Question

The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 20 calls per hour.a. What is the probability that there are exactly 5 calls in one hour? b. What is the probability that there are 3 or fewer calls in one hour? c. What is the probability that there are exactly 15 calls in two hours? d. What is the probability that there are exactly 5 calls in 30 minutes?

Answer 1

A) 0.000055

B) 320.348 x10^(-8)

C) 3.49 x 10^(-6)

D) 0.0377

Explanation:

A) First of all, let X be the Poisson random variable where the mean(μ) = 20

Thus, the probability mass function of X is;

f(x) = P(X=k) = [(20^(k))(e^(-20))]/k! with condition of k ∈ No

So probability of exactly 5 calls an hour;

P(X=5) = [(20^(5))(e^(-20))]/5!

= 0.0066/120 = 0.000055

B) Probability of 3 or fewer calls in an hour; P(X≤3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

P(X=0) = [(20^(0))(e^(-20))]/0! = 0.206 x 10^(-8)

P(X=1) = [(20^(1))(e^(-20))]/1! = 4.122 x 10^(-8)

P(X=2) = [(20^(2))(e^(-20))]/2! = 41.22 x 10^(-8)

P(X=3) = [(20^(3))(e^(-20))]/3! = 274.8 x 10^(-8)

So, P(X≤3) = [0.206 x 10^(-8)] + [4.122 x 10^(-8)] + [41.22 x 10^(-8)] + [274.8 x 10^(-8)] = 320.348 x10^(-8)

C) since it is now per 2 hours as against an hour we based our initial calculation, we have to take this 2 hours into account and thus mean(μ) = 2 x 20 = 40,to get;

f(x) = P(X=k) = [(40^(k))(e^(-40))]/k!

Thus, P(15 calls/2 hours) =

[(40^(15))(e^(-40))]/15! = 3.49 x 10^(-6)

D) Since 5 calls in 30 minutes, it's different from our initial derivation thus we have to take this 30 minutes or 0.5 hour into consideration and so, mean(μ) = 0.5 x 20 = 10,to get

Thus;

P(5 calls/half hour) = [(10^(5))(e^(-10))]/5!

= 4.54/120 = 0.0378

Answer 2

Answer: a) 0.000055, b) 0 c) 0.0000035, d) 0.0378

Step-by-step explanation: the probability mass function that defines a possion probability distribution is given below as

P(x=r) = e^-u × u^r / r!

Where u = fixed rate at which the event is occurring = 20

A) p(x = 5)

p(x=5) = e^-20 × 20^5 / 5!

p(x=5) = 0.0066 / 120

p(x=5) = 0.000055

B)

p(x≤3). This can be gotten by using a cumulative binomial probability distribution table.

Where u = 20 and n =3.

By checking the table, we have that

p(x≤3) = 0

C)

If in one hour, 20 calls come in, then in 2 hours, the number of calls will be 40

P(x=15) =? When u = 40

P(x=15) = e^-40 × 40^15 / 15!

P(x=15) = 0.0000035.

D)

If in 60 minutes, 20 calls came in, in 30 minutes an average of 10 calls will come in.

Hence u = 10 and p(x=5) =?

p(x=5) = e^-10 × 10^5 / 5!

p(x=5) = 4.53999/ 120 = 0.0378