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
