Respuesta :
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
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:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
In this problem:
Mean of 100,000 hours, hence, they decay parameter is [tex]\mu = \frac{1}{100000}[/tex]
Item a:
[tex]P(X \leq 10000) = 1 - e^{-\frac{10000}{100000}} = 1 - e^{-0.1} = 0.0952[/tex]
0.0952 = 9.52% probability that a television set fails in less than 10,000 hours.
Item b:
[tex]P(X > 120000) = e^{-\frac{120000}{100000}} = e^{-1.2} = 0.3012[/tex]
0.3012 = 30.12% probability that a television set lasts more than 120,000 hours.
Item c:
[tex]P(X \leq 60000) = 1 - e^{-\frac{60000}{100000}} = 1 - e^{-0.6} = 0.4512[/tex]
[tex]P(X \leq 100000) = 1 - e^{-\frac{100000}{100000}} = 1 - e^{-0.6} = 0.6321[/tex]
[tex]P(60000 \leq X \leq 100000) = P(X \leq 100000) - P(X \leq 60000) = 0.6321 - 0.4512 = 0.1809[/tex]
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:
[tex]P(X > x) = 0.1[/tex]
Hence:
[tex]e^{-\mu x} = 0.1[/tex]
[tex]e^{-0.00001 x} = 0.1[/tex]
[tex]\ln{e^{-0.00001 x}} = \ln{0.1}[/tex]
[tex]-0.00001 x = \ln{0.1}[/tex]
[tex]x = -\frac{\ln{0.1}}{0.00001}[/tex]
[tex]x = 230259[/tex]
The 90th percentile is of 230,259 hours.
A similar problem is given at https://brainly.com/question/25645425
