Formula to weigh an average (currently is 61.76%; needs to be 75%)

[rmart100]

I have an average in a project of 61.76% and I need it to be at least 75%. I had supposed that just getting the quotient of the two (82.3%) would get me where I need to be, but I imagine that this is only true if n = 3. I don't know how many more instances I'll have in order to raise this average, so how would I go about creating a formula or data set around this?

 

[Dr.Peterson]

I have an average in a project of 61.76% and I need it to be at least 75%. I had supposed that just getting the quotient of the two (82.3%) would get me where I need to be, but I imagine that this is only true if n = 3. I don't know how many more instances I'll have in order to raise this average, so how would I go about creating a formula or data set around this?

We'll need to know more! How is the average calculated, and what can you change to increase the average? For example, is the current average based on 3 equally weighted parts out of 5, and you can get up to 100% on each of the remaining two? I don't even know what you mean by n. It sounds like maybe you don't have enough information, either.

 

[rmart100]

We'll need to know more! How is the average calculated, and what can you change to increase the average? For example, is the current average based on 3 equally weighted parts out of 5, and you can get up to 100% on each of the remaining two? I don't even know what you mean by n. It sounds like maybe you don't have enough information, either.

The current average is based on my efficiency over the current year. I could get a count of that but I don't know how many more opportunities I'll have to raise that efficiency by the end of the year. I was aiming for 75% efficiency. What I had initially done to try and solve the efficiency problem was take the quotient of my actual average and my intended efficiency and got the 82.34%. But I had realized that 82.34% would only be adequate if all 3 variables were of equal weight.

Ultimately what I'm looking for is, if my average is 61.76%, and I want it to be 75%, how efficient do I need to be, in order to raise that average to 75%? Like you said, I may not have enough information, but I could try gathering it if I knew exactly what I should be getting. I tried putting together a formula below, but I'm not sure if that's going to give me the result I need. Or if maybe I should be looking at this from another angle (I don't remember much about matrices, but I feel like it might be applicable here).

a: Efficiency of work
b: 75%
n: Instances of work
x: Instances of work remaining
y: Average efficiency needed


[(a)₁+ (a)_{2 ...} (a)_n / n] / [(y)₁+ (y)_{2 ...} (y)_x / x = b]

I realize this has turned into an algebra problem now, but I had hoped it was simpler than its turned out to be.

 

[Dr.Peterson]

The current average is based on my efficiency over the current year. I could get a count of that but I don't know how many more opportunities I'll have to raise that efficiency by the end of the year. I was aiming for 75% efficiency. What I had initially done to try and solve the efficiency problem was take the quotient of my actual average and my intended efficiency and got the 82.34%. But I had realized that 82.34% would only be adequate if all 3 variables were of equal weight.

Ultimately what I'm looking for is, if my average is 61.76%, and I want it to be 75%, how efficient do I need to be, in order to raise that average to 75%? Like you said, I may not have enough information, but I could try gathering it if I knew exactly what I should be getting. I tried putting together a formula below, but I'm not sure if that's going to give me the result I need. Or if maybe I should be looking at this from another angle (I don't remember much about matrices, but I feel like it might be applicable here).

a: Efficiency of work
b: 75%
n: Instances of work
x: Instances of work remaining
y: Average efficiency needed


[(a)₁+ (a)_{2 ...} (a)_n / n] / [(y)₁+ (y)_{2 ...} (y)_x / x = b]

I realize this has turned into an algebra problem now, but I had hoped it was simpler than its turned out to be.

It can turn out to be very simple algebra, or not even require that, if you can get the right information.

If we suppose that the 61.76% represents your average over m observations, and there are n more, then you can think of the overall average as the sum of m instances of 61.76 plus n instances of x (assuming you will be consistently better from now on), divided by m+n. So you just have to solve

(61.76m + nx)/(m + n) = 75

You would know m and n, so you'd plug them in and solve for x. Or, you could solve it as is and get a formula so you could plug in m and n when you find them.

Using your variables, the equation would be

(an + xy)/(n + x) = b

and you want to solve for y. (Hint: multiply by the denominator, then isolate y on one side.)

If you want to gradually increase your efficiency rather than jump to a new level immediately, then we could take that into account, but it would add more unknowns.

