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 X 1 , and X 2 be the lifetimes of the two components. Assume that X 1 and X 2 are independent and that each has the Weibull distribution with α = 2 and β = 0.2. a. Find P ( X 1 > 5). b. Find P ( X 1 > 5 and X 2 > 5). c. Explain why the event T > 5 is the same as the event { X 1 > 5 and X 2 > 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?

Question

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Answer 1

Textbook Question

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?

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

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Answer 2

Textbook Question

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?

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

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Answer 3

Textbook Question

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?

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

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Answer 4

Textbook Question

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?

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

Answer 8

a.

Expert Solution

To determine

Find the value of P(X1>5).

Answer to Problem 13E

The value of P(X1>5)=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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the value of P(X1>5).

Answer 13

Answer to Problem 13E

The value of P(X1>5)=0.3679.

Answer 14

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 X1and X2 are defined as the lifetimes of the two components.

Assume that the lifetimes are independent. The random variables X1 and X2 has Weibull distribution with α=2and β=0.2.

Calculation:

The random variables X1 and X2 are defined as the lifetimes of the two components, which has Weibull distribution with α=2 and β=0.2.

Weibull distribution:

The probability density function of the Weibull distribution with parameters α and β is,

f(t)={αβαtα−1e−(βt)α t>00 t≤0

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>5)=1−P(X1≤5)=1−(1−e−((0.2)(5))2)=e−1=0.3679

Thus, the value of P(X1>5)=0.3679.

Answer 15

b.

Expert Solution

To determine

Find the value of P(X1>5 and X2>5).

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the value of P(X1>5 and X2>5).

Answer 20

Answer to Problem 13E

The value of P(X1>5 and X2>5)=0.1353.

Answer 21

Explanation of Solution

Calculation:

Here, the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>5)=P(X1>5)P(X2>5)

From part (a), P(X1>5)=e−1

Then,

P(X1>5 and X2>5)=(e−1)(e−1)=e−2=0.1353

Thus, the value of P(X1>5 and X2>5)=0.1353.

Answer 22

c.

Expert Solution

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Explain the reason behind the event T>5 is the same as {X1>5 andX2>5}.

Answer 27

Explanation of Solution

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

The random variables X1 and X2 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 {X1>5 andX2>5}.

Answer 28

d.

Expert Solution

To determine

Find the value of P(T≤5).

Answer to Problem 13E

The value of P(T≤5)=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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

Answer 29

d.

Expert Solution

Answer 30

d.

Expert Solution

Answer 31

Expert Solution

Answer 32

To determine

Find the value of P(T≤5).

Answer 33

Answer to Problem 13E

The value of P(T≤5)=0.8647.

Answer 34

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(X1>5 and X2>5)=0.1353.

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤5)=1−P(T>5)=1−P(X1>5 and X2>5)

Substitute P(X1>5 and X2>5)=0.1353 in the above formula.

P(T≤5)=1−P(T>5)=1−0.1353=0.8647

Thus, the value of P(T≤5)=0.8647.

Answer 35

e.

Expert Solution

To determine

Find the cumulative distribution function of T.

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

Answer 36

e.

Expert Solution

Answer 37

e.

Expert Solution

Answer 38

Expert Solution

Answer 39

To determine

Find the cumulative distribution function of T.

Answer 40

Answer to Problem 13E

The cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2.

Answer 41

Explanation of Solution

Calculation:

Here the random variables X1and X2 are independent.

Then the joint probability density function is the product of the marginal, each of which is Weibull with parameters α=2 and β=0.2. Therefore,

f(x1,x2;2,0.2)=∏i=12P(Xi>t)=P(X1>t)P(X2>t)

Substitute α=2and β=0.2 in the formula for cumulative distribution.

P(X1>t)=1−P(X1≤t)=1−(1−e−((0.2)(t))2)=e−((0.2)(t))2

Then,

P(X1>t and X2>t)=(e−((0.2)(t))2)2

From, part(c), it is clear that the event T>5 is the same as {X1>5 andX2>5}.

Then,

P(T≤t)=1−P(T>t)=1−P(X1>t and X2>t)=1−(e−((0.2)(t))2)2=1−(e−(0.04t2))2

=1−e−(0.08t2)=1−e−(0.08t)2

Thus, the value of P(T≤t)=1−e−(0.08t)2.

Answer 42

f.

Expert Solution

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 α=2and β=0.08.

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.

Thus, the random variable T follows Weibull distribution with parameter α=2and β=0.08.

Answer 43

f.

Expert Solution

Answer 44

f.

Expert Solution

Answer 45

Expert Solution

Answer 46

To determine

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

Answer 47

Answer to Problem 13E

The random variable T follows Weibull distribution with parameter α=2and β=0.08.

Answer 48

Explanation of Solution

The cumulative distribution function of the Weibull distribution with parameters α and β is,

F(t)=P(T≤t)={1−e−(βt)α t>00 t≤0

From (e), the cumulative distribution function of T is, P(T≤t)=1−e−(0.08t)2, which is same as the cumulative distribution function of Weibull with parameter α=2and β=0.08.