The problem involves a binomial distribution with $n = 510$ trials and success probability $p = 0.26$. Let's calculate the requested values:

1. Mean ($\mu$):

The mean of a binomial distribution is given by: $\mu = n \cdot p$ Substitute $n = 510$ and $p = 0.26$: $\mu = 510 \cdot 0.26 = 132.6$

2. Variance ($\sigma^2$):

The variance of a binomial distribution is given by: $\sigma^2 = n \cdot p \cdot (1 - p)$ Substitute $n = 510$, $p = 0.26$, and $1 - p = 0.74$: $\sigma^2 = 510 \cdot 0.26 \cdot 0.74 = 510 \cdot 0.1924 = 98.124$

3. Standard Deviation ($\sigma$):

The standard deviation is the square root of the variance: $\sigma = \sqrt{\sigma^2}$ Substitute $\sigma^2 = 98.124$: $\sigma = \sqrt{98.124} \approx 9.9$

Final Answers:

• Mean ($\mu$): $132.6$
• Variance ($\sigma^2$): $98.124$
• Standard Deviation ($\sigma$): $9.9$

Would you like me to provide further explanations or explore related concepts?

Related Questions:

1. How does the value of $p$ affect the variance in a binomial distribution?
2. What happens to the standard deviation if $n$ is doubled?
3. Can a binomial distribution be approximated to a normal distribution in this case? Why?
4. How do the mean and standard deviation change if $p$ increases to 0.5?
5. How do you interpret the standard deviation in the context of this problem?

Tip:

For large values of $n$, a binomial distribution can often be approximated by a normal distribution with the same mean and variance. This can simplify certain calculations.