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 = 52.0 kg and standard deviation = 7.1 kg. Suppose a doe that weighs less than 43 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 2550 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 = 40 does should be more than 49 kg. If the average weight is less than 49 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight x (xbar) for a random sample of 40 does is less than 49 kg (assuming a healthy population)? (Round your answer to four decimal places.)

(d) Compute the probability that x (xbar) < 53.6 kg for 40 does (assume a healthy population). (Round your answer to four decimal places.)

(e) Suppose park rangers captured, weighed, and released 40 does in December, and the average weight was x (xbar) = 53.6 kg. Do you think the doe population is undernourished or not? Explain.

-Since the sample average is below the mean, it is quite likely that the doe population is undernourished.

-Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.    

-Since the sample average is above the mean, it is quite likely that the doe population is undernourished.

-Since the sample average is below the mean, it is quite unlikely that the doe population is undernourished.

Explanation / Answer

Let X be the weight of a randomly selected doe. Then, X is normally distributed with mean = 52.0 kg and standard deviation = 7.1 kg, sample size n = 40

Formulas needed: s (sample standard deviation) = /n z = (x - ) / s

s = 7.1 / 40 = 1.1226

a) Probability that a single doe captured (weighed and released) at random in December is undernourished is

P[X<43] = P[ Z<(43-52)/7.1] = P[Z< -1.2676] = 0.1025 (refer to z table or use a software)

b) The expected number is N*Prob. = 2550*0.1025 = 261.375 = 261(This is binomial distribution formula for expected value)

c) Z = (X-)/(/sqrtn) = (49-52)/1.1226 = -2.672368
Using Excel or a z-score table => p = 0.0038

d) Z= (53.6 - 52) / 1.1226 => 1.4253

Using Excel or a z-score table => p = 0.9230

e) Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.

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