Question

Suppose that the weight of an newborn fawn is Uniformly distributed between 1.6 and 3.7 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible.

a. The mean of this distribution is  

b. The standard deviation is  

c. The probability that fawn will weigh exactly 3.2 kg is P(x = 3.2) =  

d. The probability that a newborn fawn will be weigh between 1.9 and 3.6 is P(1.9 < x < 3.6) =  

e. The probability that a newborn fawn will be weigh more than 3.12 is P(x > 3.12) =

f. P(x > 2.4 | x < 3.1) =  

g. Find the 68th percentile.

Answer 1

a) mean of this distribution=(3.7+1.6)/2=2.65

b)

standard deviation=(3.7-1.6)/sqrt(12)=0.6062

c)

probability that fawn will weigh exactly 3.2 kg is P(x = 3.2) = 0

d)P(1.9 < x < 3.6) =(3.6-1.9)/(3.7-1.6)=0.8095

e)

P(x > 3.12)=(3.7-3.12)/(3.7-1.6)=0.2762

f)P(x > 2.4 | x < 3.1) =(3.1-2.4)/(3.1-1.6)=0.4667

g) 68th percentile =1.6+0.68*(3.7-1.6)=3.0280