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# L so normality may very well be an excellent

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l So, normality may very well be an excellent approximation 8

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9 Standard normal l Recall Z scores l Z = ( X - μ )/ σ l In general this standardization yields a rv with zero l How are X related? Z = a + bX (where a = -( / ) and b = 1/ ) l So Z is a linear function (linear transformation) of X
Standard normal… l An important theorem tells us the following l Linear combinations of normal rv’s are also normal l When X ~ N ( μ , σ 2), Z is called a standard normal random variable l This result assists calculations of probabilities l Any probability calculation involving X ~ N ( , 2) translates into an equivalent statement using Z ~ N (0,1) 10

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11 Calculating normal probabilities l Normal rv’s are continuous  probabilities need to be calculated as integrals l Suppose time (in minutes) taken to assemble a computer assumed X ~ N (50,100) l What is probability that a computer is assembled in 45 to 60 minutes? l Need tables as no “closed form” solution for these integrals
12 Calculating normal probabilities… l Strategy for calculations l Standardize to yield equivalent probability statement for a standard normal l Use standard normal tables l Be careful as these can come in different forms l “BES” tables provide P (0 < Z < z ) l Keller, P(-∞<Z<z) l What is P (45 < X < 60) in computer assembly example?

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13 Calculating normal probabilities… ) 1 5 . 0 ( 10 50 60 10 50 45 ) 60 45 ( < < - = - < - < - = < < Z P X P X P σ μ l OK but only have tables for P (0 < Z < z ) l Solution: can always manipulate probabilities into this form l Helpful to recall normal distribution is l Symmetric around its mean (0 for standard normal) l Area under pdf equals 1 l P (-0.5 < Z < 1)= P (-0.5 < Z < 0)+ P (0 < Z < 1) & P (-0.5 < Z < 0)= P (0 < Z < 0.5) by symmetry
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Calculating normal probabilities… l Required probabilities are: l P (0 < Z < 1) = .3413 l P (0 < Z < 0.5) = .1915 l P (-0.5 < Z < 1) = .3413+.1915 = .5328 l  Probability of assembly time being between 45 & 60 minutes is .5328 l What is l P ( Z >1) = l P (-1 < Z < 1) = l P ( Z ≥1) = 15
16 Calculating normal percentiles l Tables can be used to solve 2 types of problems l Given a particular z find P (0 < Z < z), or l Given a particular probability A find z A such that P ( Z > zA) = A or P ( Z < zA) = 1 - A l Note that z A is the 100(1-A)th percentile of a standard normal

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l So normality may very well be an excellent approximation...

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