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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

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