Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time in minutes for customers arriving during a 1-hour period. What is the probability that the total service time will exceed 2.5 hours?

Step-by-step explanation:

60 minutes = 1 hour

The statement would be

"Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per 60 minutes

If it takes approximately ten minutes to serve each customer, then the mean service time for in minutes for customers arriving during a 1-hour period is

10/60 × 7 = 1.67

In Poisson distribution, variance = mean

Therefore,

Variance = 1.67

The probability that the total service time will exceed 2.5 hours is expressed as P(x > 2.5)

Converting 2.5 hours to minutes, it becomes 2.5 × 60 = 150 minutes

From the Poisson distribution calculator,

P(x > 150) = 0

answer: sine

step-by-step explanation:

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %