Min, max of a series from median, average and sample size

Question

I have value of $25\text%$, $75\text%$ percentile of a series as well as median, average and sample size.

Is there any way that I can get Min, max of the series from this data? I remember some formula that helps us get a range but can't seem to remember that now.

Thanks You

PS: My current approach is to assume my distribution as Normal (even though its not) and calculate IQR and average standard deviation. Using Standard deviation I am calculating min-max within a confidence interval.

Answer 1

$25$ and $75$ percentile and the median are respectively $Q_1,Q_3,Q_2$ ($1^{st},3^{rd},2^{nd}$ quartile).
Let the arithmetic mean or the average be $A$ and sample size be $N$.

Given this much data alone, you can't find the Min ($m$) and Max ($M$) of the data.

Consider the following:

$1,2,3,4,5,6,100$
Here, $Q_1=2, Q_2=4,Q_3=6,A=\frac{121}7,N=7.$
You can't know the values of $m(=1)$ or $M(=100)$ from the above values as they could be anything.

Iff you are provided more information such as:

Its a Poisson distribution.
The data follows Geometric Progression.
The distribution is approximately normal.
The data is symmetric.
The data follows some other distribution or progression. Etc.

Provided above information you may find the min/max values of the data.

You said:

I remember some formula that helps us get a range but can't seem to remember that now.

Here, you could be talking about Inter-Quartile Range or perhaps The Box Plot.
These still can't help you find the min/max values of the data. The latter actually uses the min/max values to give other types of information regarding the data while the former, as explained above, may help finding them under certain conditions.

Note: I have used "data" in place of "series" which you have used in your question for purposes you should now understand.

Example:

Given data/series follows Geometric Progression and $Q_1=2, Q_2(\text{Median})=8, Q_3=32, A(\text{Mean})=\frac{127}7,N=7$. Find the min and the max values of the data/series.

$Q_1=2=a_{\frac{N+1}4}=a_2,\qquad Q_2=8=a_{\frac{N+1}2}=a_4,\qquad Q_3=32=a_{\frac{3(N+1)}4}=a_6$.
Thus, common ratio, $r=2$
So, min value, $m=1^{st} \text{ term}=a=a_2/r=1$
And max value, $M=\text{Last term}=a_7=(a_6)(r)=64$
Here, $A$ is extra (not-need/unwanted) information.