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

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SUMMARY

This discussion focuses on reverse calculating a data set with a specified mean of 100 and a standard deviation of 15, using Excel. The participants confirm that while it is possible to generate random numbers that approximate a normal distribution using the function =NORMINV(RAND(), 100, 15), achieving the exact mean and standard deviation for a small sample size (n=3) requires additional manipulation. A system of equations is established to derive the values, where one value can be chosen freely, and the others are calculated accordingly. This method effectively addresses the challenge of creating a valid data set under the given constraints.

PREREQUISITES
  • Understanding of basic statistics, specifically mean and standard deviation
  • Familiarity with Excel functions, particularly NORMINV and RAND
  • Knowledge of solving systems of equations
  • Basic understanding of normal distribution concepts
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  • Explore advanced statistical functions in Excel, such as NORM.DIST and NORM.S.DIST
  • Study the properties of normal distributions and their applications in data analysis
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This discussion is beneficial for statisticians, data analysts, and Excel users who need to generate or manipulate data sets to meet specific statistical criteria, particularly in educational or analytical contexts.

fieldmusic123
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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.
 
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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.
 

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