one sitting. Do not leave the test before clicking Save and Submit Remaining Tim
ID: 2931549 • Letter: O
Question
one sitting. Do not leave the test before clicking Save and Submit Remaining Time: 1 hour, 16 minutes, 31 seconds Question Completion Status: stanc ity Moving to another question will save this response. Question 3 of 17 Question 3 9 points The number of power failures experienced by the Columbia Power Company in a day has a Poisson distribution with an average of 0.2 power surges per day Find the probability (use 4DP in your answers 1, there will be between 3 and 5 power surges(inclusive!) on a gven day. Your answer P(3·Explanation / Answer
Solution:-
= 0.2 per day
a) The probability that there will be between 3 and 5 power surges on a given day is 0.0115.
x1 = 3
x2 = 5
By applying poisons distribution:-
P(x; ) = (e-) (x) / x!
P(3 < x < 5) = P(x > 3) - P(x > 5)
P(3 < x < 5) = 0.001148 - 0.00000008
P(3 < x < 5) = 0.00115
b) The probability that there will be between 3 and 5 power failures within three days is 0.0231
Expected surge for 3 days = = 0.2 × 3 = 0.60
P(x; ) = (e-) (x) / x!
P(3 < x < 5) = P(x > 3) - P(x > 5)
P(3 < x < 5) = 0.02312 - 0.0000389
P(3 < x < 5) = 0.0231
c) The probability that will be no more than 3 failures within three days is 0.9966.
Expected surge for 3 days = = 0.2 × 3 = 0.60
P(x; ) = (e-) (x) / x!
P(x < 3) = 0.9966
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.