QUESTION 6 Suppose a random variable X has probability distribution function 2kr q²+I? for x = 2, 3 kx P(X = x) = for x = 4,5 %3| r² –1 ' 0, otherwise. where k is a numerical constant. Determine the expected value of X, using the value of k found. 6.228 2.938 8.452 3.586
QUESTION 6 Suppose a random variable X has probability distribution function 2kr q²+I? for x = 2, 3 kx P(X = x) = for x = 4,5 %3| r² –1 ' 0, otherwise. where k is a numerical constant. Determine the expected value of X, using the value of k found. 6.228 2.938 8.452 3.586
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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![**Question 6**
Suppose a random variable \( X \) has a probability distribution function
\[
P(X = x) =
\begin{cases}
\frac{2kx}{x^2 + 1}, & \text{for } x = 2, 3 \\
\frac{kx}{x^2 - 1}, & \text{for } x = 4, 5 \\
0, & \text{otherwise.}
\end{cases}
\]
where \( k \) is a numerical constant.
**Determine the expected value of \( X \), using the value of \( k \) found.**
- ○ 6.228
- ○ 2.938
- ○ 8.452
- ○ 3.586](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fae8ee7c4-79f3-4258-9a28-9911a1447d64%2F1e1bf336-b99c-40a8-ab88-2bb5976069d4%2Fao2nheg_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 6**
Suppose a random variable \( X \) has a probability distribution function
\[
P(X = x) =
\begin{cases}
\frac{2kx}{x^2 + 1}, & \text{for } x = 2, 3 \\
\frac{kx}{x^2 - 1}, & \text{for } x = 4, 5 \\
0, & \text{otherwise.}
\end{cases}
\]
where \( k \) is a numerical constant.
**Determine the expected value of \( X \), using the value of \( k \) found.**
- ○ 6.228
- ○ 2.938
- ○ 8.452
- ○ 3.586
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