South Dakota School of Mines & Technology Data Analysis Industrial Engineering

Data Analysis Statistical Measures Industrial Engineering

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

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

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

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 ( )! !

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

Binomial Measures   Mean: Variance:   xp ( x ) = np     ( ) x 2    ( ) x p = np(1-p)

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

Measures of Centrality Mean Median Mode

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

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

Measures of Centrality Failure Data X 1 . 19 =

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 )

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

Measures of Centrality Measure Student Gpa Failure Data Mean 3.00 19.10 Median 3.04 14.40 Mode --- 5.00 Sample mean X is a blue estimator of true mean m. X m E[ X ] = m u.b.

Measures of Dispersion Range Sample Variance

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 = 3.9 - 2.4 = 1.5

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

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

Measures of Dispersion Exercise: Compute the variance for failure time data s2 = 302.76

An Aside For Failure Time data, we now have three measures for the data Expontial ?? s2 = 302.76 X 1 . 19 =

An Aside 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 = 302.76 X 1 . 19 = s2 = 302.76 l 0524 . ˆ = l 0575 . ˆ =