Weighted Average

[Explain this!]

What is the weighted average for the following using the weighted average calculation as shown below

W₁X₁ + W₂X₂ + W_nX_n

W = relative weight (%)

X = value

This is seen at the following: https://www.financeformulas.net/Weighted_Average.html

For example, can the above formula be used for the following to determine the weighted average?

2 boxes @ $0.37
1 box @ $0.38
4 boxes @ $0.42

(1) Do the amounts for boxes represent the value and do the dollar amounts represent the relative weight or is it the opposite?

$0.87 X 2 + $0.38 X1 + $0.42 X 4 divided by 2 + 1 + 4

Or is it more accurate as shown below?

2 X $0.87 + 1 X $0.38 + 4 X $0.42 divided by 2 + 1 + 4


(2) Does the relative weight (W) need to be a percent?

Answer is $0.40 per box

I am not a high school student or a college student.

 

[Deleted member 4993]

What is the weighted average for the following using the weighted average calculation as shown below

W₁X₁ + W₂X₂ + W_nX_n

W = relative weight (%)

X = value

This is seen at the following: https://www.financeformulas.net/Weighted_Average.html

For example, can the above formula be used for the following to determine the weighted average?

2 boxes @ $0.37
1 box @ $0.38
4 boxes @ $0.42

(1) Do the amounts for boxes represent the value and do the dollar amounts represent the relative weight or is it the opposite?

$0.87 X 2 + $0.38 X1 + $0.42 X 4 divided by 2 + 1 + 4

Or is it more accurate as shown below?

2 X $0.87 + 1 X $0.38 + 4 X $0.42 divided by 2 + 1 + 4


(2) Does the relative weight (W) need to be a percent?

Answer is $0.40 per box

I am not a high school student or a college student.

2 boxes @ $0.37
1 box @ $0.38
4 boxes @ $0.42

Your question is incomplete hence cannot be answered as posted. Please answer the following before we begin to investigate:

Why are you trying to calculate price per box?

From where did the data come - why does those prices differ?

Do those boxes hold same thing (e.g. 12 loaves of bread or 10 packets of 5 hot cars, etc.)?

 

[Explain this!]

Why are you trying to calculate price per box? I want to determine the average cost per box

From where did the data come - why does sic (do) those prices differ? They differ because they represent different types of cereal, as an example.

Do those boxes hold same thing (e.g. 12 loaves of bread or 10 packets of 5 hot cars, etc.)? They represent boxes of cereal, but the boxes can represent whatever you want.

I do not understand why you have asked these questions.

 

[Dr.Peterson]

For example, can the above formula be used for the following to determine the weighted average?

2 boxes @ $0.37
1 box @ $0.38
4 boxes @ $0.42

Weighted average (in this application) is just a shortcut for an ordinary average. What you need to keep in mind is the intended meaning of the number you want to calculate. (That, I think, is the reason for the questions.)

So rather than ask for a "weighted average" (for which the "weights" could be anything you can imagine), you need to focus on the specific goal, the average price of a box of cereal.

Line up all the boxes, marked with their prices:

$0.37 $0.37 $0.38 0.42 0.42 0.42 0.42​

What is the total price? Just add them up; or, as a shortcut, multiply each price by the number of boxes:

2*$0.37 + 1*$0.38 + 4*$0.42​

How many boxes are there? Count them up, or, as a shortcut, add the quantities:

2 + 1 + 4​

Now divide to find the average.

That's all a weighted average is. In this case, the "weights" are the number of boxes of each, because the prices are what you want to average.

 

[Explain this!]

I wan to thank you for the reply. I have this follow-up question.

Is this then correct? The "value" represents the prices of the boxes of cereal and the "weights" represents the number of boxes.
To correspond with the weighted average formula, as shown below, where X equals prices of cereal and W equals the number of boxes, would this calculation be correct?

2 X $0.87 + 1 X $0.38 + 4 X $0.42 divided by 2 + 1 + 4

(W₁X₁ + W₂X₂ + [/SUB] W₂V_$0.87 + W₁V_$0.38 + W₄V_$0.42 divided by W₂ + W₁ +W₄

W = relative weight (%)

X = value

 

[Dr.Peterson]

I wan to thank you for the reply. I have this follow-up question.

Is this then correct? The "value" represents the prices of the boxes of cereal and the "weights" represents the number of boxes.
To correspond with the weighted average formula, as shown below, where X equals prices of cereal and W equals the number of boxes, would this calculation be correct?

2 X $0.87 + 1 X $0.38 + 4 X $0.42 divided by 2 + 1 + 4

(W₁X₁ + W₂X₂ + W₂V_$0.87 + W₁V_$0.38 + W₄V_$0.42 divided by W₂ + W₁ +W₄

W = relative weight (%)

X = value

We never use subscripts as values. I'd write this:

[MATH]\frac{W_1X_1+W_2X_2+W_3X_3}{W_1+W_2+W_3} = \frac{2 \cdot $0.87 + 1 \cdot $0.38 + 4 \cdot $0.42 }{2+1+4}[/MATH]​

Also, in this application, the weights are not relative; they are amounts, not percentages. There is no need to use percentages here.

 

[Harry_the_cat]

W1X1 + W2X2 + WnXn

W = relative weight (%)

X = value

If you want to use this "formula", then W is best described, in your context as proportion of boxes.

