So, the brewery selected it’s place and now needs to hire folks. After some testing, they expect to have approximately 2,000 customers per week. Since they will be open 7 days a week, they need to hire people to help serve their beer. They know that Friday and Saturday will drive sales. Between these 2 days, they will get 50% of their weekly sales. So, Friday has 25% of sales and Saturday has 25% of the sales. As for Sunday to Thursday, they assume they will get 10% of sales on each of those days.
Q1) How many people do they expect to get each day?
Sunday |
Monday |
Tuesday |
Wednesday |
Thursday |
Friday |
Saturday |
200 |
200 |
200 |
200 |
200 |
500 |
500 |
Q2) Each day of the week, they have set hours. Based upon those hours and the number of customers per day, determine the average number of customers per hour.
Day |
Open |
Close |
Avg Cust/hr |
Sunday |
12:00pm |
10:00pm |
20 |
Monday |
4:00pm |
10:00pm |
33.33 |
Tuesday |
4:00pm |
10:00pm |
33.33 |
Wednesday |
4:00pm |
10:00pm |
33.33 |
Thursday |
4:00pm |
10:00pm |
33.33 |
Friday |
4:00pm |
12:00am |
62.50 |
Saturday |
12:00pm |
12:00am |
41.667 |
Q3) Using the “Multiple” file from the author, use the average customers per hour data above and a server can handle 12 customers per hour, add servers such that the wait time for a customer to order their drink is under: 5mins, 3mins, 1min (NOTE: There is a change in time units.)
Day |
5 min |
3 min |
1 min |
Sunday |
3 |
N/A |
N/A |
Monday |
9 |
N/A |
N/A |
Tuesday |
9 |
N/A |
N/A |
Wednesday |
9 |
N/A |
N/A |
Thursday |
9 |
N/A |
N/A |
Friday |
13 |
N/A |
N/A |
Saturday |
10 |
N/A |
N/A |
****I NEED HELP WITH QUESTION 4****
Q4) We know that the values you just calculated are based upon the
average number of people entering the brewery. Use a simulation and
an appropriate probability distribution (cough cough Poisson Cough
cough) to determine the 80th and 90th
percentile for people entering each day.
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