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 = 7.3 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.)


(b) If the park has about 2650 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 = 45 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 45 does is less than 67 kg (assuming a healthy population)? (Round your answer to four decimal places.)


(d) Compute the probability that

x

< 71.2 kg for 45 does (assume a healthy population). (Round your answer to four decimal places.)

Answer 1

Answer :

Given data is :

Mean =70.0 kg

Standard deviation = 7.3kg

a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished?

mean = 70

Standard deviation = 7.3

P(X < 61) = P(X <(61 - 70)/7.3)

= P (Z<-9 / 7.3)

=P (Z<-1.2329)

=0.1093

p value = 0.1093

b) If the park has about 2650 does, what number do you expect to be undernourished in December?

Given n = 2650

so ,

expected number = n * p value

= 2650 * 0.1093

= 289.645

= 290

expected number = 290

c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 45 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 45 does is less than 67 kg.

Given n = 45 , E(Xbar) = 70 ,

sd(Xbar) = 7.3/sqrt(n)

= 7.3/sqrt(45)

= 7.3 / 6.708

=1.09

p =P(Z < 67 - 70/sd)

= P (Z<−3 / 1.09)

= P (Z<−2.75)

=0.0030

P Value = 0.0030

d) Compute the probability that x < 71.2 kg for 45 does

P(Xbar < 71.2) = P(Z < 71.2 - 70 / 1.09)

=P(Z < 1.2 / 1.09)

= P(Z < 1.10)

= 0.8643

P value = 0.8643

so here, the sample average is greater the mean, it is quite not liked ,that the doe population is undernourished.