Assume in a ICU room, the arrival rate of the infect follows a Poisson distribution with mean = 3 / hour. (a) What is the probability of the infected is more than 8 in a hour.

The probability of the infected is more than 8 in a hour is 0.00384

Step-by-step explanation:

Given that the Mean of the arrival rate of the infect follows a Poisson distribution =   x`= 3 / hour

The Poisson distribution formula is given by

P(X) = e-ˣ` x`ˣ/ x!

The mean is 3 and we have to find the probability of 8 or more which means

1 -X  where X takes the values of 0,1,2,3,,8.

P( more than 8 ) = 1- P( X ≤ 8 ) =1- {e-³ (3)⁸/8! +e-³ (3)⁷/7! +e-³ (3)⁶/6! +e-³ (3)⁵/5! +e-³ (3)⁴/4! +e-³ (3)³/3! +e-³ (3)²/2!+ e-³ (3)¹/1! +e-³ (3)⁰/0!}

Putting the Values

P( more than 8 ) = 1- P( X ≤ 8 ) =1- [ 0.04979*6561 / 40320 +0.04979*2187 / 5040 + 0.04979*729 / 720 +0.04979*243 / 120+0.04979*81 / 24 + 0.04979 *27 / 6 + 0.04979 *9 /2 + 0.04979*3/ 1 + 0.04979*1/1 }

Solving

P( more than 8 ) = 1- P( X ≤ 8 ) =1- [ 0.0081 + 0.0216 + 0.0504 + 0.1008 + 0.1680 + 0.2240 + 0.2241+ 0.14937 + 0.04979]

P( more than 8 ) = 1- P( X ≤ 8 ) =1-0.99616

P( more than 8 ) = 1- P( X ≤ 8 ) =0.00384

question:

could we have options?