Question

Suppose that the weight of an newborn fawn is Uniformly distributed between 2.5 and 4 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 3.7 kg is P(x = 3.7) = d. The probability that a newborn fawn will be weigh between 2.9 and 3.5 is P(2.9 < x < 3.5) = e. The probability that a newborn fawn will be weigh more than 3.3 is P(x > 3.3) = f. P(x > 2.9 | x < 3.7) = g. Find the 59th percentile.

Answer 1

X ~ U (2.5 , 4)

a) mean = (4 + 2.5) / 2 = 3.25

b) standard deviation = (4 - 2.5) / sqrt(12) = 0.433

c) P(X = 3.7) = 0

d) P(2.9 < X < 3.5) = (3.5 - 2.9) / (4 - 2.5) = 0.4

e) P(X > 3.3) = (4 - 3.3) / (4 - 2.5) = 0.4667

f) P(X > 2.9 | X < 3.7) = P(X > 2.9 and X < 3.7) / P(X < 3.7) = P(2.9 < X < 3.7) / P(X < 3.7) = [(3.7 - 2.9) / (4 - 2.5)] / [(3.7 - 2.5) / (4 - 2.5)] = 0.6667

g) P(X < x) = 0.59

or, (x - 2.5) / (4 - 2.5) = 0.59

or, x = 3.385