The number of events of the binomial distribution for which mean and standard deviation are 10 and `sqrt5` respectively is

Options

A. 15
B. 12
C. 16
D. 20

Answer

To solve the problem, we need to find the number of events \( n \) in a binomial distribution given that the mean \( \mu \) is 10 and the standard deviation \( \sigma \) is \( \sqrt{5} \).

### Step-by-Step Solution:

1. **Understanding the Mean and Standard Deviation of a Binomial Distribution:**
- The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p \]
- The standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \]

2. **Setting Up the Equations:**
- From the problem, we know:
\[ n \cdot p = 10 \quad \text{(1)} \]
\[ \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{5} \quad \text{(2)} \]

3. **Squaring the Standard Deviation Equation:**
- Squaring equation (2):
\[ n \cdot p \cdot (1 - p) = 5 \quad \text{(3)} \]

4. **Substituting Equation (1) into Equation (3):**
- From equation (1), we can express \( n \) in terms of \( p \):
\[ n = \frac{10}{p} \]
- Substitute \( n \) into equation (3):
\[ \frac{10}{p} \cdot p \cdot (1 - p) = 5 \]
- Simplifying this:
\[ 10(1 - p) = 5 \]
\[ 10 - 10p = 5 \]
\[ 10p = 5 \]
\[ p = \frac{1}{2} \]

5. **Finding \( n \):**
- Now substitute \( p = \frac{1}{2} \) back into equation (1):
\[ n \cdot \frac{1}{2} = 10 \]
\[ n = 10 \cdot 2 = 20 \]

6. **Conclusion:**
- The number of events \( n \) in the binomial distribution is:
\[ n = 20 \]

### Final Answer:
The number of events of the binomial distribution is \( n = 20 \).