Chapter 6.8, Problem 10E

(a).

To determine

To Calculate: The probability that the value of a random variable with an exponential probability density function is less than a certain number.

Answer to Problem 10E

A bulb fails within the first $200$ hours with a probability of $0.18$ and a bulb burns for more than $800$ hours with a probability of $0.45$ .

Explanation of Solution

Given Information: The average lifetime of the bulb is $1000$ hrs.

Concept Used:

• If the mean of a random variable $\left(X\right)$ with an exponential probability density function $f\left(x\right)$ is $\text{μ}$ , then
$f\left(x\right)=\frac{1}{u}{e}^{-\frac{x}{u}}$ if $x\ge 0$ and

  $f\left(x\right)=0$ if $x<0$ .

• If $f\left(x\right)$ is a probability density function of a random variable $X$ , then the probability that $X$ lies between $a$ and $b$ , where $a\le b$, is given by
$P\left(a\le X\le b\right)=\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

Calculation:

The mean given is $\text{μ}=1000$ , $f\left(t\right)$ is given by

  $f\left(t\right)=\frac{1}{1000}{e}^{-\frac{t}{1000}}$ if $t\ge 0$ and

  $f\left(t\right)=0$ if $t<0$ .

(i). To find the probability that a bulb fails before $200$ hours, put $a=-\infty$ and b $=200$ in the integral for probability to write

  $P\left(-\infty \le T\le 200\right)=\underset{0}{\overset{200}{{\displaystyle \int }}}\frac{1}{1000}{e}^{-\frac{t}{1000}}dt$

  $=\frac{1}{1000}\times {\frac{e^{-\frac{t}{1000}}}{\left(\frac{-1}{1000}\right)}]}_0^{200}$

  $=-\left(e^{-\frac{200}{1000}}-1\right)$

  $=0.18$

(ii). To find the probability that a bulb burns for more than $800$ hours, put $a=800$ and b $=\infty$ in the above integral for probability to write

  $\text{P}\left(800\le T\le \infty \right)=\underset{800}{\overset{\infty }{{\displaystyle \int }}}\frac{1}{1000}{e}^{-\frac{t}{1000}}\text{dt}$

  $=\frac{1}{1000}\times {\frac{e^{-\frac{t}{1000}}}{\left(\frac{-1}{1000}\right)}]}_{800}^{\infty }$

  $=-\left(0-e^{\frac{-800}{1000}}\right)$

  $=0.45$

Therefore, a bulb fails within the first $200$ hours with a probability of $0.18$ and a bulb burns for more than $800$ hours with a probability of $0.45$ .

(b).

To determine

To Calculate: The median value of a random variable with exponential probability density function

Answer to Problem 10E

The median lifetime of the bulbs is approximately $693$ hours.

Explanation of Solution

Given Information: From the previous sub-part, the probability density function is

  $f\left(t\right)=\frac{1}{1000}{e}^{-\frac{t}{1000}}$ if $t\ge 0$ and

  $f\left(t\right)=0$ if $t<0$ .

Formula Used: The median value of a random variable with an exponential probability density function having mean $\text{μ}$ is

  $medianvalue=\mu \ln 2$

Calculation: Put the given value of mean in the above equation to get

  $median=1000\times \ln 2=693$

Therefore, the median lifetime of the bulbs is approximately $693$ hours.