Chapter 4.7, Problem 11E

A light fixture contains five lightbulbs. The lifetime of each bulb is exponentially distributed with mean 200 hours. Whenever a bulb burns out, it is replaced. Let T be the time of the first bulb replacement. Let X_i, i = 1,…, 5, be the lifetimes of the five bulbs. Assume the lifetimes of the bulbs are independent.

a.    Find P(X₁ > 100).

b.    Find P(X₁ > 100 and X₂ > 100 and … and X₅ > 100).

c.    Explain why the event T > 100 is the same as {X₁, > 100 and X₂ > 100 and … and X₅ > 100}.

d.    Find P(T ≤ 100).

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

f.    Does T have an exponential distribution?

g.    Find the mean of T.

h.    If there were n lightbulbs, and the lifetime of each was exponentially distributed with parameter λ, what would be the distribution of T?

To determine

Find the value of $P\left(X_1>100\right)$.

Answer to Problem 11E

The value of $P\left(X_1>100\right)=0.6065$.

Explanation of Solution

Given info:

Total number of lightbulb is 5. The lifetime of each bulb is exponentially distributed with mean 200 hours. The random variable T is defined as the time of the first bulb replacement. The random variables $X_i,i=1,\mathrm{...},5$ are defined as the lifetimes of the five bulbs. Assume that the lifetimes of the bulbs are independent.

Calculation:

The random variables $X_1,\mathrm{...},X_5$ are defined as the lifetimes of the five bulbs which are exponentially distributed with mean 200 hours.

Exponential distribution:

The probability density function of the exponential distribution with parameter $\lambda >0$ is,

$f\left(x\right)=\{\begin{array}{l}\lambda e^{-\lambda x}x>0\\ 0x\le 0\end{array}$

Mean of anExponentialrandom variable:

The random variable X has Exponential distribution with parameter $\lambda >0$, the expected value of X is given by:

${\mu }_X=\frac{1}{\lambda }$

Substitute 200 for ${\mu }_X$ in the above formula.

$\begin{array}{c}\lambda =\frac{1}{200}\\ =0.005\end{array}$

Thus, the parameter is 0.005.

The cumulative distribution function of the exponential distribution with parameter $\lambda >0$ is,

$F\left(x\right)=1-\lambda e^{-\lambda x}$

Substitute 0.005 for $\lambda$ and 100 for x in the formula for cumulative distribution.

$\begin{array}{c}P\left(X_1>100\right)=1-P\left(X_1\le 100\right)\\ =1-\left(1-e^{-\left(0.005\right)\left(100\right)}\right)\\ =e^{-0.5}\\ =0.6065\end{array}$

Thus, the value of $P\left(X_1>100\right)=0.6065$.

To determine

Find the value of $P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)$.

Answer to Problem 11E

The value of $P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)=0.0821$.

Explanation of Solution

Calculation:

Here, the random variables $X_1,\mathrm{...},X_5$ are independent.

Then the joint probability density function is the product of the marginal, each of which is an $\text{Exp}\left(0.005\right)$ probability density function. Therefore,

$f\left(x_1,\mathrm{...},x_5;0.05\right)={\displaystyle \prod _{i=1}^{5}P\left(X_i>100\right)}$

That is,

$\begin{array}{c}P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)={\displaystyle \prod _{i=1}^{5}P\left(X_i>100\right)}\\ ={\left(e^{-\lambda x}\right)}^5\end{array}$

Substitute 0.005 for $\lambda$ and 100 for x in the formula for cumulative distribution.

$\begin{array}{c}P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)={\left(1-e^{-\left(0.005\right)\left(100\right)}\right)}^5\\ ={\left(e^{-0.5}\right)}^5\\ ={\left(0.6065\right)}^5\\ =0.0821\end{array}$

Thus, the value of $P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)=\underset{\_}{0.0821}$.

To determine

Explain the reason behind the event $T>100$ is the same as $\{X_1>100\text{ and }{X}_2>\mathrm{100...}\text{and}{X}_5>100\}$.

Explanation of Solution

The random variable T is defines as the time of the first bulb replacement.

The random variables $X_i,i=1,\mathrm{...},5$ are defined as the lifetimes of the five bulbs. Assume that the lifetimes of the bulbs are independent.

Thus, the time of the first replacement will be greater than 100 hours if and only if each of the bulb lasts longer than 100 hours.

Hence, the event $T>100$ is the same as $\{X_1>100\text{ and }{X}_2>\mathrm{100...}\text{and}{X}_5>100\}$

To determine

Find the value of $P\left(T\le 100\right)$.

Answer to Problem 11E

The value of $P\left(T\le 100\right)=0.9179$.

Explanation of Solution

Calculation:

The random variable T is defined as the time of the first bulb replacement.

From part (b), the value of $P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)=0.0821$.

From, part(c), it is clear that the event $T>100$ is the same as $\{X_1>100\text{ and }{X}_2>\mathrm{100...}\text{and}{X}_5>100\}$.

