0
$\begingroup$

Two forest patches have, respectively, $100$ and $200$ teak trees of the same age. In a given season, all trees shed some of their leaves at random. The daily total collections of the leaf litter from the two patches are expected to have

$(1)$ nearly equal means, standard deviations and coefficients of variation

$(2)$ different means, nearly equal standard deviations and coefficients of variation

$(3)$ different means, nearly equal standard deviation and different coefficients of variation

$(4)$ different means, and standard deviations but nearly equal coefficients of variation

What conclusion can we make from the above data? What I know is that the mean $\mu = \dfrac {\sum x} {n}, \ $ standard deviation $\sigma = \sqrt {\dfrac {\sum x^2} {n} - \left (\dfrac {\sum x} {n} \right )^2}$ and coefficients of variation $\nu = \dfrac {\sigma} {\mu},$ where $n =$ sample size (in this case the number of trees in each of the forest patch) and $\sum x$ represents the total number of leaves collected from each forest patch in a day. Now how do I proceed? Any help in this regard will be appreciated.

Thanks for your time.

$\endgroup$
5
  • $\begingroup$ Start with the means. Do you expect the average number of leaves shed in the grove of $100$ trees to be the same as the number shed in the grove of $200$ trees or not? $\endgroup$
    – saulspatz
    Nov 21, 2020 at 6:09
  • $\begingroup$ @saulspatz I think they are same if the number of observation increases. $\endgroup$ Nov 21, 2020 at 6:11
  • $\begingroup$ How can that be? The trees are of the same species, and the same age, so we expect them each to shed about the same number of leaves on any given day. Surely $200$ trees would shed around twice as many leaves as $100$ trees. $\endgroup$
    – saulspatz
    Nov 21, 2020 at 6:15
  • $\begingroup$ Oh! I see @saulspatz. I counted the mean for every tree of the either patch. They are same. What about the standard deviations? Will the standard deviation also be twice as large? $\endgroup$ Nov 21, 2020 at 6:18
  • $\begingroup$ (4) Assuming (approx) Poisson distribution of leaves shed $\endgroup$
    – BruceET
    Nov 21, 2020 at 17:19

1 Answer 1

1
$\begingroup$

The point of the problem is that the number of leaves shed by each tree on a given day is to be considered a random variable, and all the variables are independent and identically distributed. So in one case, we have the sum of $100$ i.i.d. random variables, and in the other case we have the sum of $200$ i.i.d. random variables. Let the mean and variance of each be $\mu$ and $\sigma^2$ respectively.

By linearity of expectation, the mean of the first patch is $100\mu$ and that of the second patch is $200\mu$. Also, the variance of the sum of independent random variables is the sum of the variances, so the variance of the first patch is $100\sigma^2$ and the standard deviation is $10\sigma$; the standard deviation of the second patch is $10\sigma\sqrt2$.

$\endgroup$
2
  • $\begingroup$ Since we are talking about mean of each random variable $X$ that means we are observing $X$ for some days. It is very less likely that all the days of observation are calm (resp. windy). So your last paragraph doesn't make much sense to me. The number of leaves shed by each tree is independent of the other because it is commonly believed that no tree makes any effect on shedding leaves to the other. Except that part your answer seems very sound to me. Thank you so much. $\endgroup$ Nov 22, 2020 at 6:05
  • $\begingroup$ @Phibetakappa Yes, I was pretty sleepy when I wrote that. $\endgroup$
    – saulspatz
    Nov 22, 2020 at 15:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .