The lifetime of a certain battery is modeled with the Weibull distribution with α= 2 and =β 0.1.

a. What proportion of batteries will last longer than 10 hours?

b. What proportion of batteries will last less than 5 hours?

c. What proportion of batteries will last longer than 20 hours?

d. The hazard function is defined in Exercise 8. What is the hazard at t = 10 hours?

Step 1 of 3:

The lifetime of a battery modeled with weibull distribution with parameters .

We have to find

The proportion of batteries will last longer than 10 hours.The proportion of batteries will last less than 5 hours.The proportion of batteries will last longer than 20 hours.If the hazard function of lifetime (T) is h(t) = , then what is the hazard at t=10 hours.Step 2 of 3:

Let T is the lifetime of batteries which follows weibull distribution with parameters .

So the pdf of T is

f(t) = ,

= (2 ) (0.1

And the density function of T is

F(t) = P(

= 1-

= 1-

1- F(t) =

Proportion of batteries will last longer than 10 hours.

P(T>10) = 1- P(T)

P() = F(10)

= 1-

= 1 -

= 0.6321.

P(T>10) = 1- 0.6321

= 0.3678

Therefore the probability P(T>10) is 0.3678.

(b) proportion of batteries will last less than 5 hours.

P(T<5) = P(

= 1-

= 1- 0.7788

= 0.2211

Therefore the probability that batteries will last less than 5 hours is 0.2211.