Question

Suppose studies indicate that the earth's vegetative mass, or
biomass for tropical woodlands, thought to be about 35 kilograms
per square meter (kg/m^{2}), may in fact be too high and
that tropical biomass values vary regionally—from about 3 to 55
kg/m^{2}. 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/m^{2}).

*σ* = kg/m^{2}

(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.

Answer #1

a)

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

b)

sample size =n= | 400 |

std
error=σ_{x̅}=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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