 

[rmart100]

It can turn out to be very simple algebra, or not even require that, if you can get the right information.

If we suppose that the 61.76% represents your average over m observations, and there are n more, then you can think of the overall average as the sum of m instances of 61.76 plus n instances of x (assuming you will be consistently better from now on), divided by m+n. So you just have to solve

(61.76m + nx)/(m + n) = 75

You would know m and n, so you'd plug them in and solve for x. Or, you could solve it as is and get a formula so you could plug in m and n when you find them.

Using your variables, the equation would be

(an + xy)/(n + x) = b

and you want to solve for y. (Hint: multiply by the denominator, then isolate y on one side.)

If you want to gradually increase your efficiency rather than jump to a new level immediately, then we could take that into account, but it would add more unknowns.

Thank you, that really helps. I don't actually know how many instances remain (m your variables, x my variables) or the efficiency I need to weigh it towards 75 (x yours, y mine). I suppose I can ballpark it and operate accordingly until I have a reason to run the equation again.

Again, thank you so much

 

[rmart100]

Thank you, that really helps. I don't actually know how many instances remain (m your variables, x my variables) or the efficiency I need to weigh it towards 75 (x yours, y mine). I suppose I can ballpark it and operate accordingly until I have a reason to run the equation again.

Again, thank you so much

I had run the numbers and got my needed efficiency to be 102.45%. I wasn't surprised by this at first, but I just recently tried increasing the number of instance I work to see if that was more feasible. Just reiterating the work I did initially in case I made an errors:

Terms:

a: 0.6176
b: 0.75
n: 141
x: 68
y: Unknown

(an + xy) / (n + x) = b

[(0.6176 * 141) + (68 * y)] / (141 + 68) = 0.75
87.0816 + 68y / 209 = 0.75
87.0816 + 68y = 156.75
87.0816/68 + y = 2.3051
1.2806 + y = 2.3051
y = 1.0245

Decided to see what it would be like with x = 85 instead
87.0816 + 85y / 226 = 0.75
87.0816 + 85y = 169.50
87.0816/85 + y = 1.9941
1.0245 + y = 1.9941
y = 0.9696

I'm sure if I keep raising x I can get a much more reasonable rate, but I was pretty surprised at how little it affected the overall rate I needed to improve by. It's not looking like an achievable goal at this point, but now I'm just interested in seeing what it would take to pull out of this rut. So I assumed y is .8, but I'm having trouble isolating x.

Plugging in all the values and then working the equation gives me something that works (373.36 additional instances for 80% efficiency going forward), but I can't work the equation this way only with variables. I could walk away with my answer, but I feel like I'm missing some fundamental math principle.

 

[Dr.Peterson]

I had run the numbers and got my needed efficiency to be 102.45%. I wasn't surprised by this at first, but I just recently tried increasing the number of instance I work to see if that was more feasible. Just reiterating the work I did initially in case I made an errors:

Terms:

a: 0.6176
b: 0.75
n: 141
x: 68
y: Unknown

(an + xy) / (n + x) = b

[(0.6176 * 141) + (68 * y)] / (141 + 68) = 0.75
87.0816 + 68y / 209 = 0.75
87.0816 + 68y = 156.75
87.0816/68 + y = 2.3051
1.2806 + y = 2.3051
y = 1.0245

Decided to see what it would be like with x = 85 instead
87.0816 + 85y / 226 = 0.75
87.0816 + 85y = 169.50
87.0816/85 + y = 1.9941
1.0245 + y = 1.9941
y = 0.9696

I'm sure if I keep raising x I can get a much more reasonable rate, but I was pretty surprised at how little it affected the overall rate I needed to improve by. It's not looking like an achievable goal at this point, but now I'm just interested in seeing what it would take to pull out of this rut. So I assumed y is .8, but I'm having trouble isolating x.

