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 μ = 58.0 kg and standard deviation σ = 7.4 kg. Suppose a doe that weighs less than 49 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 2400 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 55 kg. If the average weight is less than 55 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 55 kg (assuming a healthy population)? (Round your answer to four decimal places.) (d) Compute the probability that x < 59.9 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 x = 59.9 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 undernourished. Since the sample average is below 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.

Answer #1

a)

Here, μ = 58, σ = 7.4 and x = 49. We need to compute P(X <= 49).
The corresponding z-value is calculated using Central Limit
Theorem

z = (x - μ)/σ

z = (49 - 58)/7.4 = -1.22

Therefore,

P(X <= 49) = P(z <= (49 - 58)/7.4)

= P(z <= -1.22)

= 0.1112

b)

Expected number = 2400 * 0.1112 = 266.88

i.e. 267

c)

sigma(x) = 7.4/sqrt(55) = 0.9978

P(X <= 55) = P(z <= (55 - 58)/0.9978)

= P(z <= -3.01)

= 0.0013

d)

P(X <= 59.9) = P(z <= (59.9 - 58)/0.9978)

= P(z <= 1.9)

= 0.9713

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

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