Answer:
0.4207149;0.7271136; 0.3063987; 0.04979 ; 0.01832
Step-by-step explanation:
For an exponential distribution:
IF Mean time until failure = 23100
λ = 1/ 23100 = 0.0000432900
What is the probability that a randomly selected fan will last at least 20,000 hours
x ≥ 20000
P(X ≥ 20000) = 1 - P(X ≤ 20000)
1 - P(X ≤ 20000) = [1 - (1 - e^(-λx))]
1 - P(X ≤ 20000) = [1 - (1 - e^(-0.0000432900*20000)
1 - P(X ≤ 20000) = [1 - (1 - 0.4207148)]
1 - P(X ≤ 20000) = 1 - 0.5792851
1 - P(X ≤ 20000) = 0.4207149
11) What is the probability that a randomly selected fan will last at most 30,000 hours?
x ≤ 30000
P(X ≤ 30000) = 1 - e^(-λx)
P(X ≤ 20000) = 1 - e^(-0.0000432900*30000)
= 1 - e^(−1.2987)
= 1 - 0.2728863
= 0.7271136
111) What is the probability that a randomly selected fan will last between 20,000 hours and 30,000 hours?
0.7271136 - 0.4207149 = 0.3063987
B) What is the probability that the lifetime of a fan exceeds the mean value by more than 2 standard deviations?
More than two standard deviation
X = 23100 + 2(23100) = 23100 + 46200 = 69300
Using the online exponential probability calculator :
P(X > 69300) = 0.04979
C) What is the probability that the lifetime of a fan exceeds the mean value by more than 3 standard deviations?
X = 23100 + 3(23100) = 23100 + 69300 = 92400
P(X > 92400) = 0.01832
