Answer:
the mean service time = 70 minutes
the variance of the total service time for customers arriving during a 1-hour period is 700
the probability that the total service time will exceed 2.5 hours is 0.0012
Step-by-step explanation:
Given that:
Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of 7 per hour.
If it takes approximately 10 minutes to serve each customer
From the given information:
The mean total service time= [tex]10 \ minutes * \lambda[/tex] (i.e the average number to turn up during the hour) .
where;
[tex]\lambda =[/tex] the average rate per hour i.e 7
Thus; the mean service time = 10 × 7 = 70 minutes
For a Poisson Distribution;
The variance of the the total service time for customers arriving during a 1-hour period can be expressed by the relation:
[tex]\sigma(aX) = a* \sigma(X). \\ \\ where ; \sigma = {\sqrt{S^2}} \\ \\ \sigma(service \ time) = 10 * \sigma(arrival \ time) \\ \\S^2 (service \ time) = (10* \sigma(arrival \ time))^2 \\ \\ S^2(service \ time) = 100 * S^2(arrival \ time) \\ \\ = 100*\lambda \\ \\ = 100*7 = 700[/tex]
where S² = variance
What is the probability that the total service time will exceed 2.5 hours?
If we convert 2.5 hours to minutes ; we have
2.5 × 60 minutes =150 minutes
The probability that the total service time will exceed 2.5 hours is as follows:
[tex]Z = \dfrac{S-70}{\sqrt {700}} \ \ \ \ follow \ N(0,1) \\ \\ P(S > 150) = \dfrac{S-70}{\frac{\sqrt{700}>(150-70)}{\sqrt{700}}} \\ \\ =P(Z > -2.5512) \\ \\ =0.0012[/tex]
