SOLUTION: Suppose that the weight of an newborn fawn is Uniformly distributed between 1.7 and 3.5 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when

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Question 1173129: Suppose that the weight of an newborn fawn is Uniformly distributed between 1.7 and 3.5 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 2.1 kg is P(x = 2.1) =
d. The probability that a newborn fawn will be weigh between 2.4 and 2.8 is P(2.4 < x < 2.8) =
e. The probability that a newborn fawn will be weigh more than 2.76 is P(x > 2.76) =
f. P(x > 2 | x < 3.1) =
g. Find the 78th percentile.


Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi  

Previously Posted

Suppose that the weight of an newborn fawn is Uniformly distributed between 1.7 and 3.5 kg. 

Uniform Distribution a = 1.7 and b= 3.5

f%28x%29+=+1%2F%28b-a%29+=+1%2F1.8+=+5%2F9, mu+=%28a%2Bb%29%2F2+=++5.2%2F2=+2.6+ and  +alpha+=+sqrt%28%28b-a%29%2F12%29+=+sqrt%28%281.8%29%2F12%29+=+.3873

Uniform and Continuous

P(x = 2.1) = 0  Continuous Distribution

P(2.4 < x < 2.8) = .4(5/9) = 2/9

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P(x > 2.76) = (3.5 - 2.76)(5/9) = .74(5/9)

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P(x > 2 | x < 3.1) = (3.5-2)(1/.4)      |f(x) = 1/(b-a) = 1/(3.5-3.1) = 1/.4 

78%

P(x < k) = %28k-1.7%29%281%2F.4%29+=+.78+

 k =   1.7 + (.4)(.78)

Wish You the Best in your Studies.