Inverse Weibull distribution for α = 9, k = 6 & P = 0.75
Inverse Weibull distribution example problem workout with steps & calculation summary for shape parameter α = 9, scale parameter k = 6 & P = 0.75 failure probability to estimate the lifetime of products or services over time, along with the estimations of mean, mode, median and sample variance.
Calculation Summary | |
---|---|
Shape Parameter | 9 |
Scale Parameter | 6 |
P | 0.75 |
x | 6.2218 |
Mean µ | 5.6818 |
Median | 5.7606 |
Mode | 5.922 |
Variance σ2 | 0.5698 |
Work with Steps for α = 9, k = 6 & P = 0.75
Question:
Find the inverse probability density function for Weibull distribution having the scale parameter k = 6, shape parameter α = 9 with failure probability P(x) = 0.75
Workout :Find the inverse probability density function for Weibull distribution having the scale parameter k = 6, shape parameter α = 9 with failure probability P(x) = 0.75
step 1 Address the formula input parameters & values
shape parameter α = 9
scale parameter k = 6
P = 0.75
step 2 Find x value using k = 6, α = 9 & P = 0.75
x = k [(-ln(1 - P))(1/α)]
= 6 x [(-ln(1 - 0.75))(1/9)]
= 6 x [(-ln(0.25))(0.1111)]
= 6 x [(-(-1.3863))(0.1111)]
= 6 x (1.037)
x = 6.2218
step 3 Find Mean using k & α values
Mean µ = k [Γ(1 + (1/α))]
= 6 [Γ(1 + (1/9))]
= 6 x ( Γ(1 + 0.1111))
= 6 x (Γ(1.1111))
= 6 x 0.947
mean (µ) = 5.6818
step 4 Fine Median using k & α
Median = k [(ln(2))1/α]
= 6 x (0.6931)(1/9)
= 6 x (0.6931)(0.1111)
Median = 5.7606
step 5 Find Mode using k & α
Mode = k(α - 1/α)(1/k)
= 6(9 - 1/9)(1/9)
= 6 x (8/9)(0.1111)
= 6 x (0.987)
Mode = 5.922
step 6 Find Variance using α and k values
Variance σ2 = k2[Γ(1 + 2/α) - [ Γ(1 + 1/α)]2 ]
= 6²[Γ(1 + 2/9) - [ Γ(1 + 1/9)]2 ]
= 36 x [ Γ(1 + 0.2222) - ( Γ(1 + 0.1111))² ]
= 36 x [ Γ(1.2222) - ( Γ(1.1111))² ]
= 36 x [(0.9126) - (0.947)² ]
= 36 x [(0.9126) - (0.8968)]
= 36 x 0.0158
Variance (σ2) = 0.5698