​32% of likely U.S. voters think that the federal government should get more involved in fighting local crime. You randomly select six likely U.S. voters and ask them whether they think that the federal government should get more involved in fighting local crime. The random variable represents the number of likely U.S. voters who think that the federal government should get more involved in fighting local crime.

This problem involves a **binomial distribution** because you're dealing with a fixed number of trials (6 voters), each trial has two possible outcomes (a voter either thinks the federal government should get more involved or they don't), and the probability of success (in this case, a voter thinking the government should get more involved) is fixed.

### Key parameters:
- **Number of trials (n)**: 6 (since you're selecting 6 voters)
- **Probability of success (p)**: 0.32 (since 32% of voters think the federal government should get more involved)
- **Probability of failure (q)**: 1 - p = 1 - 0.32 = 0.68

The random variable \(X\) represents the number of likely U.S. voters (out of 6) who think the federal government should get more involved.

To calculate probabilities or moments related to this distribution, we can apply the **binomial probability formula**:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
\]

Where:
- \(P(X = k)\) is the probability that exactly \(k\) voters think the government should get more involved,
- \(\binom{n}{k}\) is the binomial coefficient representing combinations, calculated as \(\frac{n!}{k!(n-k)!}\),
- \(p^k\) represents the probability of success raised to the number of successes,
- \((1 - p)^{n - k}\) is the probability of failure raised to the number of failures.

### 1. Mean and variance of the binomial distribution:
- **Mean** \( \mu \) of a binomial distribution: 
  \[
  \mu = n \cdot p = 6 \cdot 0.32 = 1.92
  \]
  So, on average, 1.92 out of 6 voters will think the government should get more involved.

- **Variance** \( \sigma^2 \) of a binomial distribution:
  \[
  \sigma^2 = n \cdot p \cdot (1 - p) = 6 \cdot 0.32 \cdot 0.68 = 1.3104
  \]

### Next Steps:
You can use the binomial distribution formula to compute the probability of different values of \(k\) (e.g., \(P(X = 2)\), \(P(X = 3)\)) or answer specific questions regarding the distribution.

Would you like to calculate specific probabilities or go into more details on any aspect?

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### Related Questions:
1. What is the probability that exactly 3 out of 6 voters think the federal government should get more involved in fighting local crime?
2. What is the probability that fewer than 2 voters think the government should get more involved?
3. What is the cumulative probability that at least 4 voters think the government should get more involved?
4. How would the mean and variance change if you surveyed 10 voters instead of 6?
5. Can we approximate this binomial distribution with a normal distribution? Under what conditions?

### Tip:
In binomial problems, when \(n\) is large and \(p\) is neither very small nor very large, the binomial distribution can be approximated by a normal distribution using the central limit theorem.

32% of likely U.S. voters think that the federal government should get more involved in fighting local crime. You randomly select six likely U.S. voters and ask them whether they think that the federal government should get more involved in fighting local crime. The random variable represents the number of likely U.S. voters who think that the federal government should get more involved in fighting local crime.

Explore a binomial distribution problem where 32% of likely U.S. voters think the federal government should get more involved in local crime. Learn to calculate the probability of outcomes for six randomly selected voters using key formulas. Understand the mean, variance, and more, suitable for Grades 10-12 students.

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