Question

# Let x be a random variable that represents the weights in kilograms (kg) of healthy adult...

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 = 61.0 kg and standard deviation σ = 8.8 kg. Suppose a doe that weighs less than 52 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 = 75 does should be more than 58 kg. If the average weight is less than 58 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 75 does is less than 58 kg (assuming a healthy population)? (Round your answer to four decimal places.) (d) Compute the probability that x < 62.2 kg for 75 does (assume a healthy population). (Round your answer to four decimal places.) Suppose park rangers captured, weighed, and released 75 does in December, and the average weight was x = 62.2 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 below the mean, it is quite unlikely 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.

#### Homework Answers

Answer #1
 for normal distribution z score =(X-μ)/σx here mean=       μ= 61 std deviation   =σ= 8.8

a)

 probability =P(X<52)=(Z<(52-61)/8.8)=P(Z<-1.02)=0.1539

b)

expected number =np=2400*0.1539 =369.36 ~ 369

c)

 sample size       =n= 75 std error=σx̅=σ/√n= 1.01614
 probability =P(X<58)=(Z<(58-61)/1.016)=P(Z<-2.95)=0.0016

d)

 probability =P(X<62.2)=(Z<(62.2-61)/1.016)=P(Z<1.18)=0.8810

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

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