Suppose that the weight of an newborn fawn is Uniformly distributed between 2.1 and 3.6 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. a. The mean of this distribution is 2.85 b. The standard deviation is 0.4330 c. The probability that fawn will weigh exactly 3.4 kg is P(x = 3.4) = 0 d. The probability that a newborn fawn will be weigh between 2.7 and 3.5 is P(2.7 3.3) = 0.2 f. P(x > 2.7 | x < 3.3) = 0.5 g. Find the 19th percentile.

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Suppose that the weight of an newborn fawn is Uniformly distributed between 2.1 and 3.6 kg.
Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when
possible.
a. The mean of this distribution is 2.85
b. The standard deviation is 0.4330
c. The probability that fawn will weigh exactly 3.4 kg is P(x = 3.4) = 0
d. The probability that a newborn fawn will be weigh between 2.7 and 3.5 is P(2.7 <x< 3.5) =
0.5333
e. The probability that a newborn fawn will be weigh more than 3.3 is P(x > 3.3) = 0.2
f. P(x > 2.7 | x < 3.3) = 0.5
g. Find the 19th percentile.
Transcribed Image Text:Suppose that the weight of an newborn fawn is Uniformly distributed between 2.1 and 3.6 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. a. The mean of this distribution is 2.85 b. The standard deviation is 0.4330 c. The probability that fawn will weigh exactly 3.4 kg is P(x = 3.4) = 0 d. The probability that a newborn fawn will be weigh between 2.7 and 3.5 is P(2.7 <x< 3.5) = 0.5333 e. The probability that a newborn fawn will be weigh more than 3.3 is P(x > 3.3) = 0.2 f. P(x > 2.7 | x < 3.3) = 0.5 g. Find the 19th percentile.
Expert Solution
Step 1

If X ~U(a,b) 

Then 

E(X)= b+a/2

V(X)= (b-a)^2/12

19th percentile is equal to 19/100

 

 

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