Solve the problem.
In a certain town, 57% of voters favor a given ballot measure. For groups of 35 voters, find the variance for the number who favor the measure.
2.9

8.6

20

73.6

Question 8
10 pts
Provide an appropriate response. Round to the nearest hundredth.
Find the standard deviation for the given probability distribution.
P(x)
0
1
2
0.19
0.20
0.10
0.29
4 0.22
0 = 2.11
0 = 1.52
0 = 1.45
о = 2.59

### Problem 1: Variance of the number who favor the measure

We are given that 57% of voters favor the measure, and we're interested in finding the **variance** for the number of voters who favor the measure in groups of 35 voters. This situation follows a **binomial distribution**, where:

- \( n = 35 \) (the number of trials or voters),
- \( p = 0.57 \) (the probability that a single voter favors the measure),
- \( q = 1 - p = 0.43 \) (the probability that a single voter does not favor the measure).

The **variance** of a binomial distribution is given by the formula:
\[
\text{Variance} = n \times p \times q
\]
Substituting the values:
\[
\text{Variance} = 35 \times 0.57 \times 0.43
\]

Let’s calculate this.

The variance for the number of voters who favor the measure is approximately **8.58**.

### Problem 2: Standard deviation for the given probability distribution

For the second problem, we are given a probability distribution with the following values for \( P(x) \):

| x  | P(x)  |
|----|-------|
| 0  | 0.19  |
| 1  | 0.20  |
| 2  | 0.10  |
| 3  | 0.29  |
| 4  | 0.22  |

We need to calculate the **standard deviation** for this distribution. To do so, we first need to find the **mean** \( \mu \) of the distribution, and then the variance, which will give us the standard deviation.

1. **Mean (Expected Value)**:
\[
\mu = \sum (x \times P(x))
\]
2. **Variance**:
\[
\text{Variance} = \sum \left( (x - \mu)^2 \times P(x) \right)
\]
3. **Standard Deviation** is the square root of the variance.

Let’s calculate these values.The results for the given probability distribution are:

- The **mean** \( \mu \) is approximately **2.15**.
- The **variance** is approximately **2.11**.
- The **standard deviation** \( \sigma \) is approximately **1.45**.

### Answers:
- **Variance**: 2.11
- **Standard deviation**: 1.45

Would you like more details on any of these steps, or is there another question you have in mind? Here are some related questions for further exploration:

1. How does the variance of a binomial distribution change with different values of \( p \) and \( n \)?
2. How would the standard deviation change if the probabilities for the values of \( x \) were altered?
3. Can you explain how the formula for variance and standard deviation works for discrete distributions?
4. What is the relationship between the mean, variance, and standard deviation in a probability distribution?
5. How would the mean and variance differ for a normal distribution with the same parameters?

**Tip**: For binomial distributions, remember that the standard deviation can be calculated as the square root of the variance: \( \sigma = \sqrt{npq} \). This helps simplify the calculation.

In a certain town, 57% of voters favor a given ballot measure. For groups of 35 voters, find the variance for the number who favor the measure.

Provide an appropriate response. Round to the nearest hundredth.
Find the standard deviation for the given probability distribution.
P(x)
0 1 2 0.19 0.20 0.10 0.29 4 0.22
0 = 2.11
0 = 1.52
0 = 1.45
о = 2.59

Learn how to calculate the variance and standard deviation for a group of 35 voters where 57% favor a measure. This includes the calculation of variance using a binomial distribution and determining the standard deviation for a given probability distribution. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation