Question

Let x be a random variable that represents the weights in kilograms (kg) of healthy aut female deer does) in December in a national park. Thenxhas a distribution that is approximately normal with mean = 69.0 kg and standard deviation - 9.0 kg. Suppose a doe that weighs less than 60 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 2800 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 rangehuse the rule that the average weight of 55 does should be more than 66 kg. If the average weight is less than 66 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 66 kg (assuming a healthy population)? (Round your answer to four decimal places.) (0) Compute the probably that < 70.9 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 70.9 kg. Do you think the doe population is undernourished or not? Explain 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 underourished. Since the sample average is below the mean, it is quite unlikely that the doe population is underourished. Since the sample average is below the mean, it is uitekely that the doe population is underourished

Answer 1

a)

μ= 69 ,σ= 9

P(x < 60)

z= (x -μ)/σ
= ( 60 - 69)/9
= -1
P(x < 60) = P(z < -1) = 0.1587

b)

np = 2800 * 0.1587 = 444.36 = 444

c)
n =55
P(x< 66)

z= (x -μ)/(σ/sqrt(55))
= ( 66 - 69)/(9/sqrt(55))
=-2.4721

P(x < 66) = P(z < -2.4721) = 0.0067

d)

n = 55
P(x< 70.9)

z= (x -μ)/(σ/sqrt(55))
= ( 70.9 - 69)/(9/sqrt(55))
=1.5656

P(x < 70.9) = P(z < 1.5656) = 0.9413

#a)Since the sample average is above the mean, it is quite likely that the doe population is undernourishe

#option a is correct