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BANG & OLUFSEN Back MySQL - Electricity QUESTION MySQL - Electricity Hard = 2 0 0 1 2 0 ( i ) The electric company

BANG & OLUFSEN
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MySQL-Electricity
QUESTION
MySQL - Electricity
Hard =200120(i)
The electric company wants to analyze customer billing patterns to identify and reward consistent on-time payers.
Our database contains two tables:
Customers Table: This table will store information about the customers of the electric company.
Bills Table: This table keeps track of the billing information.
Tables Schema as Below:
\table[[Customers,dataType],[columnName,INT],[CustomerID,VARCHAR(100)],[Name,VARCHAR(200)],[Address,VARCHAR(15)],[PhoneNumber,VARCHAR(100)],[Email,dataType],[Bills,INT],[columnName,INT],[BillID,DATE],[CustomerID,DATE],[BillingDate,VARCHAR(50)],[DueDate,Undefined],[TotalAmount,],[Status,],[FOREIGN KEY (CustomerID),],[REFERENCES,]]
Write a SQL query that:
For each paid bill of a customer, calculate the running average considering the last 2 paid bills and the paid bill
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