Question

A new iPhone user receives an average of 4.2 phone calls per hour. The calls occur randomly in time so if X = the number of calls received in an hour, then we can assume that X has a Poisson distribution with parameter λ = 4.2. Let Y = the number of phone calls received in a day. Assume all necessary independence in the following.

a. Y also has a Poisson distribution. What is the λ for Y?

b. What is the standard deviation of X?

c. What is the standard deviation of Y?

d. What is the probability that no calls are received in an hour?

e. What is the probability that more than 120 calls are received in a day?

f. What is the probability that Y is between 90 and 110 inclusive? (i.e. 90 ≤ Y≤110)

g. What is the variance of square root of 2Y?

h. Assume this distribution holds for all hours and days and that one call is independent from another. What is the probability that 120 or fewer calls are received for every day of this week? (Assume appropriate independence

Add any comments below.

Answer 1

Part A:

Given that Y is the number of phone calls received in a day and that a new iPhone user receives an average of 4.2 phone calls per hour.

Thus, the average number of phone calls received in a day is 4.2 * 24 = 100.8

Therefore, the λ for Y is 100.8



Part B:

The standard deviation of a Poisson distribution is given by


`\sigma=√(\lambda)`

Thus, the standard deviation of X is given by


`\sigma=√(4.2)=2.05`

Therefore, the standard deviation of X is 2.05.



Part C:

The standard deviation of Y is given by


`\sigma=√(100.8)=10.04`

Therefore, the standard deviation of X is 10.04.



Part D:

The probability of a Poisson distribution is given by


`P(X=x)=e^(-\lambda) (\lambda^x)/(x!)`

where, x is the number of occurence of the event in the given period.

Thus, the probability that no calls are received in an hour is given by


`P(X=0)=e^(-4.2) ((4.2)^0)/(0!) =0.015`

Therefore, the probability that no calls are received in an hour is 0.015



Part E:

We approximate the Poisson distribution using Normal distribution by noting that for a Poisson distribution


`\mu=\lambda \\ \\ \sigma^2=\lambda \\ \\ \sigma=√(\lambda)`

The probability of a normal distribution is given by


`P(X\ \textless \ x)=P\left(z\ \textless \ (x-\mu)/(\sigma) \right)`

Thus, the probability that more than 120 calls are received in a day is given by


`P(X\ \textgreater \ 120)=1-P(X\leq120) \\ \\ =1-P\left(z\leq (120-100.8)/(10.04) \right)=1-P(z\leq1.912) \\ \\ =1-0.97208=0.0279`

Therefore, the probability that more than 120 calls are received in a day is 0.0279



Part F:

We also approximate this using Normal distribution

The probability of a normal distribution between two value (a, b) is given by


`P(a\ \textless \ X\ \textless \ b)=P\left(z\ \textless \ (b-\mu)/(\sigma) \right)-P\left(z\ \textless \ (a-\mu)/(\sigma) \right)`

Thus, the probability that Y is between 90 and 110 inclusive is given by


`P(90\leq X\leq 110)=P\left(z\leq \ (110-100.8)/(10.04) \right)-P\left(z\leq (90-100.8)/(10.04) \right) \\ \\ =P(z\leq0.916)-P(-1.076)=0.82025-0.14103=0.67922`

Therefore, the probability that Y is between 90 and 110 is 0.67922



Part G:

The variance of aY is given by


`Var(aY)=a^2Var(Y)`

Thus, the variance of
`√(2) Y` is given by


`Var( √(2) Y)=( √(2) )^2Var(Y)=2(100.8)=201.6`

Therefore, the variance of
`√(2) Y` is 201.6



Part H:

We approximate this using Normal distribution

The probability of a normal distribution is given by


`P(X\ \textless \ x)=P\left(z\ \textless \ (x-\mu)/(\sigma) \right)`

Thus, the probability that 120 or fewer calls are received in a day is given by


`P(X\leq120)=P\left(z\leq (120-100.8)/(10.04) \right) \\ \\ =P(z\leq1.912)=0.97208`

Since, this distribution holds for all hours and days and one call is independent from another. Assuming that there are seven days in a week.

The probability that 120 or fewer calls are received for every day of this week is given by


`(0.97208)^7=0.8202`

Therefore, the probability that 120 or fewer calls are received for every day of this week is 0.8202