Question:
In a Binomial distribution \(B(n,p)\), if the mean and variance are \(15\) and \(10\) respectively, then the value of the parameter \(n\) is
In a Binomial distribution \(B(n,p)\), if the mean and variance are \(15\) and \(10\) respectively, then the value of the parameter \(n\) is
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For a Binomial distribution, remember the standard formulas: \(\text{Mean}=np\) and \(\text{Variance}=npq\), where \(q=1-p\).
Updated On: Jun 22, 2026
- \(28\)
- \(16\)
- \(45\)
- \(25\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the formulas for mean and variance of Binomial distribution.
For a Binomial distribution \(B(n,p)\),
\[ \text{Mean}=np \] and
\[ \text{Variance}=npq \] where
\[ q=1-p \]
Step 2: Use the given mean.
It is given that the mean is \(15\).
Therefore,
\[ np=15 \]
Step 3: Use the given variance.
It is given that the variance is \(10\).
So,
\[ npq=10 \] Using \(np=15\), we get
\[ 15q=10 \] \[ q=\frac{10}{15} \] \[ q=\frac{2}{3} \]
Step 4: Find the value of \(p\).
Since,
\[ p+q=1 \] we have
\[ p=1-\frac{2}{3} \] \[ p=\frac{1}{3} \]
Step 5: Find the value of \(n\).
Using
\[ np=15 \] Substitute \(p=\dfrac{1}{3}\),
\[ n\left(\frac{1}{3}\right)=15 \] \[ n=15\times 3 \] \[ n=45 \]
Step 6: Final conclusion.
Hence, the value of the parameter \(n\) is
\[ \boxed{45} \]
For a Binomial distribution \(B(n,p)\),
\[ \text{Mean}=np \] and
\[ \text{Variance}=npq \] where
\[ q=1-p \]
Step 2: Use the given mean.
It is given that the mean is \(15\).
Therefore,
\[ np=15 \]
Step 3: Use the given variance.
It is given that the variance is \(10\).
So,
\[ npq=10 \] Using \(np=15\), we get
\[ 15q=10 \] \[ q=\frac{10}{15} \] \[ q=\frac{2}{3} \]
Step 4: Find the value of \(p\).
Since,
\[ p+q=1 \] we have
\[ p=1-\frac{2}{3} \] \[ p=\frac{1}{3} \]
Step 5: Find the value of \(n\).
Using
\[ np=15 \] Substitute \(p=\dfrac{1}{3}\),
\[ n\left(\frac{1}{3}\right)=15 \] \[ n=15\times 3 \] \[ n=45 \]
Step 6: Final conclusion.
Hence, the value of the parameter \(n\) is
\[ \boxed{45} \]
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