Let x be a random variable that represents the weights in kilograms (kg) of heal

Question

Let x be a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in a national park. Then x has a distribution that is approximately normal with mean = 50.0 kg and standard deviation = 8.7 kg. Suppose a doe that weighs less than 41 kg is considered undernourished.

(a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (Round your answer to four decimal places.)


(b) If the park has about 2100 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.)
_______does

(c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 45 does should be more than 47 kg. If the average weight is less than 47 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight x bar for a random sample of 45 does is less than 47 kg (assuming a healthy population)? (Round your answer to four decimal places.)

(d) Compute the probability that x bar < 51.2 kg for 45 does (assume a healthy population). (Round your answer to four decimal places.)

Explanation / Answer

Let X be the weight of a randomly selected doe. Then, X is normally distributed with mean = 50.0 kg and standard deviation = 8.7 kg
Let Z = (X-)/ = (X-50)/8.7 is a Standard normal variable.

(a)Probability that a single doe captured (weighed and released) at random in December is undernourished
is P[X<41] = P[ Z<(41-50)/8.7] = P[Z<-1.08] = 0.1515

b) The expected number is N*Prob. = 2100*0.1515 = 318.5.

c) If xbar is average weight of n=45 does, it is normal with mean = 50.0 kg and standard deviation /sqrtn = 8.7/sqrt45 kg
Z = (X-)/(/sqrtn) = (X-50)/1.298
P[xbar<47] = P[Z<-2.31] = 0.1044

d) P[xbar<51.2] = P[Z<0.92] = 0.8212

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