Find the​ mean, variance, and standard deviation of the binomial distribution with the given values of n and p.

n equals 128​, p equals 0.33

To find the mean, variance, and standard deviation of a binomial distribution, we use the following formulas:

1. **Mean** (\(\mu\)): 
   \[
   \mu = n \times p
   \]
   
2. **Variance** (\(\sigma^2\)): 
   \[
   \sigma^2 = n \times p \times (1 - p)
   \]
   
3. **Standard Deviation** (\(\sigma\)): 
   \[
   \sigma = \sqrt{n \times p \times (1 - p)}
   \]

Given:
- \( n = 128 \)
- \( p = 0.33 \)

Let's calculate each one step by step.

### 1. Mean (\(\mu\))
\[
\mu = 128 \times 0.33 = 42.24
\]

### 2. Variance (\(\sigma^2\))
\[
\sigma^2 = 128 \times 0.33 \times (1 - 0.33) = 128 \times 0.33 \times 0.67 \approx 28.2816
\]

### 3. Standard Deviation (\(\sigma\))
\[
\sigma = \sqrt{128 \times 0.33 \times 0.67} \approx \sqrt{28.2816} \approx 5.319
\]

### Summary:
- **Mean (\(\mu\))**: 42.24
- **Variance (\(\sigma^2\))**: 28.2816
- **Standard Deviation (\(\sigma\))**: 5.319

Would you like any additional details or have any questions?

Here are some related questions:
1. How does the value of \(p\) affect the mean in a binomial distribution?
2. What is the impact of increasing \(n\) on the standard deviation?
3. How would you interpret the standard deviation in this context?
4. Can a binomial distribution have a variance larger than its mean?
5. What conditions must be met for a distribution to be considered binomial?

**Tip:** The standard deviation provides insight into how much variation exists around the mean in a binomial distribution.

Learn how to calculate the mean, variance, and standard deviation of a binomial distribution with parameters n=128 and p=0.33. Understand the formulas involved and their application step-by-step.

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