If the mean of a binomial distribution is `25,` then its standard deviation lies in the interval

Options

A. `[0,5]`
B. `(0,5]`
C. `[0,25)`
D. `(0,25]`

Answer

To solve the problem of finding the interval in which the standard deviation of a binomial distribution lies, given that the mean is 25, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Binomial Distribution**:
A binomial distribution is characterized by two parameters: \( n \) (the number of trials) and \( p \) (the probability of success). The mean \( \mu \) of a binomial distribution is given by:
\[ \mu = np \]
The standard deviation \( \sigma \) is given by:
\[ \sigma = \sqrt{npq} = \sqrt{np(1-p)} \]
where \( q = 1 - p \).

2. **Set the Mean**:
From the problem, we know that the mean is 25:
\[ np = 25 \]

3. **Express the Standard Deviation**:
Substitute \( np \) into the formula for standard deviation:
\[ \sigma = \sqrt{np(1-p)} = \sqrt{25(1-p)} = 5\sqrt{1-p} \]

4. **Determine the Range of \( p \)**:
Since \( p \) is a probability, it must satisfy:
\[ 0 < p < 1 \]

5. **Find the Range of \( 1 - p \)**:
From the inequality for \( p \):
- If \( p < 1 \), then \( 1 - p > 0 \).
- If \( p > 0 \), then \( 1 - p < 1 \).
Thus, we can conclude:
\[ 0 < 1 - p < 1 \]

6. **Calculate the Range for \( \sigma \)**:
Now, we can express the standard deviation in terms of \( 1 - p \):
\[ 0 < \sqrt{1 - p} < 1 \]
Multiplying through by 5 gives:
\[ 0 < 5\sqrt{1 - p} < 5 \]
Therefore, the standard deviation \( \sigma \) lies in the interval:
\[ (0, 5) \]

### Conclusion:
The standard deviation of the binomial distribution lies in the interval \( (0, 5) \).

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