The mean of a Binomial Distribution is 20, and the standard deviation 4. Calculate n.

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.

The mean of a Binomial Distribution is 20, and the standard deviation is 4. Calculate n.

Learn how to find n in a binomial distribution with a mean of 20 and standard deviation of 4. This step-by-step solution demonstrates the use of binomial distribution formulas for mean and standard deviation, suitable for Grades 10-12.

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