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: 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

A) 0.000055

B) 320.348 x10^ (-8)

C) 3.49 x 10^ (-6)

D) 0.0377

Step-by-step 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