Question

Acceptance sampling of a product. An essential tool in the monitoring of the quality of a manufactured product is acceptance sampling. An acceptance sampling plan involves knowing the distribution of the life length of the item produced and determining how many items to inspect from the manufacturing process. The Journal of Applied Statistics (April 2010) demonstrated the use of the exponential distribution as a model for the life length x of an item (e.g., a bullet). The article also discussed the importance of using the median of the lifetime distribution as a measure of product quality since half of the items in a manufactured lot will have life lengths exceeding the median. For an exponential distribution with a mean $\mathit{\theta }$, give an expression for the median of the distribution. (Hint: Your answer will be a function of $\mathit{\theta }$.)

Short answer

The median of the distribution is $0.6931\times \theta$.

Step-by-step solution

Given information

The life length of a product is exponentially distributed with a mean $\theta$.

Let x represents the life length of a product.

The probability distribution function of a random variable x is:

$\mathit{F}\left(x\right)\mathbf{=}\mathbf{1}\mathbf{-}{\mathit{e}}^{\frac{\mathbf{-}\mathbf{x}}{\mathbf{\theta }}}\mathbf{;}\mathit{x}\mathbf{>}\mathbf{0}$.

Obtaining the median of an exponential random variable

The median of a probability distribution is a value below that half of the observations lie.

The median is obtained as:

$\begin{array}{rcl}F\left(x\right)& =& 0.5\\ 1-e^{\frac{-x}{\theta }}& =& 0.5\\ -e^{\frac{-x}{\theta }}& =& 0.5-1\\ -e^{\frac{-x}{\theta }}& =& -0.5\\ e^{\frac{-x}{\theta }}& =& 0.5\end{array}$

Taking the natural logarithm of both sides,

$\begin{array}{rcl}\ln \left(e^{\frac{-x}{\theta }}\right)& =& \ln \left(0.5\right)\\ \frac{-x}{\theta }\ln \left(e\right)& =& -0.6931\\ \frac{x}{\theta }& =& 0.6931\\ x& =& 0.6931\times \theta \end{array}$.

Therefore, the median of the distribution is $0.6931\times \theta$.