# Getting an average figure with ±

• Monocerotis
In summary, the person is asking for help in finding the average value for a series of measurements with varying errors. They attempted to use Wolfram Alpha but were unsuccessful. Another person suggested finding the weighted mean, but the original person believes it is a common mean. They are measuring slope figures on a v(t) graph and are trying to find the average value and error. The conversation ends with the original person still unsure of how to find their desired result.
Monocerotis
Gold Member

## Homework Statement

Quick question, working on a lab

Just wondering how I could go about getting an average value for a series of figures with +- values.

For instance
1) 1.5 ± 0.02
2) 1.7 ± 0.84
3) 1.7 ± 0.08

## The Attempt at a Solution

I tried enclosing the figures within parenthesis and then trying to find an average with wolfram alpha but no success.

I was just reading something on the net

would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?

Monocerotis said:
would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?
Incorrect, and very much so. Assuming the measurements are statistically independent (are they? There's no telling from the original post) then

1. That 1.5 measurement is considerably more precise than either of the other two measurements. You should get something closer the 1.5 than 1.7 for your final estimate.

2. You combined error is larger than the largest error. It should be smaller than any of individual errors.Google "weighted mean".

k I've been reading and it seems as though what I'm dealing with is a common mean rather than a weighted mean

I'm dealing with slope figures of a v(t) graph and trying to find g avg +- delta g avg as to compare it with my theoretical results

the values I gave above were ambigous but the figures I'm actually dealing with from my perspective appear to have equal weight so

still confused as to what to do however because even the common mean will only give me a single value

What exactly are you measuring here? Without details it is a bit hard to help you.

Yes, your approach is correct. To find the average value when dealing with uncertainties, you should use the weighted average method. This takes into account the individual uncertainties of each measurement. In your example, the weighted average would be 1.63 ± 0.844, as you calculated. Just make sure to include the units of the uncertainties in your final result.

## 1. What is the purpose of using ± when calculating an average figure?

The ± symbol, also known as the plus-minus symbol, is used to indicate the range of values that are within a certain distance from the calculated average figure. This allows for a more accurate representation of the data and accounts for any potential errors or variations in the data.

## 2. How do you calculate an average figure with ±?

To calculate an average figure with ±, first find the mean of the data set. Then, determine the standard deviation of the data. Finally, add and subtract the standard deviation from the mean to get the ± range of values.

## 3. Can ± be used for any type of data?

Yes, ± can be used for any type of data as long as the data is numerical and follows a normal distribution. This means that the data should be symmetrical, with most of the values falling near the mean and the remaining values spread out evenly on both sides.

## 4. What does the size of ± indicate about the data?

The size of ± indicates the level of variability in the data. A larger ± range indicates a higher level of variability, meaning that the data points are more spread out and there is more uncertainty in the average figure. A smaller ± range indicates a lower level of variability and a more precise average figure.

## 5. How is ± useful in interpreting data?

± is useful in interpreting data because it provides a more complete picture of the data by accounting for any potential errors or variations. It also allows for a more accurate comparison of data sets as it takes into account the variability of the data. Additionally, ± can help identify outliers or anomalies in the data that may affect the overall average figure.

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