Question

Now Ann also had some brass rods of 32 m length which have on average 7 cracks (i.e. 8 gaps between cracks and the ends). The length of these gaps is exponentially distributed. (so what is the average length between each crack = average length of the gaps = ?) She is interested in the distance from one end of the rod to the 5th crack. Hence, this distance will be gamma distribution with α = 5 (think why 5?) and β = average length between each crack (think why?) Find the probability that the distance from one end of the rod to the 5th crack is less than 12 m. (You will need the gamma distribution again)

Answer 1