Question

The mean and variance of a binomial distribution are \(x\) and \(5\) respectively. If \(x\) is an integer, then the possible values for \(x\) are

• A. \(6, 10, 30 \)
• B. \(8, 12, 28 \)
• C. \(10, 15, 25 \)
• D. \(9, 18, 24 \)

Hint:

For problems involving parameters of distributions like binomial, always remember the constraints on these parameters (e.g., \(0<p<1\), \(n\) must be a positive integer). Algebraic manipulation, especially polynomial division, can help identify integer conditions for \(n\).

Answer 1

The Correct Option is A

Step 1: Write the equations for mean and variance of a binomial distribution. 
For a binomial distribution, let \(n\) be the number of trials and \(p\) be the probability of success. 
The mean (\(\mu\)) is given by \(\mu = np\). 
The variance (\(\sigma^2\)) is given by \(\sigma^2 = np(1-p)\). 
Given: 
Mean = \(x\) \(\implies np = x\) 
Variance = \(5\) \(\implies np(1-p) = 5\) 
Step 2: Solve for \(p\) in terms of \(x\). 
Substitute \(np = x\) into the variance equation: 
\[ x(1-p) = 5 \] Since \(x\) is given to be an integer and the variance is 5 (a positive value), \(x\) must be positive. 
From the definition of probability, \(0 < p < 1\). This implies \(0 < 1-p < 1\). 
From \(x(1-p) = 5\), we can write \(1-p = \frac{5}{x}\). 
Since \(0 < 1-p < 1\), we must have: \[ 0 < \frac{5}{x} < 1 \] Since \(5 > 0\), for \(\frac{5}{x} > 0\), we must have \(x > 0\). 
For \(\frac{5}{x} < 1\), and knowing \(x > 0\), we can multiply by \(x\) without changing the inequality direction: 
\[ 5 < x \] So, \(x\) must be an integer greater than 5. Now, we can find \(p\): \[ p = 1 - \frac{5}{x} = \frac{x-5}{x} \] Since \(x > 5\), \(x-5 > 0\), so \(p > 0\). Also, \(x-5 < x\), so \(p < 1\). Thus, \(0 < p < 1\) is satisfied for \(x > 5\). 
Step 3: Solve for \(n\) in terms of \(x\). 
We know \(np = x\), so \(n = \frac{x}{p}\). 
Substitute the expression for \(p\): 
\[ n = \frac{x}{\frac{x-5}{x}} = \frac{x^2}{x-5} \] For a binomial distribution, \(n\) must be a positive integer. 
We need \(\frac{x^2}{x-5}\) to be an integer. 
We can perform polynomial division or algebraic manipulation: \[ n = \frac{x^2 - 25 + 25}{x-5} = \frac{(x-5)(x+5) + 25}{x-5} \] \[ n = (x+5) + \frac{25}{x-5} \] For \(n\) to be an integer, \((x-5)\) must be a divisor of 25. 
The positive divisors of 25 are 1, 5, 25. 
Since \(x\) is an integer and \(x > 5\), \(x-5\) must be a positive integer. 
We consider the possible values for \(x-5\): 
1. If \(x-5 = 1 \implies x = 6\). 
In this case, \(p = \frac{6-5}{6} = \frac{1}{6}\), and \(n = 6+5 + \frac{25}{1} = 11 + 25 = 36\). (Valid) 
2. If \(x-5 = 5 \implies x = 10\). 
In this case, \(p = \frac{10-5}{10} = \frac{5}{10} = \frac{1}{2}\), and \(n = 10+5 + \frac{25}{5} = 15 + 5 = 20\). (Valid) 
3. If \(x-5 = 25 \implies x = 30\). 
In this case, \(p = \frac{30-5}{30} = \frac{25}{30} = \frac{5}{6}\), and \(n = 30+5 + \frac{25}{25} = 35 + 1 = 36\). (Valid) 
Thus, the possible integer values for \(x\) are 6, 10, and 30. 
The final answer is \(\boxed{6, 10, 30}\).