1. Data Analysis Statistical Measures Industrial Engineering

2. Aside: Mean, Variance Mean: Variance:     xp x discrete ( ) ,  2    ( ) x p

3. 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

4. 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

5. å Binomial Mean  p ÷ ø ö ç è æ ) 1 ( )! !   xp x ( )
= 1p(1) + 2p(2) + 3p(3) np(n) x n p - = ÷ ø ö ç è æ å ) 1 ( )! !

6. å Binomial Mean  p ÷ ø ö ç è æ ) 1 ( )! !   xp x ( )
= 1p(1) + 2p(2) + 3p(3) np(n) x n p - = ÷ ø ö ç è æ å ) 1 ( )! !
Miracle 1 occurs
= np

7. Binomial Measures Mean: Variance:     xp ( x ) = np     ( ) x2    ( ) x p
= np(1-p)

8. Binomial Distribution
0.0
0.1
0.2
0.3
0.4
0.5 1 2 3 4 5 x
P(x)
0.0
0.1
0.2
0.3
0.4
0.5 1 2 3 4 5 x
P(x)
n=5, p=.3
n=8, p=.5
n=20, p=.5
n=4, p=.8
0.0
0.1
0.2
0.3
0.4
0.5 2 4 x
P(x)
0.0
0.1
0.2
0.3
0.4
0.5 1 2 3 4 5 6 7 8
P(x) x

9. Measures of Centrality
Mean
Median
Mode

10. Measures of Centrality
Mean    xp x
discrete ( ) ,      xf x dx
continuous ( ) ,
Sample Mean å = n i x X 1

11. Measures of Centrality
Exercise: Compute the sample mean for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

12. Measures of Centrality
Exercise: Compute the sample mean for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 å = n i x X 1 10 9 . 3 5 1 2 8 7 4 + =
= 3.06

13. Measures of Centrality
Failure Data X 1 . 19 =

14. Measures of Centrality
Median
Compute the median for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 . 3 2 = + X

15. Measures of Centrality
Mode
Class mark of most frequently occurring interval
For Failure data, mode = class mark first interval ( . 5 = X

16. Measures of Centrality
Measure Student Gpa Failure Data
Mean
Median
Mode

17. Measures of Dispersion
Range
Sample Variance

18. Measures of Dispersion
Range
Compute the range for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9

19. Measures of Dispersion
Range
Compute the range for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9
Min = 2.4 Max = 3.9
Range = = 1.5

20. Measures of Dispersion
Variance   2    ( ) x p   2      ( ) x f dx
Sample variance x 1 2 - = å n s i

21. Measures of Dispersion
Exercise: Compute the sample variance for the student Gpa data
2.4, 2.7, 2.8, 2.9, 3.0, 3.0, 3.1, 3.3, 3.5, 3.9 x 1 2 - = å n s i

22. Measures of Dispersion
Sample Variance x 1 2 - = å n s i ( )
185 . 1 10 06 3 95 2 = -

23. Measures of Dispersion
Exercise: Compute the variance for failure time data
s2 =

24. An Aside
For Failure Time data, we now have three measures for the data
s2 = X 1 . 19 =

25. An Aside
For Failure Time data, we now have three measures for the data
Exponential ??
s2 = X 1 . 19 =

26. An Aside X 1 . 19 = Recall that for the exponential distribution
m = 1/l
s2 = 1/l2
If E[ X ] = m and E [s2 ] = s2, then
1/l = 19.1 or
1/l2 = X 1 . 19 =
s2 = l
0524 . ˆ = l
0575 . ˆ =

28. Introduction to Probability & Statistics The Central Limit Theorem

29. 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

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

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

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

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

34. 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

35. 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

36. Distribution of X X lim Normal  n = 2 n = 10 n = 15 n   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 X
lim n
Normal   

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

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

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

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

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

42. 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

43. 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,

44. 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?

45. 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

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

47. Example P
not
breakeven { }

48. 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