# Weibull Distribution Probability for α = 3, k = 11 & x = 9 Weibull distribution example problem workout with steps & calculation summary for shape parameter α = 3, scale parameter k = 11 & x = 9 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) 3
Scale Parameter (k)11
Random variable (x)9
P 0.1056
Mean µ9.8228
Median 9.735
Mode 9.6094
Variance σ212.7451

## Work with Steps for α = 3, k = 11 & x = 9

Question:
Find the probability of failure for random variable x=9 which follows the Weibull distribution with parameters α = 3 and k = 11
Workout :
step 1 Address the formula input parameters & values
shape parameter α = 3
scale parameter k = 11
x = 9 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))
= (3/11) x (9/11)(3 - 1) x (e(-(9/11)3))
= (0.2727) x (0.6695) x e(-(0.8182)3)
= (0.2727) x (0.6695) x e-(0.5477)
= (0.2727) x (0.6695) x (0.5783)
PDF = 0.1056

step 3 Find Mean using k & α values
Mean µ = k [Γ(1 + (1/α))]
= 11 [Γ(1 + (1/3))]
= 11 x ( Γ(1 + 0.3333))
= 11 x (Γ(1.3333))
= 11 x 0.893
mean (µ) = 9.8228

step 4 Fine Median using k & α
Median = k [(ln(2))1/α]
= 11 x (0.6931)(1/3)
= 11 x (0.6931)(0.3333)
Median = 9.735

step 5 Find Mode using k & α
Mode = k(α - 1/α)(1/k)
= 11(3 - 1/3)(1/3)
= 11 x (2/3)(0.3333)
= 11 x (0.8736)
Mode = 9.6094

step 6 Find Variance using α and k values
Variance σ2 = k2[Γ(1 + 2/α) - [ Γ(1 + 1/α)]2 ]
= 11²[Γ(1 + 2/3) - [ Γ(1 + 1/3)]2 ]
= 121 x [ Γ(1 + 0.6667) - ( Γ(1 + 0.3333))² ]
= 121 x [ Γ(1.6667) - ( Γ(1.3333))² ]
= 121 x [(0.9028) - (0.893)² ]
= 121 x [(0.9028) - (0.7974)]
= 121 x 0.1053
Variance (σ2) = 12.7451 