In large statistical data sets, such as consumer behavior data mining or a population census, randomized data trees are used to determine the importance of a variable in a data set. Teachers often weigh tests and papers more heavily than quizzes and homework, for example. You determine the weight of your data points by factoring in which numbers are most important. You can calculate the weighted average of a set of numbers by multiplying each value in the set by its weight, then adding up the products.įor a more in-depth explanation of the weighted average formula above, follow these steps: 1. Weighted average differs from finding the normal average of a data set because the total reflects that some pieces of the data hold more “weight,” or more significance, than others or occur more frequently. This equals a weighted average cost of $1.18 per unit. Using the units as the weight and the total number of units as the sum of all weights, we arrive at this calculation: Other costing methods include last in, first out and first in, first out, or LIFO and FIFO respectively.Ī manufacturer purchases 20,000 units of a product at $1 each, 15,000 at $1.15 each and 5,000 at $2 each. This number goes into the calculation for the cost of goods sold. In some industries where quantities are mixed or too numerous to count, the weighted average method is useful. Weighted average is one means by which accountants calculate the costs of items. It is an important tool in accounting for stock fluctuations, uneven or misrepresented data and ensuring similar data points are equal in the proportion represented. Weighted averages are commonly used in statistical analysis, stock portfolios and teacher grading averages. There are exceptions, but to stay away from causing confusing I will not go into them.A weighted average is the average of a data set that recognizes certain numbers as more important than others. See what I mean? An average of percentages almost always yields unwanted answers. This makes sense because there are 2500 females in 5020 people. Now you can see that there is the logical just-below-one-half answer. To solve this and get a good answer, you add up the total amount of females and divide by the total amount of people: From a percentage standpoint, there are 100 things in each. Percentages work on a scale of 0 to 100- they will give no heed to the fact that there are 4980 more people in Minnesota's parade than in Pennsylvania's. But wait, that doesn't make sense- and here's why: You find that only 25 percent of participants in the two parades are female. You want to find out the percentage of females amongst the two parades. You know that 2500 out of those 5000 Minnesotans are female, but the 20 Pennsylvanians in the other parade are all male. Imagine there are 5000 people in a parade in Minnesota and 20 in Pennsylvania. What the poster is trying to say is that it is POSSIBLE to average percentages but it is almost never (actually, practically always) not what you want to do. The above answer is true, but is not very well explained. Compare this to his 40/100 on the other test, and you get 100/200, or 50%. Put another way, by Harry scoring 30 out of 50 on two tests, it would be as if he scored 60/100 on one test. His final percentage for the class would be 50%, not 53.3%. Add up his actual scores, and divide by the total possible points. I say this is not meaningful, because to get his final percentage score for the class, you would have to use weighted averages. So, specifically answering "what is is average percentage score," the answer would be D, 53.3 percent ((.6+.6+.4)/3). On two of his tests, Harry scored 30 out of a maximum of 50 and on the third he scored 40 out of a maximum of 100? What is his average percentage score on the three tests? An example here would be helpful, so I am borrowing from this message board. But it is important to note, that the average of percentages is not necessarily meaningful. To the simple question, "Can you average percentages," the answer is, yes. My answer comes with the disclaimer that I am in no way an expert, but here goes:
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