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%.

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pedrocarratore pedrocarratore · Answer 1 · 2019-11-24T02:37:30+01:00

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]