A restaurant gat an average of 14 calls in a 2 hr time period. What is the probability that at most 2 calls in 45 min period

0.10512

Step-by-step explanation:

Given the following :

Mean number of calls(μ) in 2 hours = 14

2 hours = 60 * 2 = 120 minutes

Average number of calls in 45 minutes :

= (45 / 120) * 14

= 0.375 * 14

= 5.25

Now find P( x ≤ 2) = p(x = 0) + p( x = 1) + p(x = 2)

Using the poisson probability formula:

P(x, μ) = [(e^-μ) * (μ^x)] / x!

Where :

e = euler's constant

μ = 5.25

x = 0, 1, 2

Using the online poisson probability calculator :

P(x, 5.25) = P( x ≤ 2) = p(x = 0) + p(x = 1) + p(x = 2)

P(x, 5.25) = P( x ≤ 2) = 0.00525 + 0.02755 + 0.07232 = 0.10512

where's the diagrams?