A number X is randomly selected from the interval [10, 15]. The PDF of X is 1 15 - 10 0 1. Find the expectation E[X] of X. 2. Find the variance Var(X) of X. (E[X], Var(X)) = 12.5000,2.0833 fx(x) = if x = [10, 15], Otherwise.
A number X is randomly selected from the interval [10, 15]. The PDF of X is 1 15 - 10 0 1. Find the expectation E[X] of X. 2. Find the variance Var(X) of X. (E[X], Var(X)) = 12.5000,2.0833 fx(x) = if x = [10, 15], Otherwise.
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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How was the correct answer of 12.5000 and 2.0833 attained?
![A number X is randomly selected from the interval [10, 15]. The PDF of X is
1
15 - 10
0
1. Find the expectation E[X] of X.
2. Find the variance Var(X) of X.
(E[X], Var(X)) = 12.5000,2.0833
fx(x) =
if x = [10, 15],
Otherwise.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1467f162-7185-45ab-925a-2589d3c8cce7%2Fb88487a2-5cd2-47c9-9639-36e19357af8b%2F8ezmxzk_processed.png&w=3840&q=75)
Transcribed Image Text:A number X is randomly selected from the interval [10, 15]. The PDF of X is
1
15 - 10
0
1. Find the expectation E[X] of X.
2. Find the variance Var(X) of X.
(E[X], Var(X)) = 12.5000,2.0833
fx(x) =
if x = [10, 15],
Otherwise.
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