I have normal distributed data but i don´t know the mean neither the stdv.
What i know is that 67.9% of my data points are below 2500g and 0.21% are over 3000g.
How can I estimate the mean and stdv of my population?
Thanks!!
6
-
$\begingroup$ Is this homework? $\endgroup$– TimCommented Mar 27, 2019 at 12:23
-
$\begingroup$ No, its for work. $\endgroup$– RamonCommented Mar 27, 2019 at 12:36
-
1$\begingroup$ I am assuming it as i am measuring fish weight $\endgroup$– RamonCommented Mar 27, 2019 at 12:42
-
2$\begingroup$ This is a very strong assumption, that would have profound effect on your results. It can lead to very wrong conclusions it the assumption does not hold. $\endgroup$– TimCommented Mar 27, 2019 at 12:46
-
1$\begingroup$ I know but is the best i can do, i am not going to take important decisions with the resultsits just to have an idea $\endgroup$– RamonCommented Mar 27, 2019 at 12:52
|
Show 1 more comment
2 Answers
$\begingroup$
$\endgroup$
While you can compute some numbers, they'll be too uncertain to be of any practical value.
Moreover, you likely won't even be able to estimate the uncertainty.
As @asdf said, you can theoretically solve this using the quantile function (which converts your fractions to $z$ values under the very strong assumption of a normal distribution) and then solving the resulting linear equation system.
However, consider
Even under the assumption of normality, this calculation isn't very stable: already considering small errors on the input numbers gives you noticeable error on the distribution estimates.
The normality assumption is really not appropriate here, and that assumption has an even stronger effect on the outcome. As animals that come in positive sizes, fish will typically have an asymmetric weight distribution with a heavy tail to the right (i.e. heavy animals).
As your 2nd data point is at the .9979 quantile, it is exactly in that region where one would expect the actual fish weight distribution to deviate from the normal distribution.
This is what really makes these calculations useless.We can give lower limits, though they are probabyl useless from a practical point of view: μ cannot be below 800 g, and σ not below 22 g.
Already your knowledge about the fish will suggest a much better range for the values.
answered Mar 27, 2019 at 16:08
$\begingroup$
$\endgroup$
I am not 100% sure about this approach, but I think it works.
2500 is the quantile 0.679 of your distribution. The same quantile in the standard normal is at 0.465
3000 is the quantile 0.9979 (=1-0.0021). The corresponding quantile in the standard normal is 2.863
Now you need an affine mapping $f(x)=\sigma x + \mu$ such that $f(0.465) = 2500$ and $f(2.863) = 3000$ We get the following two equation system:
$0.465 \sigma + \mu = 2500$
$2.863 \sigma + \mu = 3000$
Now we solve, so that $\sigma = 209.2$ and $\mu=2401.6$ (with a lot of rounding errors, since I am doing this by hand)
Anyway, pay close attention to the comments posted, as the normality assumption may be way off from reality
