Mathematics, 09.12.2020 21:30, genyjoannerubiera

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?

Answers: 2

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answered: Mypasswordishotdog11

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:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

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

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

In this problem:

Mean of 100,000 hours, hence, they decay parameter is $\mu = \frac{1}{100000}$

Item a:

$P(X \leq 10000) = 1 - e^{-\frac{10000}{100000}} = 1 - e^{-0.1} = 0.0952$

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

Item b:

$P(X 120000) = e^{-\frac{120000}{100000}} = e^{-1.2} = 0.3012$

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

Item c:

$P(X \leq 60000) = 1 - e^{-\frac{60000}{100000}} = 1 - e^{-0.6} = 0.4512$

$P(X \leq 100000) = 1 - e^{-\frac{100000}{100000}} = 1 - e^{-0.6} = 0.6321$

$P(60000 \leq X \leq 100000) = P(X \leq 100000) - P(X \leq 60000) = 0.6321 - 0.4512 = 0.1809$

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:

P(X x) = 0.1

Hence:

$e^{-\mu x} = 0.1$

$e^{-0.00001
x} = 0.1$

$\ln{e^{-0.00001
x}} = \ln{0.1}$

$-0.00001
x = \ln{0.1}$

$x = -\frac{\ln{0.1}}{0.00001}$

x = 230259

The 90th percentile is of 230,259 hours.

A similar problem is given at link

answered: audrey435

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

answered: Guest

answer. d.289 hope that

Alarge asteroid crashed into a moon of another planet causing several boulders from the moon to be p

answered: Guest

answer:

it is 28 degrees colder in detroit than philadelphia.

step-by-step explanation:

what number adds to 21

-7 + ? = 21

-7 + 28 = 21

hope it

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