a. assume that the tropical monsoon forest tree tectonia grandis is randomy dist

Question

a. assume that the tropical monsoon forest tree tectonia grandis is randomy distributed at a density of 159 trees per hectare. if two 10x10 meter (100m2) plots are randomly chosen, then what is the probability that one of these plots has one of these treese while the other plot has tow or more of these trees?

b. also, from the data above: if one 100 m2 plot is randomly chosen, what is the probability of finding at least one of the teak trees in such a plot . and what is the standard deviation for the distributution in this problem

-please show work

Explanation / Answer

A)

As there are 159 trees/ha, and 1 ha = 10000 m^2, then there is an average of 159*100/10000 = 1.59 trees per 100m^2.

For the first plot:

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    1.59      
          
x = the number of successes =    1      
          
Thus, the probability is          
          
P (    1   ) =    0.324241723

For the second plot:

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    1.59      
          
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   1   ) =    0.528167334
          
Thus, the probability of at least   2   successes is  
          
P(at least   2   ) =    0.471832666

Hence,

P(first has one tree, second has 2 or more) = 0.324241723*0.471832666 = 0.152987837 [ANSWER]

***********************************

b)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    1.59      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.203925612

Thus, P(at least 1) = 1 - P(0) = 0.796074388 [ANSWER]

As for Poisson variables,

standard deviation = sqrt(mean)

standard deviation = sqrt(1.59) = 1.260952021 [ANSWER]

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