Question:

\( N \) independent trials of a binomial random variable yield an expectation value of 16 and variance of 12. What is the value of \( N \)?

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To find the number of trials \( N \) in a binomial distribution, use the relationships between expectation and variance. Solve the system of equations for \( N \) and \( p \).
Updated On: Apr 14, 2025
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The Correct Option is D

Solution and Explanation

For a binomial random variable, the expectation value (mean) \( E[X] \) and variance \( {Var}[X] \) are given by: \[ E[X] = Np \] \[ {Var}[X] = Np(1-p) \] where:
\( N \) is the number of trials,
\( p \) is the probability of success on a single trial.
Step 1: Using the given expectation and variance.
From the problem, we know:
\( E[X] = 16 \),
\( {Var}[X] = 12 \).
Thus, we have the system of equations: \[ N p = 16 \quad {(1)} \] \[ N p(1-p) = 12 \quad {(2)} \] Step 2: Solving the system.
From equation (1), we can express \( p \) as: \[ p = \frac{16}{N} \] Substitute this into equation (2): \[ N \times \frac{16}{N} \times \left(1 - \frac{16}{N}\right) = 12 \] \[ 16 \left( 1 - \frac{16}{N} \right) = 12 \] \[ 16 - \frac{256}{N} = 12 \] \[ \frac{256}{N} = 4 \] \[ N = 64 \] Step 3: Conclusion.
The value of \( N \) is 64.
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