SOLUTION: 1(b) The average number of mosquitoes in a stagnant pond is 70 per square meter with a standard deviation of 8. If 36 square meters are chosen at random for a mosquito count, find
Algebra.Com
Question 923903: 1(b) The average number of mosquitoes in a stagnant pond is 70 per square meter with a standard deviation of 8. If 36 square meters are chosen at random for a mosquito count, find the probability that the average of those counts is more than 70.9 mosquitoes per square meter. Assume that the variable is normally distributed, and place your final answer in decimal form.
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
mean = 70 with a standard deviation of 8
sample: n = 36
z(70.9) = = .675
P(avg > 70.9) = P(z > .675) = normalcdf(.675, 100) = .2498
Using TI 0r similarly a Casio fx-115 ES plus
RELATED QUESTIONS
The average number of mosquitoes in a stagnant pond is 70 per square meter with a... (answered by Mmacri10)
The average number of mosquitoes in a stagnant pond is 80 per square meter with a... (answered by ewatrrr)
The average number of mosquitos in a stagnant pond is 80 per square meter with a standard (answered by Boreal,stanbon)
What is the critical value for these? How do I work them out to get the answer?
1.... (answered by ewatrrr)
A scientist is worried about a new disease being spread by mosquitoes in an area. The... (answered by Boreal)
The population of a colony of mosquitoes obeys the law of uninhibited growth. If there... (answered by Boreal)
Suppose that the average number of billable hours for attorneys in large firms is 36 with (answered by ikleyn)
1.
Convert x = 70 to z-score if Normal Distribution has mean = 50 and Standard Deviation (answered by ewatrrr)
Anna has a rectangular shape of fish pond in her backyard. The pond is in the shape of a... (answered by josmiceli)