Question

The lifetime of LCD TV sets follows an exponential distribution with a mean of 100,000 hours. Compute the probability a television set:
a. Fails in less than 10,000 hours.
b. Lasts more than 120,000 hours.
c. Fails between 60,000 and 100,000 hours of use.
d. Find the 90th percentile. So 10 percent of the TV sets last more than what length of time?

Answer 1

Answer:

0.9

0.3012

0.1809

230258.5

Step-by-step explanation:

Given that:

μ = 100,000

λ = 1/μ = 1 / 100000 = 0.00001

a. Fails in less than 10,000 hours.

P(X < 10,000) = 1 - e^-λx

x = 10,000

P(X < 10,000) = 1 - e^-(0.00001 * 10000)

= 1 - e^-0.1

= 1 - 0.1

= 0.9

b. Lasts more than 120,000 hours.

X more than 120000

P(X > 120,000) = e^-λx

P(X > 120,000) = e^-(0.00001 * 120000)

P(X > 120,000) = e^-1.2

= 0.3011942 = 0.3012

c. Fails between 60,000 and 100,000 hours of use.

P(X < 60000) = 1 - e^-λx

x = 60000

P(X < 60,000) = 1 - e^-(0.00001 * 60000)

= 1 - e-^-0.6

= 1 - 0.5488116

= 0.4511883

P(X < 100000) = 1 - e^-λx

x = 100000

P(X < 60,000) = 1 - e^-(0.00001 * 100000)

= 1 - e^-1

= 1 - 0.3678794

= 0.6321205

Hence,

0.6321205 - 0.4511883 = 0.1809322

d. Find the 90th percentile. So 10 percent of the TV sets last more than what length of time?

P(x > x) = 10% = 0.1

P(x > x) = e^-λx

0.1 = e^-0.00001 * x

Take the In

−2.302585 = - 0.00001x

2.302585 / 0.00001

= 230258.5