The average number of mosquitoes in a stagnant pond is 80 per square meter with a standard deviation of 12. if 36 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 81.8 mosquitoes per square meter. assume that the variable is normally distributed.
a) 18.4%
b) 81.6%
c) 0.3%
d) 31.6%.

a) 18.4%

Step-by-step explanation:

Assuming a normal distribution, the z-score (z) for the probability of the average of the mosquitoes count being 'X'  is given by:

Where 'μ' is the distribution mean, 'σ' is the standard deviation and 'n' is the number of square meters chosen.

For X = 81.8 and n=36:

A z-score of 0.9 is equivalent to the 81.59 th percentile in a normal distribution.

Therefore, the probability (P) that the average of those counts is more than 81.8 mosquitoes per square meter is:

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: