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 = 70.0 kg and standard deviation = 8.9 kg. Suppose a doe that weighs less than 61 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 2200 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 = 55 does should be more than 67 kg. If the average weight is less than 67 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight

x

for a random sample of 55 does is less than 67 kg (assuming a healthy population)? (Round your answer to four decimal places.)


(d) Compute the probability that x < 71.8 kg for 55 does (assume a healthy population). (Round your answer to four decimal places.)

Suppose park rangers captured, weighed, and released 55 does in December, and the average weight was x = 71.8 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 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 below the mean, it is quite unlikely that the doe population is undernourished.

n of does might be undernourished. what is the probbaty that the average weght x for a fandem sample of 55 does is ie rangers captured, weighed, and released S5 does in Decembat, and the average weght was 71.8 kg. Do you think the doe population is he sample verage i5 below the mean, it i5 gat likely that the population is undernourished. thù sampio average i5 above the meer·ttgat" urikely that the doe population is undernourished.

Explanation / Answer

(a) P(X<61) = P((X-mean)/s <(61-70)/8.9)

=P(Z<-1.02) = 0.1539 (from standard normal table)

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(b)2500*0.1539 = 385

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(c)P(xbar<52) = P((xbar-mean)/(s/vn) <(52-70)/(8.9/sqrt(65)))

=P(Z<-2.75) =0.0030 (from standard normal table)

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(d)P(xbar<64.6) = P((xbar-mean)/(s/vn) <(64.6-70)/(8.8/sqrt(45)))

=P(Z<1.83) =0.9664 (from standard normal table)

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

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