In: Statistics and Probability
An entrepreneur opened a small hardware store in a street mall.
During the first few
weeks, business was slow with the store averaging only one customer
arrived every 20
minutes. Assume that the random arrival of customers is Poisson
distributed.
(a) What is the probability that exactly 5 customers arrive during
one-hour period?
(b) If the average revenue from each customer is MYR 12, what is
the probability
that the entrepreneur’s hourly earnings are at least MYR 24?
(c) If the hardware shop is open for 8 hours, what is the
entrepreneurs mean daily
earnings if each customer were to spend on average about MYR
12.
( show steps please)
The random arrival of the customers follows a Poisson distribution.
Let x be the number of customers that arrive in 20 mins.
a) We are to find the probability that 5 customers arrive in one-hour.
The mean number of customers arriving in one hour = 3 customers (with one customer arriving in 20 minutes).
Therefore, we have
Therefore, the probability of this occuring is 0.1.
b) The average revenue for each customer is 12MYR. And on an average 3 customers are arriving in an hour. Therefore, the average revenue obtained in an hour will be 12*3 = 36MYR. Since the customer arrival follows a Poisson distribution, so will the revenue, with mean 36 MYR.
c) The mean per hour is 36MYR. The shop is open for 8 hours.
Hence the mean daily earning will be: 36*8 = 288 MYR.