Distribution with a mean of 4 a find the probability that...

Distribution with a mean of 4. a. Find the probability that there is no power outage in a given year. (6 pts) x ∼ Poi(4) λ = 4 P(x=0) = (e^(-4)*4^(0))/(0!) = 0.0183 p(x) = (0) = (e^(-4)*4^(0))/(0!) = 0.0183 b. Given that this town has 2 power outages during this year already, what is the probability that we will have at most 4 power outages by the end of this year? (10 pts) This is the answer = 0.5914 I just don't know how to get that number Part A is correct, just need part B Distribution with a mean of 4. X ∼ Poi(4) = 4 xPo = = 0.0183 to 3 wowth a caaPhoonowob that we will have at most 4 power outages by the end of this year? (10 pts) 0.5914 P(x4|x2 = P(2x4/p(x2) P2 = 0.1465 P(X4 = 0.2931

Transcript

00:02 As stated in the problem, we have a person distribution with a mean of 6 outages per year.

00:09 For the first question, you are to find the probability that there will be exactly 3 power outages.

00:16 Now we can write that as the probability that x is equal to 3.

00:24 The density function for a person distribution is given as p of x or the probability of x is equal to the mean value based the power x times the exponent of negative or the mean value divided by x factorial so for our question here we simply going to substitute x as 3 and then the mean value at 6 so we have the mean value which is 6 raised the power x that is 3 you have the exponent of negative 6 that is a mean value divided by x factorial therefore we have 3 factor we can go ahead and compute this on our calculators and that gives us 0 .0892.

01:22 For the next question, you have to find the probability that there will be more than 3 power outages in a year.

01:30 You can also write that as the probability that x is greater than 3.

01:36 So what this problem really means is that in a single year, the probability that there will be, say, 4, 5, 6, 7, 9, 11 or an infinite number of power outages in that same exact.

01:53 Year so to make the computation very simple we can rewrite this as one minus the probability that x is less than or equal to three that is one minus the probability that there will be three or less outages in the exact same year and this is come written as one minus the probability that let me put this in square brackets the probability that x is equal to zero plus the probability that x is equal to 1 plus the probability that x is equal to 2 and then x is equal to 3 we are going to solve this before we have 1 minus for this problem we are going to substitute x to be 0 and then 6 to be the mean value therefore we are going to have the mean value 6 gives the power 0 they have the exponent of negative of the mean value 6 divided by 0 factorial 0 factor is equal to 1 plus we have 6 raise the power 1 exponent of negative 6 divided by 1 factorial plus 6 with the power 2 exponent negative 6 divided by 1 factorial plus 6 with the power 2 exponent negative 6 divided by two factorial then the last expression we have six to the third power and then e to the power negative six i'm going to divide that by three factorial so you have one minus we're going to compute this individually so i have 0 .002479 plus 0 .0 .0 .0 114873 plus 0 .04461824.

05:17 0 .0 .04618 plus 0 .0 .0 .8924.

05:29 So we add the expression here and we have 0 .15121.

05:41 1 minus this is equal to 0 .8...

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