The lognormal distribution has two parameters, μ and σ. These are not the same as mean and standard deviation, which is the subject of another post, yet they do describe the distribution, including the reliability function.

$latex \displaystyle R(t)=1-\Phi \left( \frac{\ln (t)-\mu }{\sigma } \right)$

Where Φ is the standard normal cumulative distribution function, and t is time.

One of the nice features of the lognormal distribution is the estimate of the parameters is similar to estimating the mean and standard deviation of the data using the same functions on our calculator or spreadsheet. There is one difference though. First calculate the natural logarithm of each data value.

Let’s say we have the time to failure times for four heater elements. We know the time to failure distribution is lognormal from previous work. We want to estimate the lognormal parameters and estimate the reliability of this type of heater elements at 365 days.

Time to Fail ln(Time to Fail)

385 5.9532

427 6.0568

490 6.1944

705 6.5582

## Calculate μ

In the table we have the time to failure data and the calculation of the natural log of each data reading. To calculate the μ we calculate the mean or average value of the four ln(time to failure) readings.

$latex \displaystyle \mu =\frac{5.9532+6.0568+6.1944+6.5582}{4}=6.1907$

## Calculate σ

The calculation of σ requires a little more math. The formula for the calculation of standard deviation includes the sum of values squared and the sum of squares of the values.

$latex \displaystyle s=\sqrt{\frac{n\sum\limits_{i=1}^{n}{t_{i}^{2}}-{{\left( \sum\limits_{i=1}^{n}{{{t}_{i}}} \right)}^{2}}}{n(n-1)}}$

We need the sum of the ln(time to failure) for the second summation term. And the sum of squares for the first summation term. Expanding the table to make the calculations we find the two summation results.

Time to Fail ln(Time to Fail) ln(Time to Fail) Squared

385 5.9532 35.4411

427 6.0568 36.6846

490 6.1944 38.3706

705 6.5582 43.0100

Sum 24.7626 153.5063

n equals four in the example, as we have four readings. Inserting the sums and n, an doing the math to find the value of σ, the second parameter for the lognormal distribution.

$latex \displaystyle s=\sqrt{\frac{4(153.5064)-{{24.7626}^{2}}}{4(4-1)}}=0.2642$

## Determine reliability at one year

Now that we have the two parameters for the lognormal distribution which describes the life distribution of heater elements based on the four readings, we can estimate the probability of successfully operating for one year. Using the reliability function of the lognormal distribution, insert 365 for t, 6.1907 for μ, and 0.2642 for σ, to find the reliability value at one year.

$latex \displaystyle R(t)=1-\Phi \left( \frac{\ln (365)-6.1907}{0.2642} \right)=1-\Phi \left( -1.1007 \right)$

The standard normal cumulative distribution function (try Excel function =normsdist(-1.1007) or for the CRE exam use a standard normal cumulative distribution table) determines the probability of failure at time, t given the lognormal parameters. Φ(-1.1007) = 0.1355.

Therefore completing the calculations for the reliability function, we have

$latex \displaystyle R(365)=1-0.1355=0.8645$

Thus, give the data, we can expect approximately 86.45% of heater elements to survive for 365 days.

## Leave a Reply