A system consists of two components connected in series. The system will fail when either of the two components fails. Let T be the time at which the system fails. Let X1 and X2 be the lifetimes of the two components. Assume that X1and X2 are independent and that each has the Weibull distribution with α=2 and β=0.2

a. Find P(X1 > 5).

b. Find P(X1 > 5 and X2 > 5).

c. Explain why the event T > 5 is the same as the event {X1 > 5 and X2 > 5}.

d. Find P(T ≤ 5).

e. Let t be any positive number. Find P(T ≤ t). which is the cumulative distribution function of T.

f. Does T have a Weibull distribution? If so, what are its parameters?

Step 1 of 5:

A system consist of two components which is connected in series.the system will fail if one of the two component fails. Let T be the lifetime of the system and X1 and X2 be the lifetime of each components. Here X1 and X2 are independent which follows weibull distribution with parameter and .

We have to find

P(X1>5)P(X1>5 and X2>5)The event T>5 is as same as the event {X1>5 and X2>5}. P(P, which is the cumulative density function of T.The distribution of T with parameters.Step 2 of 5:

The probability P(X1>5):Since it is given that the lifetime of each component follows weibull distribution with parameter and , the cdf of X1 and X2 is:

And

P( = 1- P(

= 1- F(5)

F(5) = 1-

= 1-

P(x1>5) =

= 0.3679

Therefore the probability that P(X1>5) is 0.3679.

Step 3 of 5:

(c) P(X1>5 and X2>5)

Since X1 and X2 follows the same distribution.

P(X1>5) =P(X2>5)

= 1- P(

= 1- (1-

=

= 0.3679

Since X1 and X2 are independent

P(X1>5 and X2>5) = )P(X2>5)

=

= 0.1353

Therefore the probability P(X1>5 and X2>5) is 0.1353.

(c)

Here T is the lifetime of the system.It is given that the system fails if any one of the two component fails.

T>5,which is the event that system the lifetime of the system is greater than 5. Since the lifetime of the system is depend on the lifetime of the both components.

We can say that the lifetime of the system will be greater than 5 if and only if the lifetime of the two components greater than 5.

T>5 = {X1>5 and X2>5}

Therefore the events T>5 is as same as the event {X1>5 and X2>5}.