under the curve is the percentage ] OPIM 303 Lecture 6 Page 20

21 This returns the Standardized value. Mean is 0, std dev is 1. Always gives the probability to the left. Using the previous weight problem as an example: = norminv(.999, 180, 20) = GIVEN THE PROBABILITY, MEAN, AND STANDARD DEVIATION THE NORMINV FUNCTION RETURNS THE X VALUE AT WHICH THIS PROBABILITY OCCURES. OPIM 303 Lecture 6 Page 21

22 Examples from text. 6.7 pg 239 a) P( X <= 25 ) = normdist(25, 36.16, 10, True) =.1322 = 13.22% or solve using the standardized table z look up in table to find P( X <= 25 ) = 13.35% 10 b) P( X >= 50 ) = 1 P( X <= 50 ) = 1 normdist(50, 36.16, 10, True) = =8.3% z look up in table to find P( X <= 50 ) = Solve for 1 P( X <= 50 ) = = 8.38% c) P( X >= 75 ) = 1 P( X <= 75 ) = 1 normdist(75, 36.16, 10, True) = d) P( 30 <= X <= 40 ) = P( X <= 40 ) - P( X <= 30 ) = normdist(40, 36.16, 10, True) - normdist(30, 36.16, 10, True) = = 38% d) 99% spend less than (?) amount? X = norminv(99, 36, 10) = $59.42 x or from table z 2.33 solve for x $ f) 80% spend more than (?) amount? X = norminv(1-.80, 36.16, 10) = OPIM 303 Lecture 6 Page 22

23 (May have to trim outliers.) OPIM 303 Lecture 6 Page 23