Answer:
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:
[tex]z(X)=\frac{X-\mu }{\frac{\sigma}{\sqrt{n}}}[/tex]
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:
[tex]z(X)=\frac{81.8 - 80}{\frac{12}{\sqrt{36}}}\\z(X)= 0.9[/tex]
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:
[tex]P= 100\% - 81.59\%\\P=18.41\%[/tex]