Type A: 2 boxes @ $0.37
Type B: 1 box @ $0.38
Type C: 4 boxes @ $0.42

So you have 2 + 1 + 4 = 7 boxes in all.
\(\displaystyle \frac{2}{7}\) of those are Type A; \(\displaystyle \frac{1}{7}\) of those are Type B; \(\displaystyle \frac{4}{7}\) of those are Type C.
(You can convert those proportions to percentages but why bother!)

So using your formula you have \(\displaystyle \frac{2}{7}*0.37 +\frac{1}{7}*0.38+\frac{4}{7}*0.42 \).

This, of course, is the same as \(\displaystyle \frac{2*0.37 + 1* 0.38 + 4 * 0.42}{7}\) as in Dr P's post above.

 

[Explain this!]

We never use subscripts as values. I'd write this:

[MATH]\frac{W_1X_1+W_2X_2+W_3X_3}{W_1+W_2+W_3} = \frac{2 \cdot $0.87 + 1 \cdot $0.38 + 4 \cdot $0.42 }{2+1+4}[/MATH]​

Also, in this application, the weights are not relative; they are amounts, not percentages. There is no need to use percentages here.

I want to thank you for the follow-up reply.

I am still somewhat confused by the subscripts 1, 2, and 3 as shown in the formula. In W₁X₁, Does W₁ represent 2 and does X₁ represent $0.87, etc.?

I think that they must since the division has W₁ + W₂ + W_3. These are the weights. Therefore the X₁, X₂, and X₃ must be the values or prices in this example.

 

[Dr.Peterson]

Yes, I thought that was reasonably clear. I told you that the weights are the quantities and the data are the prices, and I showed W₁ replaced by 2, etc. The subscripts 1, 2, 3 mean "first, second, third". Are you unfamiliar with subscripts?

 

[Explain this!]

Yes, I thought that was reasonably clear. I told you that the weights are the quantities and the data are the prices, and I showed W₁ replaced by 2, etc. The subscripts 1, 2, 3 mean "first, second, third". Are you unfamiliar with subscripts?

Yes, I am unfamiliar with subscripts, but it is now clear in this example what they represent.

Thanks again!

 

[Explain this!]

I want to thank you for the follow-up reply.

I am still somewhat confused by the subscripts 1, 2, and 3 as shown in the formula. In W₁X₁, Does W₁ represent 2 and does X₁ represent $0.87, etc.?

I think that they must since the division has W₁ + W₂ + W_3. These are the weights. Therefore the X₁, X₂, and X₃ must be the values or prices in this example.

If a sigma sign is used in front of 2⋅$0.87+1⋅$0.38+4⋅$0.422+1+4 the numerator and 2+1+4 the denominator, what would the top and bottom indices look like?

 

[Deleted member 4993]

If a sigma sign is used in front of 2⋅$0.87+1⋅$0.38+4⋅$0.422+1+4 the numerator and 2+1+4 the denominator, what would the top and bottom indices look like?

Are you sure that you want ....+1+4 there?

 

[Dr.Peterson]

If a sigma sign is used in front of 2⋅$0.87+1⋅$0.38+4⋅$0.42 the numerator and 2+1+4 the denominator, what would the top and bottom indices look like?

It would be [MATH]\frac{\sum_{i=1}^{3}W_iX_i}{\sum_{i=1}^{3}W_i}[/MATH].

In this notation, you can't include the specific values of [MATH]W_i[/MATH] and [MATH]X_i[/MATH], if that is what you are asking for.

 

[Explain this!]

Are sure that you want ....+1+4 there?

Sorry about that ! It should be 2 X $0.87 + 1 X $0.38 + 4 X $0.42 divided by 2 + 1 + 4

It would be [MATH]\frac{\sum_{i=1}^{3}W_iX_i}{\sum_{i=1}^{3}W_i}[/MATH].

In this notation, you can't include the specific values of [MATH]W_i[/MATH] and [MATH]X_i[/MATH], if that is what you are asking for.

Yes, but why can't the specific values be used with a sigma sign?

 

[Dr.Peterson]

How would you write it? Give me an example of what you think you could do!

The purpose of Sigma is to write a long sum briefly, without writing an actual list of terms.

 

[Explain this!]

How would you write it? Give me an example of what you think you could do!

The purpose of Sigma is to write a long sum briefly, without writing an actual list of terms.

I have no idea how to write it! It is too complicated for me to express it with a Sigma sign.

Thanks for the reply!

 

[Explain this!]

It would be [MATH]\frac{\sum_{i=1}^{3}W_iX_i}{\sum_{i=1}^{3}W_i}[/MATH].

In this notation, you can't include the specific values of [MATH]W_i[/MATH] and [MATH]X_i[/MATH], if that is what you are asking for.

I have another follow-up question. In you Sigma summation example, you used 3 as the top index. Does the 3 indicate the number of addends or the weights 1, 2, and 4. The index 3 cannot represent 1, 2, and 4. There is not a weight of 3. I think that the top index should be n.

 

[Deleted member 4993]

I have another follow-up question. In you Sigma summation example, you used 3 as the top index. Does the 3 indicate the number of addends or the weights 1, 2, and 4. The index 3 cannot represent 1, 2, and 4. There is not a weight of 3. I think that the top index should be n.

In your problem, W₃ = 4, and i = 3

 

[Dr.Peterson]

The index i represents the subscripts; it ranges from 1 to 3. This is not the same as the values of the quantities.

Have you read about sigma notation, in order to understand what it means? Try here:

Sigma Notation

 

[Explain this!]

In your problem, W₃ = 4, and i = 3

I'm not sure what you are referring to. There is no weight of 3.