Question

1. Number of patients that arrive in a hospital emergency center between 6 pm and 7 pm is modeled by a Poisson distribution with λ=3.5. Determine the probability that the number of arrivals in this time period will be

1. Exactly four
2. At least two
3. At most three

Answer 1

Solution :

Given that ,

mean = = 3.5

a)

Using poisson probability formula,

P(X = x) = (e^- * ^x ) / x!

P(X = 4) = (e^-3.5 * 3.5⁴⁾ / 4! = 0.1888

Probability = 0.1888

b)

P(X ≥ 2) = 1 − P(X < 2)

= 1−P( X ≤ 1)

= 1 - (P(X = 0) + P(X = 1))

= 1 - ( (e^-3.5 * 3.5⁰⁾ / 0! +  (e^-3.5 * 3.5¹⁾ / 1!)

= 1 - (0.0302 + 0.1057)

=1−0.1359

= 0.8641

Probability = 0.8641

c)

P( X ≤ 3) = P(X = 0)+P(X = 1)+P(X =2)+P(X = 3)

=  e^-3.5 * 3.5⁰ / 0! + e^-3.5 * 3.5¹ +e^-3.5 * 3.5² / 2! + e^-3.5 * 3.5³ /3!

⁼ 0.0302+0.1057+0.185+0.2158

= 0.5366

Probability = 0.5366