Well, first, let's solve the question for y so you don't have to keep doing that over and over. I'll just do essentially the same work you did, but on symbols:

(an + xy) / (n + x) = b

(an + xy) = b * (n + x)

xy = bn + bx - an

xy = n(b - a) + bx

y = (n(b - a) + bx)/x

y = n(b - a)/x + b

Now, for your given numbers, this is

y = 141(0.75 - 0.6176)/68 + 0.75 = 1.0245

y = 141(0.75 - 0.6176)/85 + 0.75 = 0.9696

Reversing this, if we take y as known and x unknown, we can solve my equation for x (a little easier than solving the original for x, because x appeared twice):

y = n(b - a)/x + b

y - b = n(b - a)/x

x(y - b) = n(b - a)

x = n(b - a)/(y - b)

For your numbers,

x = 141(0.75 - 0.6176)/(0.8 - 0.75) = 373.4

Yes, it can take a lot of time to increase an average! If you examine the formulas closely, you can see why. The "distance" you have to go (b-a) is almost three times the "distance" you've come (y-b), so it will take almost three times as long.

 

[rmart100]

Well, first, let's solve the question for y so you don't have to keep doing that over and over. I'll just do essentially the same work you did, but on symbols:

(an + xy) / (n + x) = b

(an + xy) = b * (n + x)

xy = bn + bx - an

xy = n(b - a) + bx

y = (n(b - a) + bx)/x

y = n(b - a)/x + b

Now, for your given numbers, this is

y = 141(0.75 - 0.6176)/68 + 0.75 = 1.0245

y = 141(0.75 - 0.6176)/85 + 0.75 = 0.9696

Reversing this, if we take y as known and x unknown, we can solve my equation for x (a little easier than solving the original for x, because x appeared twice):

y = n(b - a)/x + b

y - b = n(b - a)/x

x(y - b) = n(b - a)

x = n(b - a)/(y - b)

For your numbers,

x = 141(0.75 - 0.6176)/(0.8 - 0.75) = 373.4

Yes, it can take a lot of time to increase an average! If you examine the formulas closely, you can see why. The "distance" you have to go (b-a) is almost three times the "distance" you've come (y-b), so it will take almost three times as long.

Your equation makes it so much simpler. When I solved for y, I left the equation as y = (bn + bx -an)/x instead of combining terms so it didn't seem like a good starting point initially. I couldn't get past how to consolidate the x's. Just for posterity I'll post what I had done and where I got stuck:

(an + xy) / (n + x) = b
an + xy = b (n + x)
xy = b (n + x) - an

and that's where I got myself stuck. No matter what I did to try and get combine the x's, they would just wind up negating each other.

 

[Dr.Peterson]

Your equation makes it so much simpler. When I solved for y, I left the equation as y = (bn + bx -an)/x instead of combining terms so it didn't seem like a good starting point initially. I couldn't get past how to consolidate the x's. Just for posterity I'll post what I had done and where I got stuck:

(an + xy) / (n + x) = b
an + xy = b (n + x)
xy = b (n + x) - an

and that's where I got myself stuck. No matter what I did to try and get combine the x's, they would just wind up negating each other.

This is a classic type of problem where students tend to get stuck. You have x in two places, so you have to first expand (remove parentheses), then get the terms with x on the same side, and then factor x from those terms, so you have only one x in the equation, and then you can solve. Give it a try!

 

[rmart100]

This is a classic type of problem where students tend to get stuck. You have x in two places, so you have to first expand (remove parentheses), then get the terms with x on the same side, and then factor x from those terms, so you have only one x in the equation, and then you can solve. Give it a try!

I really appreciate your help. Just tried it out as you suggested. Had a little trouble because instead of multiplying x back over to the left I decided to move the rest of the equation over and was left with 1/x (Though I didn't realize that and had assumed I isolated the x on the right). I wound up figuring out my mistake as I tried to explain it here.

 

[Dr.Peterson]

I really appreciate your help. Just tried it out as you suggested. Had a little trouble because instead of multiplying x back over to the left I decided to move the rest of the equation over and was left with 1/x (Though I didn't realize that and had assumed I isolated the x on the right). I wound up figuring out my mistake as I tried to explain it here.

Good. That often happens.

In case you did something a little different from mine, and for the sake of other readers, here is how I would do it:

(an + xy) / (n + x) = b
an + xy = b(n + x)
an + xy = bn + bx
xy - bx = bn - an
x(y - b) = (b - a)n
x = (b - a)n/(y - b)