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?

Using the exponential distribution, it is found that:

a) There is a 0.0952 = 9.52% probability that a television set fails in less than 10,000 hours.

b) There is a 0.3012 = 30.12% probability that a television set lasts more than 120,000 hours.

c) There is a 0.1809 = 18.09% probability that a television set fails between 60,000 and 100,000 hours of use.

d) The 90th percentile is of 230,259 hours.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:  

In which is the decay parameter.

The probability that x is lower or equal to a is given by:

Which has the following solution:

The probability of finding a value higher than x is:

In this problem:

Mean of 100,000 hours, hence, they decay parameter is

Item a:

0.0952 = 9.52% probability that a television set fails in less than 10,000 hours.

Item b:

0.3012 = 30.12% probability that a television set lasts more than 120,000 hours.

Item c:

0.1809 = 18.09% probability that a television set fails between 60,000 and 100,000 hours of use.

Item d:

This is x for which:

Hence:

The 90th percentile is of 230,259 hours.

A similar problem is given at link

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