MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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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.](https://content.bartleby.com/qna-images/question/281e2029-de70-4b7c-8ebc-bf0179904e66/cba4387d-f997-4930-8619-2c8ffcd96eae/ujabtn_processed.png)
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.
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