Question

Product failure behavior. An article in Hotwire (December 2002) discussed the length of time till the failure of a product produced at Hewlett Packard. At the end of the product’s lifetime, the time till failure is modeled using an exponential distribution with a mean of 500 thousand hours. In reliability jargon, this is known as the “wear-out” distribution for the product. During its normal (useful) life, assume the product’s time till failure is uniformly distributed over the range of 100 thousand to 1 million hours.

a. At the end of the product’s lifetime, find the probability that the product fails before 700 thousand hours.

b. During its normal (useful) life, find the probability that the product fails before 700 thousand hours.

c. Show that the probability of the product failing before 830 thousand hours is approximately the same for both the normal (useful) life distribution and the wear-out distribution.

Short answer

a. At the end of the product’s lifetime, the probability that the product fails before 700 thousand hours is 0.7534.

b. During its normal life, the probability that the product fails before 700 thousand hours is 0.6667.

c. The probabilities are approximately the same for the normal (useful) life distribution and the wear-out distribution.

Step-by-step solution

Given information

The lifetime of a product produced at Hewlett-Packard is modeled using an exponential distribution with a mean of 500 thousand hours. During the normal (useful) life, the product time till failure is uniformly distributed 0ver the range of 100 thousand to 1 million hours.

Define the probability density functions

Let X denotes the length of time till the failure of a product.

The probability density function of a random variable X is:

$f\left(x\right)=\frac{1}{500}{e}^{\frac{-x}{500}};x>0$.

Let Y denotes the product’s time till failure during its normal life

The probability density function of Y is:

$\begin{array}{rcl}f\left(y\right)& =& \frac{1}{1000-100};100⩽x⩽1000\\ & =& \frac{1}{900};100⩽x⩽1000\end{array}$.

Calculating the probability using exponential distribution

a.

At the end of the product’s lifetime, the probability that the product fails before 700 thousand hours is obtained as:

$\begin{array}{rcl}P\left(X⩽700\right)& =& 1-P\left(X>700\right)\\ & =& 1-e^{\frac{-700}{500}}\\ & =& 1-e^{-1.4}\\ & =& 1-0.2466\\ & =& 0.7534\end{array}$

For the exponential distribution:$\mathit{P}\left(X⩾a\right)\mathbf{=}{\mathit{e}}^{\frac{\mathbf{-}\mathbf{a}}{\mathbf{\theta }}}$.

Therefore, the required probability is 0.7534.

Calculating the probability using the uniform distribution

b.

During its normal life, the probability that the product fails before 700 thousand hours is obtained as:

$\begin{array}{rcl}P\left(Y⩽700\right)& =& \frac{700-100}{900}\\ & =& \frac{600}{900}\\ & =& 0.6667\end{array}$

For the uniform distribution: $\mathit{P}\left(\mathit{a}<\mathit{x}<\mathit{b}\right)\mathbf{=}\frac{\mathbf{b}\mathbf{-}\mathbf{a}}{\mathbf{d}\mathbf{-}\mathbf{c}}\mathbf{,}\mathit{c}\mathbf{⩽}\mathit{a}\mathbf{<}\mathit{b}\mathbf{⩽}\mathit{d}$.

Therefore, the required probability is 0.6667.

Showing the probability is approximately the same in both ways

c.

The probability of the product failing before 830 thousand hours is:

$\begin{array}{rcl}P\left(X⩽830\right)& =& 1-P\left(X>830\right)\\ & =& 1-e^{\frac{-830}{500}}\\ & =& 1-e^{-1.6}\\ & =& 1-0.2019\\ & =& 0.7981\end{array}$

Also,

$\begin{array}{rcl}P\left(Y⩽830\right)& =& \frac{830-100}{900}\\ & =& \frac{730}{900}\\ & =& 0.8111\\ & & \end{array}$

Therefore, the probabilities are approximately the same for the normal (useful) life distribution and the wear-out distribution.