Question:
A random variable \(X\) follows the binomial distribution and
\[
X\sim B(n,0.3).
\]
If the mean of \(X\) is three times as large as the standard deviation of \(X\), then \(n=\)
A random variable \(X\) follows the binomial distribution and
\[
X\sim B(n,0.3).
\]
If the mean of \(X\) is three times as large as the standard deviation of \(X\), then \(n=\)
Show Hint
For binomial distribution, always remember:
\[
\text{Mean}=np,\qquad
\text{Variance}=npq,\qquad
\text{S.D.}=\sqrt{npq}.
\]
Updated On: Jun 18, 2026
- \(9\)
- \(21\)
- \(27\)
- \(3\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Write mean and standard deviation.
For \[ X\sim B(n,p), \] mean is \[ \mu=np \] and standard deviation is \[ \sigma=\sqrt{npq}, \] where \[ q=1-p. \] Given \[ p=0.3,\qquad q=0.7. \]
Step 2: Use the given condition.
The mean is three times the standard deviation. Hence, \[ np=3\sqrt{npq}. \] Substituting \(p=0.3\) and \(q=0.7\), \[ 0.3n = 3\sqrt{0.21n}. \]
Step 3: Square both sides.
\[ 0.09n^2 = 9(0.21n). \] \[ 0.09n^2 = 1.89n. \] \[ 0.09n = 1.89. \] \[ n = \frac{1.89}{0.09}. \] \[ n=21. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{21} \]
For \[ X\sim B(n,p), \] mean is \[ \mu=np \] and standard deviation is \[ \sigma=\sqrt{npq}, \] where \[ q=1-p. \] Given \[ p=0.3,\qquad q=0.7. \]
Step 2: Use the given condition.
The mean is three times the standard deviation. Hence, \[ np=3\sqrt{npq}. \] Substituting \(p=0.3\) and \(q=0.7\), \[ 0.3n = 3\sqrt{0.21n}. \]
Step 3: Square both sides.
\[ 0.09n^2 = 9(0.21n). \] \[ 0.09n^2 = 1.89n. \] \[ 0.09n = 1.89. \] \[ n = \frac{1.89}{0.09}. \] \[ n=21. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{21} \]
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