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 µ | 4.4311 |
| Median | 4.1628 |
| Mode | 3.5355 |
| Variance σ2 | 5.365 |
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 :Find the probability of 11th failure by using Weibull distribution with parameters α = 2 and k = 5
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
