Question

When success is not an impossible event, then the mean of Binomial distribution is

• A. always more than its variance
• B. always equal to its variance
• C. always less than its variance
• D. always equal to its standard deviation

Hint:

For binomial distribution: \[ \text{Mean}=np \] and \[ \text{Variance}=npq \] Since \[ 0\leq q\lt 1, \] variance is always less than the mean.

Answer 1

The Correct Option is A

Step 1: Recall the mean of binomial distribution.
For a binomial distribution \[ B(n,p), \] the mean is \[ \mu=np \] where \[ 0\lt p\leq 1 \]

Step 2: Recall the variance of binomial distribution.
The variance of a binomial distribution is \[ \sigma^2=npq \] where \[ q=1-p \]

Step 3: Compare mean and variance.
We have \[ \mu=np \] and \[ \sigma^2=npq \] Since \[ q=1-p \] and success is not impossible, we have \[ 0\lt p\leq 1 \] Thus, \[ 0\leq q\lt 1 \]

Step 4: Use the inequality for \(q\).
Because \[ q\lt 1, \] multiplying both sides by \(np\gt 0\), \[ npq\lt np \] That is, \[ \sigma^2\lt \mu \]

Step 5: Conclude the relation.
Hence, the mean is always greater than the variance.

Step 6: Verify extreme cases.
If \[ p=1, \] then \[ q=0 \] So, \[ \text{Variance}=0 \] while \[ \text{Mean}=n \] Thus, mean is still greater than variance.

Step 7: Final conclusion.
Therefore, \[ \boxed{\text{Mean is always more than its variance}} \]