Reverse Exponential

Now let's flip the exponential problem around.

Instead of asking,

• for a given \(\mu\), what's the probability of getting X = 0 successes during time t?

We can ask,

• for a given \(\mu\), what time t will give us a specified probability of not getting any successes (X = 0)?

To solve for t, given \(\mu\) and the probability, we rearrange the exponential formula \(prob = e^{-\mu t}\) to isolate t and get:

 

\(t = \dfrac{ln(prob)}{-\mu}\),

 

where ln is the natural logarithm and prob is the probability of getting X = 0 occurrences of what you're looking for. 

We use this for warranty purposes. Let's say we're willing to service warranty claims on 25 percent of our products sold. That means, we expect 75 percent of them to not fail during the warranty period.

 

What warranty period, t, will gives us a probability (percentage) of 0.75 during this period?

Let's say the mean failure rate for this product is one failure every 230 days. That's an expectation of \( \mu \) = 1/230th of a failure every day.

 

The warranty period in days that gives us prob = 0.75 is 

 

\( t = \dfrac{ln(0.75)}{-1/230} \) = 66.16688 days

Fin

Oh, so you want to know where I got

\( t = \dfrac{ln(prob)}{-\mu} \).

OK. Here you go:

\( prob = e^{-\mu t} \)

\( ln(prob) = ln(e^{-\mu t}) \)

\( ln(prob) = {-\mu t} \times ln(e) \)

\( ln(prob) = {-\mu t} \)

\( \dfrac{ln(prob)}{- \mu} = t \)

 

Tah dah!

Fin Fin