# Weibull Distribution Probability for α = 2, k = 5 & x = 11 Weibull distribution example problem workout with steps & calculation summary for shape parameter α = 2, scale parameter k = 5 & x = 11 products or services to estimate the probabilty of failure or failure rate of products or services over time, along with the estimations of mean, mode, median, sample variance.

Calculation Summary
Shape Parameter (α) 2
Scale Parameter (k)5
Random variable (x)11
P 0.007
Mean µ4.4311
Median 4.1628
Mode 3.5355
Variance σ25.365

## Work with Steps for α = 2, k = 5 & x = 11

Question:
Find the probability of 11th failure by using Weibull distribution with parameters α = 2 and k = 5
Workout :
step 1 Address the formula input parameters & values
shape parameter α = 2
scale parameter k = 5
x = 11 products or services

step 2 Find P value using k,α & x values
f(x) = (α/k) (x/k)(α - 1)(e(-(x/k)α))
= (2/5) x (11/5)(2 - 1) x (e(-(11/5)2))
= (0.4) x (2.2) x e(-(2.2)2)
= (0.4) x (2.2) x e-(4.84)
= (0.4) x (2.2) x (0.0079)
PDF = 0.007

step 3 Find Mean using k & α values
Mean µ = k [Γ(1 + (1/α))]
= 5 [Γ(1 + (1/2))]
= 5 x ( Γ(1 + 0.5))
= 5 x (Γ(1.5))
= 5 x 0.8862
mean (µ) = 4.4311

step 4 Fine Median using k & α
Median = k [(ln(2))1/α]
= 5 x (0.6931)(1/2)
= 5 x (0.6931)(0.5)
Median = 4.1628

step 5 Find Mode using k & α
Mode = k(α - 1/α)(1/k)
= 5(2 - 1/2)(1/2)
= 5 x (1/2)(0.5)
= 5 x (0.7071)
Mode = 3.5355

step 6 Find Variance using α and k values
Variance σ2 = k2[Γ(1 + 2/α) - [ Γ(1 + 1/α)]2 ]
= 5²[Γ(1 + 2/2) - [ Γ(1 + 1/2)]2 ]
= 25 x [ Γ(1 + 1) - ( Γ(1 + 0.5))² ]
= 25 x [ Γ(2) - ( Γ(1.5))² ]
= 25 x [(1) - (0.8862)² ]
= 25 x [(1) - (0.7854)]
= 25 x 0.2146
Variance (σ2) = 5.365 