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.

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.
These values are the probabilities for which the data values will be calculated.
0.05 0.95 0.025 0.975 - Choose .
- Select Inverse cumulative probability.
- In Mean, enter
`1000`. - In Standard deviation, enter
`300`. - In Input column, enter Probabilities.
- Click OK.

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.

P( X ≤ x ) | x |
---|---|

0.050 | 506.54 |

0.950 | 1493.46 |

0.025 | 412.01 |

0.975 | 1587.99 |