d. The probability that a newborn fawn will be weigh between 2.7 and 3.5 is P(2.7 3.3) = f. P(x > 2.7 | x< 3.3) =

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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please help me with part D E F G thank you.

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) =
e. The probability that a newborn fawn will be weigh more than 3.3 is P(x > 3.3) =
f. P(x > 2.7 | x< 3.3) =
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) = e. The probability that a newborn fawn will be weigh more than 3.3 is P(x > 3.3) = f. P(x > 2.7 | x< 3.3) = g. Find the 19th percentile.
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d)Given that x follows a uniform distribution between 2.1 and 3.6 kg.So the probability that newborn fawn will be weigh between 2.7 and 3.5 isP(2.7<x<3.5)=(3.5-2.7)(3.6-2.1)=0.5333

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