# Why is my weighted average much higher than the normal average?

For my profession I have to calculate the weighted average number of days it takes a business to review contracts. The count of each contract types reviewed is the weight, and the sum of days it takes to review is the number. The weighted average for this below data set is 35.6 days, however the raw data indicates that not a single contract took more than 25 days to review. Should my days be calculated as the average of days instead of sum? If that's the case, my weighted average is 8 days, which seems more accurate. I don't know why the weighted average using the Sum of Days is so much higher than what the raw data would tell me. I can't find my error - is using the Sum of the Days incorrect and I should be using the Average of days instead?

weighted average = Sum(contract count*Days in Review)/Sum(Contract Count)

Contract Type Contract Count Sum of Days in Review Average of Days in Review
Contract A 2 16 8
Contract B 2 8 4
Contract C 2 16 8
Contract D 1 0 0
Contract E 8 65 8.125
Contract F 1 13 13
Contract G 1 5 5
Contract H 2 29 14.5
Weighted Average with Sum of Days 35.6
Weighted Average with Average of Days 8
• Not sure what you've calculated but your sum of days is $172$ and your sum of contracts is $19$ so the correct mean contract days in review is $172\div19=9.05$. The weighted mean is always the correct mean. Where you calculated $8$ (the unweighted average), that's something slightly different. It's still a mean of sorts, but of something else, it's not the mean contract time in review. Apr 24 at 15:39
• On the Contract G row, is "5 average days" a typo and supposed to be 25? Every other row is correctly "average" = "sum"/"count"... Apr 24 at 15:44
• Contract G sum of days should have been 5, not 25, I have corrected the table. Apr 24 at 15:50
• Note that the question has been edited to replace a 25 by a 5, which explains a bit of discrepancy between answers, some answers counting a total of 172 and others counting a total of 152.
– Stef
Apr 25 at 9:33

The value $$35.6$$ is not a correct reflection of the weighted average of the number of days of review per contract.

In fact, I would avoid using the term "weighted average" here entirely. You have $$n = 19$$ contracts in your table. You want the average number of days of review per contract. So you would take the total number of days that these $$19$$ contracts were in review, then divide by $$19$$ to get the average. The total number of days of review is $$16+8+16+0+65+13+25+29 = 172$$. So the average number of days of review per contract is approximately $$\frac{172}{19} = 9.05263$$.

• Thanks, I'm being asked to provide a weighted average, and they want the weight to be the count of contract types. They want to contract types that have the highest volume to have higher weight. Apr 24 at 15:54
• @CaitlinB That's perfectly fine, but if your column "Contract Count" represents the volume of contracts of that particular type, then by taking the average in the way I have shown, you are doing the correct "weighting." The "naive" average is just the average of the last column. That value represents the average number of days in review where each contract type is granted the same "weight." In short, the calculation I showed is what you want to do; whether you want to call it "weighted" or "unweighted" is a semantic issue. Apr 24 at 15:59
• thank you for clarifying! Apr 24 at 16:05
• @Stef, there was an edit to one number in the post, not too long after the initial version, but after this answer was posted. Apr 25 at 9:50

Three approaches (the second and the third are equivalent):

• The simple average across contract types: the average of $$8,4,8,0,8.125,13,5,14.5$$ is about $$7.58$$ days per contract

• The overall average: a total of $$152$$ days to review a total of $$19$$ contracts to give an average of $$8$$ days per contract

• A weighted average across contract types, weighted by the number of contracts per type: $$8 \times \frac{2}{19}+4\times \frac{2}{19} + 8\times \frac{2}{19} + 0\times \frac{1}{19} + 8.125\times \frac{8}{19} + 13\times \frac{1}{19}+ 5\times \frac{1}{19}+ 14.5\times \frac{2}{19}$$ is again $$8$$ days per contract

There are more approaches, including looking at contracts per day though this would be infinite for Contract D. You could say a total of $$19$$ contracts reviewed in a total of $$152$$ days for an average of $$0.125$$ contracts per day, the reciprocal of the second approach above.

You want to compute the average number of review days per contract.

The label on your third column is confusing. Does the last row represent $$2$$ contracts each of which took $$29$$ days, or $$2$$ contracts each of which took $$14.5$$ days? I think you mean the former. In that case the (weighted) average for the last two rows would be

$$\frac{(1 \times 25 + 2 \times 29) \text{ review days}} {(1 + 2) \text{ contracts }} = 27.7 \frac{ \text{ review days}} { \text{ contracts }} .$$

• See my comment... Apr 24 at 15:43
• @it'sahirecarbaby That is indeed the correct solution with the second interpretation for the numbers in the third column.That's consistent with the fourth column. The OP must clarify. Apr 24 at 15:46
• Thanks, the last row represents 2 contracts that took a combined total of 29 days to review (second column) or an average of 14.5 days each to review (third column) Apr 24 at 15:59
• Then @it'sahirecarbaby has answered your question. (PS It's awkward for us when your edit changes the numbers in the question but not the substance.) Apr 24 at 16:04

The weighted average is $$\frac{\Sigma(group\ average)(number\ in\ group)}{\Sigma(number\ in\ group)}$$
In your case this equates to $$\frac{total\ days}{total\ contracts}$$
You could simply table these results in that manner for you presentation.
The result is, as already pointed out, $$\frac{172}{19} = 9.05$$

It looks like all you want to know is how long it takes on average for a typical contract to take.

Ideally, if we knew the individual times of each contract, we'd calculate the average very simply: $$T_{avg}=\frac{T_{total}}{N_{total}}=\frac{(A_1+A_2+A_3...)+(B_1+B_2+B_3...)+...}{N_A+N_B+...}=\frac{\sum T_{i}}{\sum N_i}$$ Where the top sums all of the individual contract times, and the bottom sums all of the typed contract counts. (I've written it out in a generic sense; your specific example doesn't have $$A_3$$..., $$B_3$$..., etc. to worry about, but you will have to account for your $$C_1$$...,$$D_1$$...,$$E_1$$..., etc.)

Unfortunately, in your example you don't know individual contract times, only group total times. But, if you look at the way I've used parentheses to group the $$A$$s above, you'll see that group totals still give us what we need. The $$A_1+A_2=16$$ you know tells you everything you need to know about the subtotal sum of $$(A_1+A_2+A_3...)$$, and likewise for your $$B$$s, etc. You essentially have: $$\frac{\sum T_{i}}{\sum N_i}=\frac{\sum A_{i}+\sum B_{i}...}{N_A+N_B...}=\frac{(16)+(8)+(16)+(0)+(65)+(13)+(5)+(29)}{2+2+2+1+8+1+1+2}$$ Which will give you a very reasonable average of 8.

None of this really involves "weighted" averages so much as regular everyday averages.