Chapter 4.8, Problem 13E

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 X₁ and 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?

To determine

Find the value of $P\left(X_1>5\right)$.

Answer to Problem 13E

The value of $P\left(X_1>5\right)=0.3679$.

Explanation of Solution

Given info:

The total number of components is 2. The random variable T is defined as the time at which the system fails. The random variables $X_1\text{and }{X}_2$ are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables $X_1\text{and }{X}_2$ has Weibull distribution with $\alpha =2\text{and }\beta =0.2$.

Calculation:

The random variables $X_1\text{and }{X}_2$ are defined as the lifetimes of the two components, which has Weibull distribution with $\alpha =2\text{and }\beta =0.2$.

Weibull distribution:

The probability density function of the Weibull distribution with parameters $\alpha \text{ and }\beta$ is,

$f\left(t\right)=\{\begin{array}{l}\alpha {\beta }^{\alpha }{t}^{\alpha -1}{e}^{-{\left(\beta t\right)}^{\alpha }}t>0\\ 0t\le 0\end{array}$

The cumulative distribution function of the Weibull distribution with parameters $\alpha \text{ and }\beta$ is,

$\begin{array}{c}F\left(t\right)=P\left(T\le t\right)\\ =\{\begin{array}{l}1-e^{-{\left(\beta t\right)}^{\alpha }}t>0\\ 0t\le 0\end{array}\end{array}$

Substitute $\alpha =2\text{and }\beta =0.2$ in the formula for cumulative distribution.

$\begin{array}{c}P\left(X_1>5\right)=1-P\left(X_1\le 5\right)\\ =1-\left(1-e^{-{\left(\left(0.2\right)\left(5\right)\right)}^2}\right)\\ =e^{-1}\\ =0.3679\end{array}$

Thus, the value of $P\left(X_1>5\right)=0.3679$.

To determine

Find the value of $P\left(X_1>5\text{ and }{X}_2>5\right)$.

Answer to Problem 13E

The value of $P\left(X_1>5\text{ and }{X}_2>5\right)=0.1353$.

Explanation of Solution

Calculation:

Here, the random variables $X_1\text{and }{X}_2$ are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters $\alpha =2\text{and }\beta =0.2$. Therefore,

$\begin{array}{c}f\left(x_1,x_2;2,0.2\right)={\displaystyle \prod _{i=1}^{2}P\left(X_i>5\right)}\\ =P\left(X_1>5\right)P\left(X_2>5\right)\end{array}$

From part (a), $P\left(X_1>5\right)=e^{-1}$

Then,

$\begin{array}{c}P\left(X_1>5\text{ and }{X}_2>5\right)=\left(e^{-1}\right)\left(e^{-1}\right)\\ =e^{-2}\\ =0.1353\end{array}$

Thus, the value of $P\left(X_1>5\text{ and }{X}_2>5\right)=0.1353$.

To determine

Explain the reason behind the event $T>5$ is the same as $\{X_1>5\text{ and}{X}_2>5\}$.

Explanation of Solution

The random variable T is defined as the time at which the system fails.

The random variables $X_1\text{and }{X}_2$ are defined as the lifetimes of the two components. Assume that the lifetimes are independent.

Thus, the lifetime of the system will be greater than 5 hours if and only if the lifetimes of both components greater than 5 hours.

Hence, the event $T>5$ is the same as $\{X_1>5\text{ and}{X}_2>5\}$.

To determine

Find the value of $P\left(T\le 5\right)$.

Answer to Problem 13E

The value of $P\left(T\le 5\right)=0.8647$.

Explanation of Solution

Calculation:

The random variable T is defined as the time at which the system fails.

From part (b), the value of $P\left(X_1>5\text{ and }{X}_2>5\right)=0.1353$.

From, part(c), it is clear that the event $T>5$ is the same as $\{X_1>5\text{ and}{X}_2>5\}$.

Then,

$\begin{array}{c}P\left(T\le 5\right)=1-P\left(T>5\right)\\ =1-P\left(X_1>5\text{ and }{X}_2>5\right)\end{array}$

Substitute $P\left(X_1>5\text{ and }{X}_2>5\right)=0.1353$ in the above formula.

$\begin{array}{c}P\left(T\le 5\right)=1-P\left(T>5\right)\\ =1-0.1353\\ =0.8647\end{array}$

Thus, the value of $P\left(T\le 5\right)=0.8647$.

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, $P\left(T\le t\right)=1-e^{-{\left(\sqrt{0.08}t\right)}^2}$.

Explanation of Solution

Calculation:

Here the random variables $X_1\text{and }{X}_2$ are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters $\alpha =2\text{and }\beta =0.2$. Therefore,

$\begin{array}{c}f\left(x_1,x_2;2,0.2\right)={\displaystyle \prod _{i=1}^{2}P\left(X_i>t\right)}\\ =P\left(X_1>t\right)P\left(X_2>t\right)\end{array}$

Substitute $\alpha =2\text{and }\beta =0.2$ in the formula for cumulative distribution.

$\begin{array}{c}P\left(X_1>t\right)=1-P\left(X_1\le t\right)\\ =1-\left(1-e^{-{\left(\left(0.2\right)\left(t\right)\right)}^2}\right)\\ =e^{-{\left(\left(0.2\right)\left(t\right)\right)}^2}\end{array}$

Then,

$P\left(X_1>t\text{ and }{X}_2>t\right)={\left(e^{-{\left(\left(0.2\right)\left(t\right)\right)}^2}\right)}^2$

From, part(c), it is clear that the event $T>5$ is the same as $\{X_1>5\text{ and}{X}_2>5\}$.

Then,

$\begin{array}{c}P\left(T\le t\right)=1-P\left(T>t\right)\\ =1-P\left(X_1>t\text{ and }{X}_2>t\right)\\ =1-{\left(e^{-{\left(\left(0.2\right)\left(t\right)\right)}^2}\right)}^2\\ =1-{\left(e^{-\left(0.04t^2\right)}\right)}^2\end{array}$

                  $\begin{array}{c}=1-e^{-\left(0.08t^2\right)}\\ =1-e^{-{\left(\sqrt{0.08t}\right)}^2}\end{array}$

Thus, the value of $P\left(T\le t\right)=1-e^{-{\left(\sqrt{0.08}t\right)}^2}$.

To determine

Check whether T has a Weibull distribution or not. Find the parameters if so.

Answer to Problem 13E

The random variable T follows Weibull distribution with parameter $\alpha =2\text{and }\beta =\sqrt{0.08}$.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters $\alpha \text{ and }\beta$ is,

$\begin{array}{c}F\left(t\right)=P\left(T\le t\right)\\ =\{\begin{array}{l}1-e^{-{\left(\beta t\right)}^{\alpha }}t>0\\ 0t\le 0\end{array}\end{array}$

From (e), the cumulative distribution function of T is, $P\left(T\le t\right)=1-e^{-{\left(\sqrt{0.08}t\right)}^2}$, which is same as the cumulative distribution function of Weibull with parameter $\alpha =2\text{and }\beta =\sqrt{0.08}$.

Thus, the random variable T follows Weibull distribution with parameter $\alpha =2\text{and }\beta =\sqrt{0.08}$.