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?

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

the system of inequity is

step-by-step explanation:

from the given graph it is noticed that we have two related lines. one is solid and another is dashed.

if a line passing through two points, then the equation of line is

the solid line passing through the points (0,4) and (4,0).

the related equation of solid line is .

similarly, find the related equation of dashed line. the dashed liune passing through the points (0,0) and (3,1).

the related equation of dashed line is

from the given graph it is noticed that the point (0,1) lies on the shaded region, therefore the point (0,1) must be satisfied by the related equations.

check the related equations by the point (0,1).

this statement is true if the sign of inequality is < instead of equality. since it is an solid line therefore the sign of inequality will be . the required inequality is

similarly, check the second related equation.

it is a dashed line, so this statement is true if the sign of inequality is > . the required inequality is

a c

step-by-step explanation: