Suppose a representative at a credit card customer service center receives a phone call every 5 minutes on average.

(a) find the probability that she receives 3 phone calls in 20 minutes?
(b) find the probability that she receives less than 3 phone calls in 20 minutes?
(c) find the expected number of phone calls she receives in 20 minutes?

a) probability = 0.195

b) probability = 0.2381

c) expected number = 4

Step-by-step explanation:

Let X be a random variable which shows the number of calls in 20 minutes. Since credit card customer service center receives a phone call every 5 minutes on average, given X has Poisson distribution with parameter λ = (1/5)*20 = 4 call per 20 minutes.

Given that t = 20

a)  

The probability that she receives 3 phone calls in 20 minutes is

      P(X=3) = [ e∧-λt (λt)3 ] / 3ḷ

      P(X-3) = [e∧-20/5 (20/5)3] / 3ḷ = 0.195

b)

The probability that she receives fewer than 3 phone calls in 20 minutes is

P(X˂3) = ∑_(x=0)^2▒ (e∧-4×4x) / xḷ = 0.2381

c)

The expected number of phone calls she receives in 20 minutes is

  E(X) = λ = 4

   i.e.

   λ = 1/5 * 20 = 4

it's 100

step-by-step explanation:

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

  n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the  14th  row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

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