Example of inverse cumulative distribution function (ICDF)

A reliability engineer for an appliance manufacturer investigates failure times for the heating element within the company's toasters. The engineer wants to determine the time at which specific proportions of heating elements fail in order to set the warranty period. Heating element failure times follow a normal distribution, with a mean of 1000 hours and a standard deviation of 300 hours.

The engineer uses the ICDF to determine the time by which 5% of the heating elements fail, the times between which 95% of all heating elements fail, and the time at which only 5% of the heating elements continue to function.

Note

This example uses the normal distribution. However, you follow the same steps for any distribution that you select.

In the column name cell of an empty worksheet column, type Probabilities.
Copy and paste, or type the following data into the Probabilities column.

0.05

0.95

0.025

0.975

These values are the probabilities for which the data values will be calculated.
Choose Calc > Probability Distributions > Normal.
Select Inverse cumulative probability.
In Mean, enter 1000.
In Standard deviation, enter 300.
In Input column, enter Probabilities.
Click OK.

Interpret the results

If the distribution of heating element failures follows a normal distribution with a mean of 1000 and a standard deviation of 300, then the following are true:

The time by which 5% of the heating elements are expected to fail is the ICDF of 0.05, or approximately 507 hours.
The middle 95% of all heating elements are expected to fail between 412 hours and 1588 hours, which is the ICDF of 0.025 and the ICDF of 0.975.
The time at which only 5% of the heating elements are expected to continue to function is the ICDF of 0.95, or 1493 hours.

Normal with mean = 1000 and standard deviation = 300

P( X ≤ x )	x
0.050	506.54
0.950	1493.46
0.025	412.01
0.975	1587.99