Answer:
A. 10 loads
B. 0.428335
C. 3.2 years
Step-by-step explanation:
Given
Mean time between occurrence = 0.4 year
A Number of loads expected to occur during a 4 year period
Period = 4
Mean time = 0.4
Formula = Period/Mean Time
Formula =
So, the number of loads expected to occur during a 4-year period = 4/0.4 =10 loads
B. Probability that more than 11 loads occur during 4 year period
Since, the expected number of loads during 4-year period = 10 (from A above)
Then the mean is 10
Using Poisson distribution,
P(k events in interval)= (λ^k * e^-k)/k!
where: k = 0, 1, 2,3,4, . . ., 11 and λ = 10.
P(k=0)= (10^0 * e^-10)/0! = 0.000045
P(k=1)= (10^1 * e^-10)/1! = 0.000454
P(k=2)= (10^2 * e^-10)/2! = 0.00227
P(k=3)= (10^3 * e^-10)/3! = 0.007567
P(k=4)= (10^4 * e^-10)/4! = 0.018917
P(k=5)= (10^5 * e^-10)/5! = 0.037833
P(k=6)= (10^6 * e^-10)/6! = 0.063055
P(k=7)= (10^7 * e^-10)/7! = 0.090079
P(k=8)= (10^8 * e^-10)/8! = 0.112599
P(k=9)= (10^9 * e^-10)/9! = 0.12511
P(k=10)= (10^10 * e^-10)/10! = 0.12511
P(k=11)= (10^11 * e^-10)/11! = 0.113736
The probability that more than 11 loads occur during a 4-year period is then given by the following Express
1 - [P(k = 0) + P(k = 1) + P(k = 2) + . . . + P(k = 11)]
= 1 - [0.000045 + 0.000454 + 0.00227 + 0.007567 + 0.018917 + 0.037833 + 0.063055 + 0.090079 + 0.112599 + 0.12511+ 0.12511 + 0.113736]
= 1 - 0.571665 = 0.428335
C. How long must a time period be so that the probability of no loads occurring during that period is at most 0.3
This is
(λ^0 * e^-λ)/0! <= 0.3
e^-λ/1 <= 0.3
e^-λ <= 0.3
-λ <= ln 0.3
-λ <= -1.204
λ >= 1.204
The time period = 4 years / 1.204
Time period = 3.2 years
