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 = 70.0 kg and standard deviation σ = 8.0 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.) Incorrect: Your answer is incorrect. (b) If the park has about 2700 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.) Incorrect: Your answer is incorrect. does (c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 60 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 60 does is less than 67 kg (assuming a healthy population)? (Round your answer to four decimal places.) Incorrect: Your answer is incorrect. (d) Compute the probability that x < 71 kg for 60 does (assume a healthy population). (Round your answer to four decimal places.) Incorrect: Your answer is incorrect.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	70
std deviation   =σ=	8.000

probability =P(X<61)=(Z<(61-70)/8)=P(Z<-1.13)=0.1292

b)

expected number =2700*0.1292 =349

c)

sample size       =n=	60
std error=σx̅=σ/√n=	1.0328

probability =P(X<67)=(Z<(67-70)/1.033)=P(Z<-2.9)=0.0019

d)

probability =P(X<71)=(Z<(71-70)/1.033)=P(Z<0.97)=0.8340