For a binomial distribution $\text{Binomial}(n, p)$, the mean and variance are given by the following formulas:

• Mean: $\mu = np$
• Variance: $\sigma^2 = np(1 - p)$

We are given that the mean is 20 and the variance is 16. Thus, we have the following two equations:

1. $np = 20$
2. $np(1 - p) = 16$

Now, let's solve this system of equations step by step.

From the first equation, we have: $np = 20 \quad \Rightarrow \quad p = \frac{20}{n}$

Substitute this value of $p$ into the second equation:

$np(1 - p) = 16$

Substitute $p = \frac{20}{n}$ into the equation:

$n \cdot \frac{20}{n} \cdot \left(1 - \frac{20}{n}\right) = 16$

Simplifying this expression:

$20 \left(1 - \frac{20}{n}\right) = 16$

$20 - \frac{400}{n} = 16$

$\frac{400}{n} = 4$

$n = 100$

Now that we have $n = 100$, substitute this value back into the first equation to find $p$:

$np = 20 \quad \Rightarrow \quad 100p = 20 \quad \Rightarrow \quad p = 0.2$

Thus, the values of $n$ and $p$ are: $n = 100, \quad p = 0.2$

Let me know if you would like any further details or clarification!

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Here are 5 related questions for further practice:

1. What is the probability of getting exactly 25 successes in a $\text{Binomial}(100, 0.2)$ distribution?
2. How would the mean and variance change if $p$ were doubled in the binomial distribution?
3. How can you compute the skewness of a binomial distribution?
4. What is the probability of getting more than 30 successes in a $\text{Binomial}(100, 0.2)$ distribution?
5. How does the binomial distribution relate to the normal distribution when $n$ is large?

Tip: When $n$ is large, the binomial distribution can often be approximated by a normal distribution using the Central Limit Theorem.