Question

For each of the following, determine whether the given values can serve as the values of a probability distribution of a random variable with the range $x=1,2,3$, and 4: (a) $f(1)=0.25, f(2)=0.75, f(3)=0.25$, and $f(4)=$ $-0.25$ (b) $f(1)=0.15, f(2)=0.27, f(3)=0.29$, and $f(4)=0.29 ;$ (c) $f(1)=\frac{1}{19}, f(2)=\frac{10}{19}, f(3)=\frac{2}{19}$, and $f(4)=\frac{5}{19}$.

Name: Problem 1, Chapter 3: John E. Freund's Mathematical Statistics with Applications
Uploaded: 2023-06-27T13:23:48-08:00
Duration: 3 min 10 s
Channel: Vysakh M
Description: For each of the following, determine whether the given values can serve as the values of a probability distribution of a random variable with the range x=1,2,3, and 4: (a) f(1)=0.25, f(2)=0.75, f(3)=0.25, and f(4)= -0.25 (b) f(1)=0.15, f(2)=0.27, f(3)=0.29, and f(4)=0.29 ; (c) f(1)=(1)/(19), f(2)=(10)/(19), f(3)=(2)/(19), and f(4)=(5)/(19).

   For each of the following, determine whether the given values can serve as the values of a probability distribution of a random variable with the range $x=1,2,3$, and 4:
(a) $f(1)=0.25, f(2)=0.75, f(3)=0.25$, and $f(4)=$
$-0.25$
(b) $f(1)=0.15, f(2)=0.27, f(3)=0.29$, and $f(4)=0.29 ;$
(c) $f(1)=\frac{1}{19}, f(2)=\frac{10}{19}, f(3)=\frac{2}{19}$, and $f(4)=\frac{5}{19}$.

John E. Freund's Mathematical Statistics with Applications

Irwin Miller,… 8th Edition

Chapter 3, Problem 1

Instant Answer

Solved by Expert Vysakh M

06/27/2023

Step 1

In this case, we have: $f(1) + f(2) + f(3) + f(4) = 0.25 + 0.75 + 0.25 + (-0.25) = 1$ However, $f(4) = -0.25$ is not between 0 and 1, so this cannot be a probability distribution. Show more…

Show all steps

Thanks for your feedback!

For each of the following, determine whether the given values can serve as the values of a probability distribution of a random variable with the range $x=1,2,3$, and 4: (a) $f(1)=0.25, f(2)=0.75, f(3)=0.25$, and $f(4)=$ $-0.25$ (b) $f(1)=0.15, f(2)=0.27, f(3)=0.29$, and $f(4)=0.29 ;$ (c) $f(1)=\frac{1}{19}, f(2)=\frac{10}{19}, f(3)=\frac{2}{19}$, and $f(4)=\frac{5}{19}$.

Play audio

Feedback

Vysakh M and 73 other subject Intro Stats / AP Statistics educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Probability Distribution

A probability distribution assigns probabilities to the outcomes of a random variable. Each probability must be between 0 and 1, and the sum of the probabilities for all outcomes must equal exactly 1. This ensures that the distribution properly represents the likelihood of all possible events.

Non-Negativity Constraint

The non-negativity condition requires that all individual probabilities are greater than or equal to zero. Negative values are not allowed since they do not make sense within the framework of probability theory, which measures the likelihood of an event occurring.

Normalization

Normalization is the requirement that the sum of the probabilities of all outcomes must be exactly 1. This condition guarantees that the probability distribution accounts for the entire sample space, ensuring that the total probability is complete and exhaustive.

Please give Ace some feedback