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* = 67.0 kg and standard
deviation *σ* = 8.0 kg. Suppose a doe that weighs less than
58 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 2850 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* = 60
does should be more than 64 kg. If the average weight is less than
64 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 64 kg (assuming a
healthy population)? (Round your answer to four decimal
places.)

(d) Compute the probability that *x*< 68.6 kg for 60 does
(assume a healthy population). (Round your answer to four decimal
places.)

(c) Suppose park rangers captured, weighed, and released 60 does in
December, and the average weight was *x*= 68.6 kg. Do you
think the doe population is undernourished or not? Explain.

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.

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.

Answer #1

a)

for normal distribution z score =(X-μ)/σ |

probability =P(X<58)=(Z<(58-67)/8)=P(Z<-1.13)=0.1292 |

b)

expect to be undernourished in December =np=2850*0.1292 =368

c)

std deviation =σ= | 8.000 |

sample size =n= | 60 |

std error=σ_{x̅}=σ/√n= |
1.03280 |

probability =P(X<64)=(Z<(64-67)/1.033)=P(Z<-2.9)=0.0019 |

d)

probability =P(X<68.6)=(Z<(68.6-67)/1.033)=P(Z<1.55)=0.9394 |

e)

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

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 = 67.0 kg and standard
deviation σ = 8.1 kg. Suppose a doe that weighs less than
58 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...

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