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 µ | 9.8228 |
| Median | 9.735 |
| Mode | 9.6094 |
| Variance σ2 | 12.7451 |
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 :Find the probability of failure for random variable x=9 which follows the Weibull distribution with parameters α = 3 and k = 11
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
