Question

In a call center that stays open all the time, calls arrive as a poisson process with a mean rate of 0.3 complaints per hour.(a) (5 points) after receiving a call (the 1st call), what is the probability that the 51st call arrives within 150 hours?(b) (5 points) after receiving a call, what is the probability that the next call arrives after 2 hours?(c) (5 points) the number of calls received each day is recorded for 50 consecutive days. find the probability that the sum of these 50 numbers is less than 356.

Answer 1

The probability the 51st call arriaves within 150hours is 0.0431, the probability the next call arrives within the next 2 hours 0.5488, the probability the sum of these 50 numbers is less than 356 is 0.4165.

Data;

• Mean rate = 0.3
• x = 50
• standard deviation = ?

Poission Rule

Using poission formula,

$P(x=x) = \frac{e^-^\lambda - \lambda^x}{x!}\\\lambda = 0.3 per minute$

Let's substitute the values into the formula.

For 50 calls in 150 hours

For 150 hours = x = 0.3 * 150 = 45

$p(x=50) = \frac{e^-^4^5 * 45^5^0}{50!} = 0.0431$

b)

The probability the next call arrives after 2 hours.

$\lambda = 0.3 * 2 = 0.6\\p(x=0) = \frac{e^-^0^.^6 * 0.6^0}{0!} = 0.5488$

c)

The number of calls recieved each day is recorded for 50 consecutive days.

for 50 days;

$\lambda = 0.3 * 50 * 24 = 360$

The mean = 360

The standard deviation is given as

$S.D = \sigma =\sqrt{360} = 18.974$

The probability the sum of these 50 number is less than 356 is

$p = (x < 356) = z = \frac{356 - 360}{18.974} = -0.2108\\p(z < -0.2108) = 0.4165$

Learn more on poission formula here;

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