The lifetime \( X \) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \( \alpha = 2 \) and \( \beta = 3 \). Compute the following. (Round your answers to three decimal places.)(a) \( E(X) \) and \( V(X) \)- \( E(X) = \) [Text Box]- \( V(X) = \) [Text Box](b) \( P(X \leq 3) \)- [Text Box](c) \( P(1 \leq X \leq 3) \)- [Text Box]

Weibull distribution is used to understand or estimate the life time or reliability of the product. It has two parameters, such as α and β. Where the mean and variance of the Weibull distribution is calculated using the formula : Ex=β×Γ1+1αVx=β2Γ1+2α-Γ1+1α2Given : The Weibull distribution parameters α=2β=3 The mean and variance of the parameters are calculated using the formula : Ex=β×Γ1+1αVx=β2Γ1+2α-Γ1+1α2 a) Mean : Using the given parameters values, Ex=β×Γ1+1α=3×Γ1+12=3×12Γ12=1.5×π=2.659 Where, Γ1+12=Γ32=12Γ12=12π Variance : Vx=β2Γ1+2α-Γ1+1α2=32Γ1+22-Γ1+122=32Γ2-Γ322=322-1!-12Γ122=91-0.5×π2=91-0.7854=1.9314 Where, Γ2=2-1! Thus, Ex=2.659Vx=1.931b) Find : Px≤3 The cumulative value of Weibull distribution PX≤x==1-e-xβα Using the above formula and the given parameters, Px≤3=F3=1-e-xβα=1-e-332=1-e-12=1-e-1=1-0.368=0.632 Thus, Px≤3=0.632c) Find : P1≤x≤3 Using the cumulative formula and the given parameters, P1≤x≤3=F3-F1=1-e-332-1-e-132=1-e-1-1-e-0.111=0.632-0.105=0.527 Thus, P1≤x≤3=0.527