Then,

$\begin{array}{c}P\left(T\le 100\right)=1-P\left(T>100\right)\\ =1-P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)\end{array}$

Substitute $P\left(X_1>100\text{ and }{X}_2>100\text{ and}\mathrm{...}\text{and }{X}_5>100\right)=0.0821$ in the above formula.

$\begin{array}{c}P\left(T\le 100\right)=1-P\left(T>100\right)\\ =1-0.0821\\ =0.9179\end{array}$

Thus, the value of $P\left(T\le 100\right)=0.9179$.

To determine

Find the cumulative distribution function of T, that is $P\left(T\le t\right)$, where t be any positive number.

Answer to Problem 11E

The cumulative distribution function of T is, $P\left(T\le t\right)=1-e^{-0.025t}$.

Explanation of Solution

Calculation:

Here the random variables $X_1,\mathrm{...},X_5$ are independent.

Then the joint probability density function is the product of the marginal, each of which is an $\text{Exp}\left(0.005\right)$ probability density function. Therefore,

$f\left(x_1,\mathrm{...},x_5;0.05\right)={\displaystyle \prod _{i=1}^{5}P\left(X_i>x\right)}$

That is,

$\begin{array}{c}P\left(X_1>x\text{ and }{X}_2>x\text{ and}\mathrm{...}\text{and }{X}_5>x\right)={\displaystyle \prod _{i=1}^{5}P\left(X_i>x\right)}\\ ={\left(e^{-\lambda x}\right)}^5\end{array}$

From, part(c), it is clear that the event $T>100$ is the same as $\{X_1>100\text{ and }{X}_2>\mathrm{100...}\text{and}{X}_5>100\}$.

Then,

$\begin{array}{c}P\left(T\le t\right)=1-P\left(T>t\right)\\ =1-P\left(X_1>x\text{ and }{X}_2>x\text{ and}\mathrm{...}\text{and }{X}_5>x\right)\\ =1-{\displaystyle \prod _{i=1}^{5}P\left(X_i>x\right)}\\ =1-{\left(e^{-\lambda x}\right)}^5\end{array}$

Substitute 0.005 for $\lambda$ and t for x in the formula for cumulative distribution.

$\begin{array}{c}P\left(T\le t\right)=1-{\left(e^{-\left(0.005\right)\left(t\right)}\right)}^5\\ =1-e^{-0.025t}\end{array}$

Thus, the value of $P\left(T\le t\right)=1-e^{-0.025t}$.

To determine

Check whether T has an exponential distribution or not.

Answer to Problem 11E

The random variable T follows exponential distribution with parameter $\lambda =0.025$.

Explanation of Solution

If the random variable X follows exponential distribution with parameter $\lambda >0$, the cumulative distribution function of X is,

$\begin{array}{c}F\left(x\right)=P\left(X\le x\right)\\ =\{\begin{array}{c}1-e^{-\lambda x}x>0\\ 0x\le 0\end{array}\end{array}$

From (e), the cumulative distribution function of T is, $P\left(T\le t\right)=1-e^{-0.025t}$, which is same as the cumulative distribution function of exponential with parameter $\lambda =0.025$.

Thus, the random variable T follows exponential distribution with parameter $\lambda =0.025$.

To determine

Find the mean of T.

Answer to Problem 11E

The mean of T is40 hours.

Explanation of Solution

Calculation:

Mean of anExponentialrandom variable:

The random variable T has Exponential distribution with parameter $\lambda >0$, the expected value of T is given by:

${\mu }_T=\frac{1}{\lambda }$

Substitute 0.025 for $\lambda$ in the above formula.

$\begin{array}{c}{\mu }_T=\frac{1}{0.025}\\ =40\end{array}$

Thus, the mean of T is40 hours.

To determine

Find the distribution of T, if there were n lightbulbs and the lifetime of each was exponentially distributed with parameter $\lambda$.

Answer to Problem 11E

The distribution of T is exponential with parameter $n\lambda$.

Explanation of Solution

Calculation:

The random variable T is defined as the time of the first bulb replacement.

The random variables $X_1,\mathrm{...},X_n$ are defined as the lifetimes of the n bulbs and which are exponentially distributed with parameter $\lambda$.

Here the random variables $X_1,\mathrm{...},X_n$ are independent.

Then the joint probability density function is the product of the marginal, each of which is an $\text{Exp}\left(\lambda \right)$ probability density function. Therefore,

$\begin{array}{c}f\left(x_1,\mathrm{...},x_n;\lambda \right)={\displaystyle \prod _{i=1}^{n}P\left(X_i>x\right)}\\ ={\left(e^{-\lambda x}\right)}^n\end{array}$

From, part(e),

$\begin{array}{c}P\left(T\le t\right)==1-{\displaystyle \prod _{i=1}^{n}P\left(X_i>x\right)}\\ =1-{\left(e^{-\lambda x}\right)}^n\\ =1-e^{-n\lambda x}\end{array}$

Thus, the cumulative distribution function of T is, $P\left(T\le t\right)=1-e^{-n\lambda x}$, which is same as the cumulative distribution function of exponential with parameter $n\lambda$.

Thus, the random variable T follows exponential distribution with parameter $n\lambda$.