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 = 55.0 kg and standard deviation σ = 8.2 kg. Suppose a doe that weighs less than 46 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 2600 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 = 70 does should be more than 52 kg. If the average weight is less than 52 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 70 does is less than 52 kg (assuming a healthy population)? (Round your answer to four decimal places.)


(d) Compute the probability that

x

< 56.9 kg for 70 does (assume a healthy population). (Round your answer to four decimal places.)


Suppose park rangers captured, weighed, and released 70 does in December, and the average weight was

x

= 56.9 kg. Do you think the doe population is undernourished or not? Explain.

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

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 55
standard deviation ( sd )= 8.2
----------------------------------------------------------------------------------------
(a)
P(X < 46) = (46-55)/8.2
= -9/8.2= -1.0976
= P ( Z <-1.0976) From Standard Normal Table
= 0.1362
probability that a single doe captured (weighed and released) at
random in December is undernourished is 13.62%
----------------------------------------------------------------------------------------
(b)
If the park has about 2600 does, number expect to be undernourished
is = 2600 * 13.62 = 354.12 ~ 324
----------------------------------------------------------------------------------------
(c)
sample size (n) = 70
mean of the sampling distribution ( x ) = 55
standard deviation ( sd )= 8.2/ Sqrt ( 70 ) =0.9801
P(X < 52) = (52-55)/8.2/ Sqrt ( 70 )
= -3/0.9801= -3.061
= P ( Z <-3.061) From Standard NOrmal Table
= 0.0011
that the entire population of does might be undernourished is 0.0011
----------------------------------------------------------------------------------------
(d)
P(X < 56.9) = (56.9-55)/8.2/ Sqrt ( 70 )
= 1.9/0.9801= 1.9386
= P ( Z <1.9386) From Standard NOrmal Table
= 0.9737
----------------------------------------------------------------------------------------
(e)
since the sample average is above the mean, it is quite likely that
the doe population is undernourished