Question:
The maximum value of the variance of Binomial distribution with parameters \(n\) and \(p\) is:
The maximum value of the variance of Binomial distribution with parameters \(n\) and \(p\) is:
Show Hint
For a Binomial distribution,
\[
\mathrm{Var}(X)=np(1-p).
\]
The expression \(p(1-p)\) attains its maximum value \(\frac14\) at \(p=\frac12\).
Updated On: Jun 26, 2026
- \(\frac{1}{2}\)
- \(\frac{n}{4}\)
- \(np(1-p)\)
- \(2n\)
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The Correct Option is B
Solution and Explanation
Step 1: Recall the variance of a Binomial distribution.
If \[ X\sim B(n,p), \] then the variance is \[ \mathrm{Var}(X)=np(1-p). \]
Step 2: Maximize the expression \(p(1-p)\).
Consider \[ f(p)=p(1-p). \] Expanding, \[ f(p)=p-p^2. \] This is a downward-opening parabola whose maximum occurs at \[ p=\frac{1}{2}. \] Substituting, \[ f\!\left(\frac12\right) = \frac12\left(1-\frac12\right) = \frac14. \]
Step 3: Obtain the maximum variance.
Substituting the maximum value of \(p(1-p)\) into the variance formula, \[ \mathrm{Var}(X)_{\max} = n\left(\frac14\right) = \frac{n}{4}. \]
Step 4: Final conclusion.
Therefore, the maximum value of the variance is \[ \boxed{\frac{n}{4}}. \]
If \[ X\sim B(n,p), \] then the variance is \[ \mathrm{Var}(X)=np(1-p). \]
Step 2: Maximize the expression \(p(1-p)\).
Consider \[ f(p)=p(1-p). \] Expanding, \[ f(p)=p-p^2. \] This is a downward-opening parabola whose maximum occurs at \[ p=\frac{1}{2}. \] Substituting, \[ f\!\left(\frac12\right) = \frac12\left(1-\frac12\right) = \frac14. \]
Step 3: Obtain the maximum variance.
Substituting the maximum value of \(p(1-p)\) into the variance formula, \[ \mathrm{Var}(X)_{\max} = n\left(\frac14\right) = \frac{n}{4}. \]
Step 4: Final conclusion.
Therefore, the maximum value of the variance is \[ \boxed{\frac{n}{4}}. \]
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