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

In summary: Hi, Thanks for your reply!There is a function in Excel that will do this for you, it's called NORMINV. You can find it in the Analysis Tools>Nonlinear Regression>Linear Regression>General Linear Models menu. Hope this helps!
  • #1
fieldmusic123
7
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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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 $x_2$.
 
  • #6
Dear Ackbach and I Like Serena, thanks very much for your help, now I understand. All the best.
 

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

How do you calculate the mean of a data set with a given mean, standard deviation, and sample size?

To calculate the mean of a data set with a given mean, standard deviation, and sample size, you can use the formula: mean = given mean + (standard deviation / square root of sample size). In this case, the mean would be 100 + (15 / √3) = 100 + 8.66 = 108.66.

What does the standard deviation represent in a data set?

The standard deviation represents the average amount of variation or dispersion of the data points from the mean. It is a measure of how spread out the data is from the mean.

How does the sample size affect the calculation of the mean in a data set?

The sample size affects the calculation of the mean by determining the weight of each data point in the calculation. The larger the sample size, the more accurate the mean will be as it takes into account a larger number of data points.

Can you have a negative standard deviation in a data set?

No, the standard deviation is always a positive value. It represents the distance of the data points from the mean, so it cannot be negative.

How can you use the mean and standard deviation to analyze a data set?

The mean and standard deviation can be used to determine the normal distribution of a data set and identify any outliers. They can also be used to compare different data sets and see how they differ in terms of central tendency and variability.

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