Question

1. The average amount of time for a particular brand of dryer to break down is 13,500 hours of use. Let D be the amount of time for this dryer to break down (i.e. stop working).

a) Identify the distribution, parameter(s), and support of D.

b) What is the probability that the dryer will stop working break before the average time to breakdown?

c) What is the probability that dryer operates over 16,600 hours before breaking down?

d) What is the probability that the dryer operates between 10,500 and 14,500 hours before breaking down?

e) Given a dryer has already operated for 10,000 hours, what’s the probability that it will operate more than 14,500 hours total?

f) Given a dryer works for operates no more than 15,000 hours, what’s the probability that it breaks before 9,500 hours?

g) What is the median amount of time to the dryer’s breakdown?

Answer 1

a)

distribution for D is exponential ; paramter =1/13500 and support of D is (0,)

b)

probability that the dryer will stop working break before the average time to breakdown

=P(X<13500)=1-e^-t =1-e^{-(1/13500)*13500} =1-e^-1 =0.6321

c)

probability that dryer operates over 16,600 hours before breaking down=P(X>16600)=e^-16600/13500

^=0.2924

^d) probability that the dryer operates between 10,500 and 14,500 hours before breaking down

=P(10500<X<14500)=(1-e^-14500/13500)-(1-e^-10500/13500)=0.1178

e)

Given a dryer has already operated for 10,000 hours, what’s the probability that it will operate more than 14,500 hours total =P(X>14500|X>10000)=e^-14500/13500/e^-10000/13500 =0.7165

f)

P(X<9500|X<15000)=(1-e^-9500/13500)/(1-e^-15000/13500)=0.7532

g)

median =-ln(1-p)/=-ln(1-0.5)/(1/13500)=9357.49