5.82 Carbon Tax. A poll commissioned by Friends of the Earth and conducted by the Mellman Group found that 72% of American vot- ers are in favor of a carbon tax. Suppose that six voters in the United States are randomly sampled and asked whether they favor a car- bon tax. Determine the probability that the number answering in the affirmative is a. exactly two. d. Determine the probability distribution of the number of American b. exactly four. c. at least two.
5.82 Carbon Tax. A poll commissioned by Friends of the Earth and conducted by the Mellman Group found that 72% of American vot- ers are in favor of a carbon tax. Suppose that six voters in the United States are randomly sampled and asked whether they favor a car- bon tax. Determine the probability that the number answering in the affirmative is a. exactly two. d. Determine the probability distribution of the number of American b. exactly four. c. at least two.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
Please answer parts d and e in he following question given parts a b and c
Transcribed Image Text:**5.82 Carbon Tax**
A poll commissioned by Friends of the Earth and conducted by the Mellman Group found that 72% of American voters are in favor of a carbon tax. Suppose that six voters in the United States are randomly sampled and asked whether they favor a carbon tax. Determine the probability that the number answering in the affirmative is:
a. exactly two.
b. exactly four.
c. at least two.
d. Determine the probability distribution of the number of American voters in a sample of six who favor a carbon tax.
e. Strictly speaking, why is the probability distribution that you obtained in part (d) only approximately correct? What is the exact distribution called?
![**Solution:**
**a) \( P(X = 2) \)**
\[
P(X = 2) = \binom{6}{2} (0.72)^2 (0.28)^{6-2}
\]
\[
= 15 \times 0.5184 \times 0.0061
\]
\[
P(X = 2) = 0.0478
\]
**b) \( P(X = 4) \)**
\[
P(X = 4) = \binom{6}{4} (0.72)^4 (0.28)^{6-4}
\]
\[
= 15 \times 0.2687 \times 0.0784
\]
\[
P(X = 4) = 0.3159
\]
**c) \( P(X \geq 2) \)**
\[
P(X \geq 2) = 1 - P(X < 2)
\]
\[
= 1 - \left[ P(X = 0) + P(X = 1) \right]
\]
\[
= 1 - \left[ \binom{6}{0} (0.72)^0 (0.28)^6 + \binom{6}{1} (0.72)^1 (0.28)^{6-1} \right]
\]
\[
= 1 - \left[ 0.00048 + 0.0074 \right]
\]
\[
= 1 - 0.00788
\]
\[
P(X \geq 2) = 0.9921
\]
**Explanation:**
The solutions provide the calculations for three different probabilities of a binomial distribution:
1. **\( P(X = 2) \)**: The probability of exactly two successes in six trials.
2. **\( P(X = 4) \)**: The probability of exactly four successes in six trials.
3. **\( P(X \geq 2) \)**: The probability of two or more successes in six trials, calculated as one minus the probability of obtaining less than two successes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F281e2029-de70-4b7c-8ebc-bf0179904e66%2Fcba4387d-f997-4930-8619-2c8ffcd96eae%2Fujabtn_processed.png&w=3840&q=75)
Transcribed Image Text:**Solution:**
**a) \( P(X = 2) \)**
\[
P(X = 2) = \binom{6}{2} (0.72)^2 (0.28)^{6-2}
\]
\[
= 15 \times 0.5184 \times 0.0061
\]
\[
P(X = 2) = 0.0478
\]
**b) \( P(X = 4) \)**
\[
P(X = 4) = \binom{6}{4} (0.72)^4 (0.28)^{6-4}
\]
\[
= 15 \times 0.2687 \times 0.0784
\]
\[
P(X = 4) = 0.3159
\]
**c) \( P(X \geq 2) \)**
\[
P(X \geq 2) = 1 - P(X < 2)
\]
\[
= 1 - \left[ P(X = 0) + P(X = 1) \right]
\]
\[
= 1 - \left[ \binom{6}{0} (0.72)^0 (0.28)^6 + \binom{6}{1} (0.72)^1 (0.28)^{6-1} \right]
\]
\[
= 1 - \left[ 0.00048 + 0.0074 \right]
\]
\[
= 1 - 0.00788
\]
\[
P(X \geq 2) = 0.9921
\]
**Explanation:**
The solutions provide the calculations for three different probabilities of a binomial distribution:
1. **\( P(X = 2) \)**: The probability of exactly two successes in six trials.
2. **\( P(X = 4) \)**: The probability of exactly four successes in six trials.
3. **\( P(X \geq 2) \)**: The probability of two or more successes in six trials, calculated as one minus the probability of obtaining less than two successes.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman