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:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

the line of