Calculating Data Set w/ MEAN=100 & STD DEV=15 & n=3

  • Context: MHB 
  • Thread starter Thread starter fieldmusic123
  • Start date Start date
  • Tags Tags
    Derive Mean Reverse
Click For Summary

Discussion Overview

The discussion revolves around the possibility of reverse calculating a data set given a specific mean (100) and standard deviation (15) with a sample size of n=3. Participants explore methods to generate such a data set, including the use of Excel functions and mathematical equations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions whether it is possible to reverse derive a data set with a known mean and standard deviation, acknowledging that multiple datasets could satisfy these conditions.
  • Another participant suggests that for larger datasets, one could assume values based on a normal distribution, but notes that for n=3, a system of equations can be set up to find the values.
  • It is proposed that generating random numbers using the Excel function =NORMINV(RAND(), mean, stdev) can yield values that appear normally distributed, but these may not meet the specified mean and standard deviation.
  • A participant elaborates on how to manipulate generated values in Excel to ensure the resulting data set meets the required statistical properties, providing a mathematical framework for adjusting the values.

Areas of Agreement / Disagreement

Participants express varying methods for generating a data set with the specified mean and standard deviation, but there is no consensus on a definitive solution. The discussion remains open with multiple approaches presented.

Contextual Notes

The discussion involves assumptions about the nature of the data set and the limitations of using random generation methods in Excel. The mathematical steps provided are not fully resolved, leaving some uncertainty in the approach.

fieldmusic123
Messages
7
Reaction score
0
If I know the mean and standard deviation, but I don't have an original data set, is it possible to reverse calculate a data set that has this known mean and std dev? Is there an excel function I could use to do this easily? I know there could be multiple correct datasets and that's fine, I'm just wondering if it's possible to reverse derive a correct one that has the known mean and std dev.

So, for a MEAN = 100 and STD. DEV= 15, can anyone tell me how to calculate a data set with n=3?

Many thanks.
 
Physics news on Phys.org
I've wondered this same problem myself. The problem is much more difficult with larger data sets (you would just have to assume certain values), although with a large enough data set, you can just go with a normal distribution, and generate a whole raft of values from a known normal distribution. That procedure might not be exact, but it would be close. The larger the data set the closer it would be.

For $n=3$, you can set up a system of equations:
\begin{align*}
100&=\frac{x_1+x_2+x_3}{3} \\
15^2&=\frac12\left[(x_1-100)^2+(x_2-100)^2+(x_3-100)^2\right]
\end{align*}
Since you have three unknowns and two equations, this is an underdetermined system. Just pick one value, and the other two will be determined.
 
fieldmusic123 said:
If I know the mean and standard deviation, but I don't have an original data set, is it possible to reverse calculate a data set that has this known mean and std dev? Is there an excel function I could use to do this easily? I know there could be multiple correct datasets and that's fine, I'm just wondering if it's possible to reverse derive a correct one that has the known mean and std dev.

So, for a MEAN = 100 and STD. DEV= 15, can anyone tell me how to calculate a data set with n=3?

Many thanks.

Hi fieldmusic123! Welcome to MHB! (Smile)

It's [m]=NORMINV(RAND(), mean, stdev)[/m].

To explain, the CDF of some x has the range 0 to 1.
When we generate a random number between 0 and 1 from the uniform distribution, we can use the inverse CDF to find the corresponding random x value.
 
I like Serena said:
Hi fieldmusic123! Welcome to MHB! (Smile)

It's [m]=NORMINV(RAND(), mean, stdev)[/m].

To explain, the CDF of some x has the range 0 to 1.
When we generate a random number between 0 and 1 from the uniform distribution, we can use the inverse CDF to find the corresponding random x value.

Hey there, thanks for the welcome and your reply! NORMINV(RAND(), 100, 15) in Excel will give me 3 random numbers in what looks like a normal distribution but the 3 numbers as a combined data set do not give a mean of 100 and std dev of 15. Is there perhaps a way to use Excel to manipulate one of these values so that the set has the required mean (eg. 100) and std dev (eg. 15)? I'm a little rusty with complicated maths. Cheers, Sean
 
fieldmusic123 said:
Hey there, thanks for the welcome and your reply! NORMINV(RAND(), 100, 15) in Excel will give me 3 random numbers in what looks like a normal distribution but the 3 numbers as a combined data set do not give a mean of 100 and std dev of 15. Is there perhaps a way to use Excel to manipulate one of these values so that the set has the required mean (eg. 100) and std dev (eg. 15)? I'm a little rusty with complicated maths. Cheers, Sean

As Ackbach explained, you can generate $n-2$ numbers this way, say $x_3...x_n$, and calculate $x_1$ and $x_2$ to match from:
$$\begin{cases}\mu &= \frac{x_1+x_2 + \sum x_i}{n} &= 100 \\
\sigma^2 &= \frac{(x_1-100)^2 + (x_2-100)^2 + \sum (x_i-100)^2}{n} &= 15^2\end{cases} \\ \Rightarrow
\begin{cases}x_2 &= 100n - \sum x_i - x_1 &= A - x_1 \\
(x_1-100)^2 + (x_2-100)^2 &= 15^2n - \sum (x_i-100)^2 &= B\end{cases} \\
\Rightarrow\begin{cases}x_2 = A - x_1 \\
(x_1-100)^2 + (A - x_1-100)^2 = B\end{cases}
$$
The last equation is a quadratic equation that can be solved with the quadratic formula yielding $x_1$, after which the first equation gives $x_2$.
 
Dear Ackbach and I Like Serena, thanks very much for your help, now I understand. All the best.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
10K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
10K
  • · Replies 2 ·
Replies
2
Views
2K