Question

Suppose studies indicate that the earth's vegetative mass, or biomass for tropical woodlands, thought to be...

Suppose studies indicate that the earth's vegetative mass, or biomass for tropical woodlands, thought to be about 35 kilograms per square meter (kg/m2), may in fact be too high and that tropical biomass values vary regionally—from about 3 to 55 kg/m2. Suppose you measure the tropical biomass in 400 randomly selected square-meter plots.

(a)

Approximate σ, the standard deviation of the biomass measurements (in kg/m2).

σ =  kg/m2

(b)

What is the probability that your sample average is within two units of the true average tropical biomass? (Use your answer from part (a) in your calculations. Round your answer to four decimal places.)

(c)

If your sample average is x = 32.75, what would you conclude about the overestimation that concerns the scientists?

Because z is less than 3 standard deviations below the mean, it is more likely that the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical woodlands.Because z is less than 3 standard deviations below the mean, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.    Because z is more than 3 standard deviations below the mean, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.Because z is more than 3 standard deviations below the mean, it is more likely that the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical woodlands.Because z is negative, it is not likely that the mean is an overestimate of the mean biomass for tropical woodlands.

You may need to use the appropriate appendix table or technology to answer this question.

Homework Answers

Answer #1

a)

from range rule : Approximate σ =range/4 =(55-3)/4=13

b)

sample size       =n= 400
std error=σ=σ/√n= 0.6500

  probability that your sample average is within two units of the true average tropical biomass:

probability =P((-2-0)/0.65)<Z<(2-0)/0.65)=P(-3.08<Z<3.08)=0.999-0.001=0.9980

c)

z score =(X-mean)/standard deviation =(32.75-35)/0.65 = -3.46

Because z is more than 3 standard deviations below the mean, it is more likely that the scientists are correct in assuming that the mean is an overestimate of the mean biomass for tropical woodlands

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