Return to Main Page A Model for Travel Time Thomas G. Hess  Most actuaries know the negative binomial distribution as a model for loss   frequency. The negative binomial distribution can also describe another familiar process. To quote Hastings and Peacock (1974), "The negative binomial variate Y:x,p is the number of failures before the xth success in a sequence of Bernoulli trials where the probability of failure is q = 1p. Parameters x, the Bernoulli success parameter, a positive integer, p the Bernoulli probability parameter, 0 < p < 1." The mean number of failures before the   xth success is xq/p. We can call a success a pass and the sum of x and y the   number of trials or sittings needed to achieve x passes. The average number of trials or sittings to achieve x passes is x(1+q/p). The variance is xq/p2. Let x = 7 and p = 1/3, then the mean number of failures before the seventh pass is   14. On average, 21 sittings are needed to achieve seven passes. If p=9/20 and x=7 then the mean number of failures before the seventh pass is 8.56 and, on average   15.56 sittings are needed to achieve seven passes. A pass rate between 1/3 and 9/20 would imply between 21 and 15 sittings to achieve Associate. For x = 10 and p =1/3 the mean number of sittings to achieve ten passes is 30. If two of the required passes are split into two parts and another is partitioned into three parts, a required seven passes has now become 11 required passes and the mean number of trials or sittings has grown to 33. This 57.1 percent increase may be mitigated by multiple trials at a sitting. To achieve Fellowship, the needed number of passes has increased to 14 and the mean number of sittings has increased to 42, an increase of 40 percent. Changing the needed number of passes to 9 and keeping   p=1/3, gives an average of 27 sittings to Fellowship. This is a decrease of 36 percent. I have set p = 1/3 for convenience; the average number of sittings to achieve x passes is 3x. For p = 1/2, the average number of sittings to achieve x passes is 2x, or a third less than the average when p=1/3. If a student can raise the probability of   passing to 2/3, then the average number of sittings needed drops to 1.5x. If the probability of success is only 1/4 then the average number of sittings to obtain x passes is 4x. The model lets one estimate the impact of various changes. If p=1/3 and three passes are needed the average number of sittings needed is nine. If the needed number of passes is changed to two then one can solve for the new p value that would   keep the average number of sittings at nine (the new p is 2/9). This initial model, as do all models, has limitations. There are at least three   limitations. First, the model focuses on how an individual candidate travels through the exams, not on how a cohort of candidates who start the exam process at the same time move through the exams. It does not consider multiple exams at a sitting, people who chose to quit taking exams, etc. Second, the model gives answers in terms of sittings, not time. Finally, the model assumes that p does not change. P equal 1/3 is lower than the average pass ratio. A nonrandom sample of 108 exams from Fall 1987 to Fall 1997 has a mean of 37.9 percent with standard deviation of 5.8. The mean plus or minus two standard deviations gives a range of 26.3 to 49.5. The average p and the standard deviation of p do appear to vary among exams, but that would be the subject of another article.