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  ( )

Introduction to Probability & Statistics Inverse Functions

Inverse Functions Actually, we’ve already done this with the normal distribution.

Inverse Normal Actually, we’ve already done this with the normal distribution. x 3.0 3.38 0.1 s m - = X Z x = m + sz = 3.0 + 0.3 x 1.282 = 3.3846

Inverse Exponential  f x e ( )   F a X ( ) Pr{ }     e dx   e 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 

Inverse Exponential F(x) x X e l - F ( x ) = 1 -

e Inverse Exponential F(x) F(a) a Suppose we wish to find a such that the probability of a failure is limited to 0.1. 0.1 = 1 - ln(0.9) = -la a e l - a = - ln(0.9)/l

Inverse Exponential a = - ln(0.9)/l = - (-2.3026)/0.005 = 21.07 hrs. Suppose a car battery is governed by an exponential distribution with l = 0.005. We wish to determine a warranty period such that the probability of a failure is limited to 0.1. a = - ln(0.9)/l = - (-2.3026)/0.005 = 21.07 hrs. F(x) F(a) x a

South Dakota School of Mines & Technology Module 3 Industrial Engineering

Introduction to Probability & Statistics Expectations

Expectations    Mean:   E X xdF x [ ] ( )   xp x discrete ( ) ,    E X xdF x [ ] ( )    xp x discrete ( ) ,      xf x dx continuous ( ) ,

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6  = E[X] = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 3.5

Expected Life   E [ x]  x  e dx   x e dx    ( ) 2  1  For a producted governed by an exponential life distribution, the expected life of the product is given by  2.0 E [ x]   x  e   x dx 1.8 1.6 1.4 1.2 f (x t )   e   x 1.0      x e dx 2 1 Density 0.8 0.6 0.4 0.2 0.0 X 0.5 1 1.5 2 2.5 3    ( ) 2 1/  1 

Variance      ( ) x dF     E x [( ) ] =     ( ) x p    2    ( ) x dF   2   E x [( ) ] =   2    ( ) x p   2      ( ) x f dx

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 2 = E[(X-)2] = (1-3.5)2(1/6) + (2-3.5)2(1/6) + (3-3.5)2(1/6) + (4-3.5)2(1/6) + (5-3.5)2(1/6) + (6-3.5)2(1/6) = 2.92

Property        ( ) x f dx    ( ) x f dx     x f dx xf ( 2      ( ) x f dx     ( ) x f dx 2      x f dx xf 2 ( ) 

Property        ( ) x f dx    ( ) x f dx     x f dx xf ( 2      ( ) x f dx     ( ) x f dx 2      x f dx xf 2 ( )     E X [ ] 2    E X [ ] 2 

Example Consider the discrete uniform die example: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 2 = E[X2] - 2 = 12(1/6) + 22(1/6) + 32(1/6) + 42(1/6) + 52(1/6) + 62(1/6) - 3.52 = 91/6 - 3.52 = 2.92

Exponential Example       E X [ ]   1   x e dx ( )   x e For a producted governed by an exponential life distribution, the expected life of the product is given by   2   E X [ ] 2.0 1.8   2 1    x e dx ( ) 1.6 1.4 1.2 f (x t )   e   x 1.0 Density 0.8      x e dx 3 1 0.6  1 2  0.4 0.2 0.0 X 0.5 1 1.5 2 2.5 3    ( ) 3 1/ 1 2   1 2  =

Properties of Expectations 1. E[c] = c 2. E[aX + b] = aE[X] + b 3. 2(ax + b) = a22 4. E[g(x)] = g(x) E[g(x)] X (x-)2 e-tx g x dF ( ) 

Property Derviation Prove the property: E[ax+b] = aE[x] + b

Class Problem Total monthly production costs for a casting foundry are given by TC = $100,000 + $50X where X is the number of castings made during a particular month. Past data indicates that X is a random variable which is governed by the normal distribution with mean 10,000 and variance 500. What is the distribution governing Total Cost?

Class Problem Soln: TC = 100,000 + 50X is a linear transformation on a normal TC ~ Normal(mTC, s2TC)

Class Problem Using property E[ax+b] = aE[x]+b mTC = E[100,000 + 50X] = 100,000 + 50(10,000) = 600,000

Class Problem Using property s2(ax+b) = a2s2(x) s2TC = s2(100,000 + 50X) = 502 s2(X) = 502 (500) = 1,250,000

Class Problem TC = 100,000 + 50 X but, X ~ N(100,000 , 500) TC ~ N(600,000 , 1,250,000) ~ N(600000 , 1118)

South Dakota School of Mines & Technology Module 3 Industrial Engineering

Introduction to Probability & Statistics Joint Distributions

Discrete Bivariate Suppose we track placement data for 1,000 recent graduates at a local university. Students are tracked by undergraduate major and are placed in one of three categories.

Discrete Bivariate If we divide the number in each cell, by 1,000 we then have defined a discrete bivariate distribution.

Discrete Bivariate

Discrete Bivariate p x y X Y , ( ) Pr{ } = p x y , ( ) . = å 1

Conditional Distribution Suppose that we wish to look at the distribution of students going to graduate school.

Conditional Distribution We are now placing a condition on the sample space that we only want to look at students in graduate school. Since the total probability of students in graduate school is only 0.21, we must renormalize the conditional distribution of students in graduate school.

Conditional Distribution

Conditional Distribution

Condition on Field Placement

Conditioning on Major Suppose, we wish to condition by major (y-axis) and look at placement for engineers only.

Conditional; Engineering

Conditional

Marginal Distribution The marginal distribution for placement by major is just the sum of joint probabilities for each major.

Marginal Distribution

Marginal Distribution We can find the marginal distribution by category in a similar fashion.

