Let's solve the given problem step-by-step.

We are tasked to find the mean ($\mu$), variance ($\sigma^2$), and standard deviation ($\sigma$) for a binomial distribution, given $n = 275$ and $p = 0.23$.

Formula Recap for a Binomial Distribution

1. Mean ($\mu$): $\mu = n \cdot p$

2. Variance ($\sigma^2$): $\sigma^2 = n \cdot p \cdot (1-p)$

3. Standard Deviation ($\sigma$): $\sigma = \sqrt{\sigma^2}$

Step 1: Calculate the Mean ($\mu$)

$\mu = 275 \cdot 0.23 = 63.25$

Step 2: Calculate the Variance ($\sigma^2$)

$\sigma^2 = 275 \cdot 0.23 \cdot (1 - 0.23)$ $\sigma^2 = 275 \cdot 0.23 \cdot 0.77 = 275 \cdot 0.1771 = 48.7525$

Step 3: Calculate the Standard Deviation ($\sigma$)

$\sigma = \sqrt{48.7525} = 6.981$

Final Answers

• Mean ($\mu$): $63.250$
• Variance ($\sigma^2$): $48.753$
• Standard Deviation ($\sigma$): $6.981$

Let me know if you'd like a more detailed breakdown of any step!

Related Questions

1. What happens to the mean, variance, and standard deviation if $p$ is increased?
2. How does the standard deviation change when $n$ is doubled?
3. Why does the variance formula include $1-p$?
4. How do you interpret the mean and standard deviation in the context of a binomial distribution?
5. What is the relationship between the mean and the mode in a binomial distribution?

Tip

Always ensure the value of $p$ lies between 0 and 1, as it represents a probability. If $p$ is too small or large, the distribution becomes highly skewed.