Question

5. For the Weibull distribution with parameters a and X, recall that for t> 0 the density function and distribution function are, respectively, f(t) = 410-1-(At) F(t) = 1 -e-(1)" Suppose that T has the Weibull distribution with parameters a = 1/2 and X = 9. (a) (4 points) Compute exactly P(1 <1 < 1.017 > 1). Show your work. Write your answer to 6 decimal places. (b) (4 points) Compute an approximation of P(1 <T < 1.01 T > 1) using the hazard rate. Show your work.

Answer 1

T is having Weibull distributon with parameters 1/2 and 9

Then

$P(1<T<1.01/T>1)=\frac{P(1<T<1.01)}{P(T>1}$

  $=\frac{P(T<1.01)-P(T<1)}{P(T>1)}$=$=\frac{P(T<1.01)-P(T<1)}{1-P(T<1)}$

$=\frac{0.950952-0.950213}{0.049787}$

=0.014843

For Weibull disttribution with parameter  $\alpha$ and $\lambda$ ,

the hazard rate

$h(t)=\alpha \lambda ^{\alpha}t^{\alpha-1}$

Here, $\alpha$ = 1/2,  $\lambda$ = 9

Then

$h(T)=\frac{1}{2} 9 ^{\frac{1}{2}}T^{\frac{1}{2}-1}=\frac{3}{2} T^{\frac{1}{2}-1}$

But we have,

$h(T)=\lim_{x\rightarrow 0}\frac{P(a<T<a+x/T>a)}{x}$

Hence,

can be approximated as $h(T) \times x$

Hence here,

can be approximated as, $h(1) \times 0.01$

$h(1)=\frac{3}{2} \times 1^{\frac{1}{2}-1}=\frac{3}{2}=1.5$

Then

$h(1) \times 0.01=1.5\times 0.01 = 0.015$

That is,       is approximated as 0.015.  

This is matching with the probability calculated directly.