Marginal Distribution

Marginal

Bivariate Uniform  f x y dxdy ( , ) Pr{X< a, Y< b} = b a fXY(x,y) x y b a Pr{X< a, Y< b} = f x y dxdy XY b a  ( , )

Bivariate Uniform   ( ) f x y dydx   f x dx y dy ( ) fXY(x,y) x y b a For X,Y independent, fXY(x,y) = fX(x)fY(y) ( ) f x y dydx X b a Y   Pr{X< a, Y< b}   f x dx y dy X a Y b ( )

Conditional Distribution fXY(x,y) x y Lets take a slice out of fXY(x,y) a particular value of x. The area under the response surface fXY(x,y) is just the conditional probability of y for a specific value of x. Or f(y|x) = fXY(x,y|x)

Marginal Distribution fXY(x,y) x y If we look at this for all values of x, we get f ( y )   f ( x , y ) dx Y x XY

Marginal Distribution fXY(x,y) x y Similarly, looking at the slice the other way ) ( f x y dy X XY ,  

South Dakota School of Mines & Technology Module 3 Industrial Engineering

Introduction to Probability & Statistics Joint Expectations

Properties of Expectations Recall that: 1. E[c] = c 2. E[aX + b] = aE[X] + b 3. 2(ax + b) = a22 4. E[g(x)] = g x dF ( ) 

Joint Expectation Let Z = X + Y E[Z] = E[X] + E[Y]  ( ) cov( , z x y 2 ( ) cov( , z x y  

Derivation   E Z zf x y dxdy [ ] ( , )    ( ) , x y f dxdy Let Z = X + Y E Z zf x y dxdy XY [ ] ( , )      ( ) , x y f dxdy XY

Derivation E[Z]      E Z zf x y dxdy [ ] ( , )    ( ) , x y f Let Z = X + Y E Z zf x y dxdy XY [ ] ( , )      ( ) , x y f dxdy XY    y XY x xf dxdy yf ( , )    x f y dydx dxdy XY ( , )    xf x dx yf y dy X Y ( ) = E[X] + E[Y]

Derivation of 2(z)   ( ) , [ ] z f x y dxdy E Z   Miracle 13 occurs = 2(x) + 2(y)

In General In general, for Z = the sum of n independent variates, Z = X1 + X2 + X3 + . . . + Xn E Z X i n [ ]   1  2 ( ) z x

Class Problem Suppose n customers enter a store. The ith customer spends some Xi amount. Past data indicates that Xi is a uniform random variable with mean $100 and standard deviation $5. Determine the mean and variance for the total revenue for 5 customers.

Class Problem Total Revenue is given by TR = X1 + X2 + X3 + X4 + X5 Using the property E[Z] = E[X] + E[Y], E[TR] = E[X1] + E[X2] + E[X3] + E[X4] + E[X5] = 100 + 100 + 100 + 100 + 100 = $500

Class Problem TR = X1 + X2 + X3 + X4 + X5 Use the property s2(Z) = s2(X) + s2(Y). Assuming Xi, i = 1, 2, 3, 4, 5 all independent s2(TR) = s2(X1) + s2(X2) + s2(X3) + s2(X4) + s2(X5) = 52 + 52 + 52 + 52 + 52 = 125

Class Problem TR = X1 + X2 + X3 + X4 + X5 Xi , independent with mean 100 and standard deviation 5 E[TR] = $500 s2(TR) = 125

South Dakota School of Mines & Technology Module 3 Industrial Engineering

Introduction to Probability & Statistics The Central Limit Theorem

The Sample Mean    x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6   Suppose, for our die example, we wish to compute the mean from the throw of 2 dice: x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6    xp x ( ) . 3 5 Estimate  by computing the average of two throws:   X   1 2

Joint Distributions x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 X1 X2 X

Joint Distributions x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 X1 X X2

Distribution of X x 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 Distribution of X Distribution of X 0.00 0.05 0.10 0.15 0.20 1 2 3 4 5 6 0.00 0.05 0.10 0.15 0.20 1 2 3 4 5 6 7 8 9 10 11

Distribution of X n = 2 n = 10 n = 15 0.0 0.1 0.2 0.3 0.4 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 0.00 0.05 0.10 0.15 0.20 1 2 3 4 5 6 7 8 9 10 11 0.0 0.1 0.2 0.3 0.4 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 n = 15

Expected Value of X   E X n [ ] .           1 n E X [ ] . 2     1 2 n E X [ ] .

Expected Value of X       E X n [ ] .           1 n E 2     1 2 n E X [ ] .     1 2 n  .    1 n 

Variance of X  ( ) . x X n                  n X . 2 1 ( ) . x X n                  2 1 n X .

Variance of X    ( ) . x X n                  n X 2 1 ( ) . x X n                  2 1 n X .           1 2 n X  ( ) .        1 2 n ( )    2 n

Distribution of x Recall that x is a function of random variables, so it also is a random variable with its own distribution. By the central limit theorem, we know that where,

Example Suppose that breakeven analysis indicates we must have average daily revenues of $500. A random sample of 10 days yields an average of only $450 dollars. What is the probability we will not breakeven this year?

Example P not breakeven x { } = < m 500 450 = - > P x { } m 500 Suppose that breakeven analysis indicates we must have average daily revenues of $500. A random sample of 10 days yields an average of only $450 dollars. What is the probability we will not breakeven this year? P not breakeven x { } = < m 500 450 = - > P x { } m 500 450

Example P not breakeven { } = - > x m 500 450 Recall that Using the standard normal transformation

Example P not breakeven { }

Example   In order to solve this problem, we need to know the true but unknown standard deviation . Let us assume we have enough past data that a reasonable estimate is s = 25. P Z          50 25 10      P Z 1 58 . Pr{not breakeven} = = 0.943