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.