Batareau et al . (2011) compared four nonlinear growth functions in modelling bo

Question

Batareau et al. (2011) compared four nonlinear growth functions in modelling body length and mass size-at-age data for the grizzly bear (Ursus arctos) in the Northwest Territories and Nunavut. The mature adult body mass for a male bear based on data from 119 bears was estimated to have a mean of 209 kg with a standard deviation of 57 kg based on the von Bertalanffy growth model. Treat these statistics as if they were population parameters. consider a hypothetical sample of 100 mature male bears in the Omineca territory northwest of Prince George and use the probability you find in part (a)

a)Find the probability that a mature male grizzly bear would have a body mass less than 225 kg.

b)Find the body mass m of mature male grizzly bears such that only 25% of the body mass falls above m for this population in northern Canada.

c)State the exact distribution of the number of mature male grizzly bears with a body mass less than 225 kg.

d)State a large-sample approximation for the distribution of the number of mature male grizzly bears with a body mass less than 225 kg in this sample.

e)Find the probability that 51 or 52 mature male bears in a sample of 100 such bears would have a body mass less than 225 kg using both the exact distribution from part (c) and the large-sample approximation in part (d). Compare your answers and discuss.

Explanation / Answer

mean = 209
std. dev. = 57
n = 100

a)
P(X < 225)
= P(z < (225 - 209)/(57/sqrt(100)) )
= P(z < 2.807)
= 0.9975

b)
We can find z-value for 25% = 0.6745

xbar = mean + z*std.dev./sqrt(n)
= 209 + 0.6745*(57/sqrt(100))
= 212.84465

c)
Distribuion:
p = 0.9975
n = 100

mean = 0.9975*100 = 99.75
std.dev. = sqrt(npq) = sqrt(100 * 0.9975 * (1-0.9975)) = 0.4994

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