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

2. Introduction to Probability & Statistics Continuous Review

3. 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

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

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

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

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

8. 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 . 5 Pr s m X
= Pr(z > 1.67) = =

9. 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

10. 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

11. 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
= x 1.282 =

12. 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 ( )   

13. 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 

14. 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 

15. 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  ( )

17. Introduction to Probability & Statistics Inverse Functions

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

19. 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
= x 1.282 =

20. 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 

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

22. 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

23. 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 = We wish to determine a warranty period such that the probability of a failure is limited to 0.1.
a = - ln(0.9)/l
= - ( )/0.005
= hrs.
F(x)
F(a) x a

25. South Dakota School of Mines & Technology Module 3 Industrial Engineering

26. Introduction to Probability & Statistics Expectations

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

28. Example Consider the discrete uniform die example: x 1 2 3 4 5 6
p(x) /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

29. 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 

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

31. Example Consider the discrete uniform die example: x 1 2 3 4 5 6
p(x) /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

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

33. 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 

34. Example Consider the discrete uniform die example: x 1 2 3 4 5 6
p(x) /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)
= 91/
= 2.92

35. 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  =

36. 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 ( ) 

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

38. 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 What is the distribution governing Total Cost?

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

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

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

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

44. South Dakota School of Mines & Technology Module 3 Industrial Engineering

45. Introduction to Probability & Statistics Joint Distributions

46. 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.

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

48. Discrete Bivariate

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

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

51. 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.

52. Conditional Distribution

53. Conditional Distribution

54. Condition on Field Placement

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

56. Conditional; Engineering

57. Conditional

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

59. Marginal Distribution

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

61. Marginal Distribution

62. Marginal

63. 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  ( , )

64. 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 ( )

65. 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)

66. 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

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

69. South Dakota School of Mines & Technology Module 3 Industrial Engineering

70. Introduction to Probability & Statistics Joint Expectations

71. 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 ( ) 

72. Joint Expectation Let Z = X + Y E[Z] = E[X] + E[Y]  ( ) cov( , z x y2 ( )
cov( , z x y  

73. 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

74. 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]

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

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

77. 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.

78. 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] =
= $500

79. 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) =
= 125

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

82. South Dakota School of Mines & Technology Module 3 Industrial Engineering

83. Introduction to Probability & Statistics The Central Limit Theorem

84. 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
p(x) /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

85. Joint Distributions x
p(x) /6 1/6 1/6 1/6 1/6 1/6 X1 X2 X

86. Joint Distributions x
p(x) /6 1/6 1/6 1/6 1/6 1/6 X1 X X2

87. Distribution of X x
p(x) / /36 3/ /36 5/36 6/ / /36 3/ / /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

88. 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

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

90. Expected Value of X       E X n [ ] .           1 n E2     1 2 n E X [ ] .     1 2 n  .    1 n 

91. Variance of X  ( ) . x X n                  n X . 21 ( ) . x X n                  2 1 n X .

92. Variance of X    ( ) . x X n                  n X2 1 ( ) . x X n                  2 1 n X .           1 2 n X  ( ) .        1 2 n ( )    2 n

93. 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,

94. 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?

95. 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

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

97. Example P
not
breakeven { }

98. 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