For example, an appliance manufacturer investigates failure times for the heating element within its toasters. They want to determine the time by which specific proportions of heating elements will fail so they can set the warranty period. Heating element failure times follow a normal distribution with a mean of 1000 hours and standard deviation of 300 hours. The probability density function (PDF) helps identify regions of higher and lower failure probabilities. The inverse CDF gives the corresponding failure time for each cumulative probability.
Use the inverse CDF to estimate the time by which 5% of the heating elements will fail, times between which 95% of all heating elements will fail, or the time at which only 5% of the heating elements remain. The inverse CDF for specific cumulative probabilities is equal to the failure time at the right side of the shaded area under the PDF curve.
The time by which 5% of the heating elements are expected to have failed is the inverse CDF of 0.05 or 506.544 hours.
The time by which 2.5% of the heating elements are expected to have failed is the inverse CDF of 0.025 or 412 hours.
Therefore, times between which 95% of all heating elements are expected to fail is the inverse CDF of 0.025 and the inverse CDF of 0.975 or 412 hours and 1588 hours.
The time at which only 5% of the heating elements are expected to remain is the inverse CDF of 0.95 or 1493 hours.
When you try to determine the inverse cumulative probability of a discrete distribution, the output in contains two sets of columns.
Suppose you have the inverse cumulative probability of a proportion, p. The first set of columns in the output lists the largest x such that P(X ≤ x) ≤ p. The second set of columns lists the smallest x such that P(X ≤ x) ≥ p.
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Now that you know the cumulative probabilities associated with the number of defects, calculate the inverse cumulative probability.
Suppose that you want to calculate the number of defects, x, such that the cumulative probability, p, is 0.50. From the previous results, you know that P(X ≤ 1 ) = 0.391619 and P(X ≤ 2 ) = 0.676941. Because the hypergeometric distribution is a discrete distribution, the number of defects cannot be between 1 and 2. In other words, you may have 1 defect or 2 defects, but not 1.4 defects. Therefore, if you choose Input constant and enter 0.50, Minitab calculates both probabilities in the output, as shown in the following example:
The first probability indicates a value of x such that P(X ≤ x) < p and the second probability indicates the smallest x such that P(X ≤ x) ≥ p. In this example, the first probability shows the largest number of defectives, x = 2, such that P(X ≤ 2) <0.5 and the 2^{nd} shows the smallest number of defectives, x = 3, such that P(X ≤ 3) ≥ 0.5.
You can use Minitab to calculate a critical value for a hypothesis test instead of looking in a table.
Suppose that you perform a chi-square test with an α = 0.02 and 12 degrees of freedom. What is the corresponding critical value? An α = 0.02 corresponds to a cumulative probability value of 1 – 0.02 = 0.98.
Minitab displays the critical value, 24.054. For the chi-square test, if the test statistic is greater than the critical value, you can conclude that there is statistical evidence to reject the null hypothesis.
This example uses the chi-square distribution. However, you follow these same steps for any distribution that you select.