South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering.

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South Dakota School of Mines & Technology Introduction to Probability & Statistics Industrial Engineering

Introduction to Probability & Statistics Continuous Review

Expectations Continuous Review Mean and Variance Cumulative and Inverse Functions Mean and Variance Expected Value properties for one variable Expected Value properties for two variables Central Limit Theorem

Continuous Distribution f(x) A x a b c d 1. f(x) > 0 , all x 2. 3. P(A) = Pr{a < x < b} = 4. Pr{X=a} = f x dx a d ( )   1 b c

Normal Distribution f x e ( )  1 2          65% 95%       1 2           65% 95% 99.7%

Std. Normal Transformation Standard Normal f(z)    Z X     f z e ( )   1 2  N(0,1)

Example Suppose the distribution of student grades for university are approximately normally distributed with a mean of 3.0 and a standard deviation of 0.3. What percentage of students will graduate magna or summa cum laude? x 3.0 3.5

Example Cont. ÷ ø ö ç è æ - ³ = 3 . 5 Pr s m X 3.0 3.5 Pr{magna or summa} = Pr{X > 3.5}} ÷ ø ö ç è æ - ³ = 3 . 5 Pr s m X = Pr(z > 1.67) = 0.5 - 0.4525 = 0.0475

Example Suppose we wish to relax the criteria so that 10% of the student body graduates magna or summa cum laude. x 3.0 ? 0.1

Example Suppose we wish to relax the criteria so that 10% of the student body graduates magna or summa cum laude. x 3.0 ? 0.1 0.1 = Pr{Z > z} z = 1.282

Example x = m + sz = 3.0 + 0.3 x 1.282 = 3.3846 But s m - = X Z 3.0 ? 0.1 But s m - = X Z x = m + sz = 3.0 + 0.3 x 1.282 = 3.3846

Example Let X = lifetime of a machine where the life is governed by the exponential distribution. determine the probability that the machine fails within a given time period a. , x > 0,  > 0 f x e ( )   

Example  f x e ( )   F a X ( ) Pr{ }     e dx   e   1 e a Exponential Life 2.0 f x e ( )    1.8 1.6 1.4 1.2 F a X ( ) Pr{ }   f(x) Density 1.0 0.8 0.6     e dx x a 0.4 0.2 0.0 0.5 1 1.5 2 2.5 3   e x a  a Time to Fail   1 e a 

Complementary  F a X ( ) Pr{ }     e dx  e Exponential Life Suppose we wish to know the probability that the machine will last at least a hrs? 2.0 1.8 1.6 1.4 1.2 f(x) Density 1.0 0.8 0.6 F a X ( ) Pr{ }   0.4 0.2 0.0     e dx x  a 0.5 1 1.5 2 2.5 3 a Time to Fail   e a 

Example Suppose for the same exponential distribution, we know the probability that the machine will last at least a more hrs given that it has already lasted c hrs. a c c+a Pr{X > a + c | X > c} = Pr{X > a + c  X > c} / Pr{X > c} = Pr{X > a + c} / Pr{X > c}    e c a  ( )