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.
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.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.CR: Chapter 13 Review
Problem 58CR
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2248ad32-7549-4acf-a044-4efbc21677c8%2F137a5f4e-9316-435c-a772-36e96b577d37%2Fiyn8bn_processed.png&w=3840&q=75)
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
![](/static/compass_v2/shared-icons/check-mark.png)
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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