Question

a. Tonda is testing a new lawnmower model that should be very fuel-efficient. They are still testing, but specifications from the engineers claim that the mower should run from between 5 hours to 8 hours on a single gallon of gas. What would be an appropriate distribution for modeling this and what would be the probability that the mower will last longer than 7 hours?

b. Tonda has now completed some testing and found that most of the time, mowers last around 5.75 hours. What would be a more appropriate distribution to use and how does the probability that a mower will last longer than 7 hours change?

Answer 1

a) here as mover life is randomly distributed between 5 hour and 8 hour and can take any values between them ; therefore it follows uniform distribution with parameter a=5 and b=8

probability that the mower will last longer than 7 hours =P(X>7)=(b-x)/(b-a)=P(8-7)/(8-5)=1/3 =0.3333

b)

here as expected life time of mower is known ; it follows exponential distribution with parameter

$\beta$=5.75

probability that a mower will last longer than 7 hours change =P(X>7)=e^-x/$\beta$ =e^-7/5.75 =0.2960