For a binomial distribution, the mean $\mu$ and standard deviation $\sigma$ are given by the formulas:

$\mu = n \cdot p$ $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Given:

• Mean, $\mu = 20$
• Standard deviation, $\sigma = 4$

Step 1: Solve for $p$

From the mean formula, we have: $20 = n \cdot p \Rightarrow p = \frac{20}{n}$

Step 2: Substitute $p$ in the standard deviation formula

Since $\sigma = 4$, we can write: $4 = \sqrt{n \cdot \frac{20}{n} \cdot \left(1 - \frac{20}{n}\right)}$

Square both sides to eliminate the square root: $16 = n \cdot \frac{20}{n} \cdot \left(1 - \frac{20}{n}\right)$

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

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

Rearrange to solve for $n$: $\frac{400}{n} = 4 \Rightarrow n = \frac{400}{4} = 100$

Answer:

$n = 100$

Would you like more details on this solution or have any questions?

---

1. How do we interpret mean and standard deviation in binomial distribution?
2. Can the same approach be used for different distributions?
3. What happens if we know $n$ but not $p$?
4. How does changing $n$ affect the standard deviation in a binomial distribution?
5. How do we verify if the values make sense in practical scenarios?

Tip: In binomial distribution, knowing any two of $n$, $p$, or the mean allows calculating the third.