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Table 5 Results of the distribution fitting of sub-processes

From: Analysis of processes and costs of alternative packaging options of sterile goods in hospitals – a case study in two German hospitals

Process

Distribution

Parameters

C01S

Wakeby

α = 10.1959

β = 0.0440

γ = 0

δ = 0

ξ = 0.8446

C02S

Johnson SB*

γ = 4.6286

δ = 2.4426

λ = 114.0524

ξ = −4.4557

 

C03S

Log-Pearson 3

α = 13.6700

β = − 0.1317

γ = 3.2264

  

W01S

Johnson SB*°

γ = 2.7028

δ = 1.4808

λ = 107.0346

ξ = −1.0275

 

W02S

Beta*°

α1 = 0.7476

α2 = 1.1590

a = 0 (Fixed)

b = 173.0000

 

W03S

Gen. Logistic*°

k = − 0.0314

σ = 5.0845

μ = 31.9906

  

C04S

Pareto 2*

α = 0.5536

β = 0.0000

   

C05S

Beta

α1 = 2.4664

α2 = 4.774

a = 0 (Fixed)

b = 35.3533

 

C06S

Wakeby

α = 820.7058

β = 70.7089

γ = 11.5924

δ = −0.3908

ξ = −5.0945

C07S

Burr*

k = 0.6447

α = 4.0719

β = 9.2797

  

W04S

Gen. Extreme Value*°

k = − 0.0385

σ = 10.6179

μ = 22.5843

  

C08S

Pearson 6

α1 = 41.7009

α2 = 4.2773

β = 0.3206

  

C09S

Log-Pearson 3*°

α = 63.1624

β = − 0.0679

γ = 7.7562

  

W05S

Dagum*°

k = 1.1123

α = 3.8903

β = 35.8406

  

C10S

Gen. Extreme Value*

k = 0.1449

σ = 3.8764

μ = 9.3659

  

C11S

Inv. Gaussian*°

λ = 110.0178

μ = 34.5505

   

W06S

Gen. Extreme Value*°

k = − 0.0255

σ = 28.8306

μ = 78.5859

  

W07S

Dagum*°

k = 0.6310

α = 3.4634

β = 17.3266

  

C12O

Log-Logistic*

α = 3.0702

β = 4.1318

   

C13O

Log-Logistic*

α = 4.2215

β = 5.9365

   

W08O

Burr*°

k = 0.6565

α = 4.4354

β = 14.3868

  

C14O

Gen. Extreme Value

k = 0.3659

σ = 1.6206

μ = 3.6036

  

W09O

Pearson 6*

α1 = 11.9523

α2 = 3.4768

β = 0.9310

  

C15O

Burr*

k = 0.5800

α = 4.2272

β = 3.4437

  

C16O

Log-Logistic*

α = 3.3153

β = 4.01482

   

C17O

Johnson SB*°

γ = 1.5759

δ = 0.7162

λ = 81.6481

ξ = 1.4134

 

W10O

Wakeby*°

α = 20.6865

β = 7.7626

γ = 4.8861

δ = −0.2130

ξ = 1.6206

W11O

Wakeby*°

α = 23.2417

β = 0.1569

γ = 0

δ = 0

ξ = 0.0504

  1. *significant at a = 0,01 (AD-test)° significant at a = 0,01 (KS-test)