Question

In a Binomial distribution \( B(n,p) \), the sum and product of the mean and the variance are 5 and 6 respectively, then \( 6(n + p - q) = \):

• A. 50
• B. 53
• C. 52
• D. 51

Hint:

When dealing with binomial distributions, remember to use the relationships for mean and variance, and solve the system of equations accordingly.

Answer 1

The Correct Option is C

To solve the problem, we use the properties of the Binomial distribution \( B(n, p) \):
- Mean (\( \mu \)): \( \mu = np \),
- Variance (\( \sigma^2 \)): \( \sigma^2 = npq \), where \( q = 1
- p \). Given: 1. The sum of the mean and variance is 5: \[ \mu + \sigma^2 = 5 \quad \Rightarrow \quad np + npq = 5. \] 2. The product of the mean and variance is 6: \[ \mu \cdot \sigma^2 = 6 \quad \Rightarrow \quad np \cdot npq = 6. \] Step 1: Simplify the equations From the first equation: \[ np + npq = 5 \quad \Rightarrow \quad np(1 + q) = 5. \] From the second equation: \[ np \cdot npq = 6 \quad \Rightarrow \quad n^2 p^2 q = 6. \] Step 2: Express \( q \) in terms of \( p \) Since \( q = 1
- p \), substitute into the first equation: \[ np(1 + 1
- p) = 5 \quad \Rightarrow \quad np(2
- p) = 5. \] Step 3: Solve for \( n \) and \( p \) From the second equation: \[ n^2 p^2 q = 6. \] Substitute \( q = 1
- p \): \[ n^2 p^2 (1
- p) = 6. \] From the first equation, solve for \( np \): \[ np = \frac{5}{2
- p}. \] Substitute \( np = \frac{5}{2
- p} \) into the second equation: \[ \left( \frac{5}{2
- p} \right)^2 p^2 (1
- p) = 6. \] Simplify: \[ \frac{25 p^2 (1
- p)}{(2
- p)^2} = 6. \] Multiply through by \( (2
- p)^2 \): \[ 25 p^2 (1
- p) = 6 (2
- p)^2. \] Expand \( (2
- p)^2 \): \[ 25 p^2 (1
- p) = 6 (4
- 4p + p^2). \] Simplify: \[ 25 p^2
- 25 p^3 = 24
- 24 p + 6 p^2. \] Rearrange: \[
-25 p^3 + 19 p^2 + 24 p
- 24 = 0. \] Step 4: Solve the cubic equation The cubic equation \(
-25 p^3 + 19 p^2 + 24 p
- 24 = 0 \) can be solved numerically or by inspection. One solution is \( p = 1 \), but this is not valid since \( p \) must be between 0 and 1. Another solution is \( p = \frac{3}{5} \). Substitute \( p = \frac{3}{5} \) into \( np = \frac{5}{2
- p} \): \[ n \cdot \frac{3}{5} = \frac{5}{2
- \frac{3}{5}} = \frac{5}{\frac{7}{5}} = \frac{25}{7}. \] Thus: \[ n = \frac{25}{7} \cdot \frac{5}{3} = \frac{125}{21}. \] Step 5: Compute \( 6(n + p
- q) \) Substitute \( n = \frac{125}{21} \), \( p = \frac{3}{5} \), and \( q = 1
- p = \frac{2}{5} \): \[ n + p
- q = \frac{125}{21} + \frac{3}{5}
- \frac{2}{5} = \frac{125}{21} + \frac{1}{5}. \] Find a common denominator: \[ n + p
- q = \frac{125 \cdot 5 + 1 \cdot 21}{105} = \frac{625 + 21}{105} = \frac{646}{105}. \] Multiply by 6: \[ 6(n + p
- q) = 6 \cdot \frac{646}{105} = \frac{3876}{105} = 36.914. \] This does not match any of the options. Let's re
-examine the problem. Step 6: Re
-examining the problem The correct approach is to use the given equations: 1. \( np + npq = 5 \), 2. \( np \cdot npq = 6 \). Substitute \( q = 1
- p \): \[ np + np(1
- p) = 5 \quad \Rightarrow \quad np(2
- p) = 5, \] \[ np \cdot np(1
- p) = 6 \quad \Rightarrow \quad n^2 p^2 (1
- p) = 6. \] Let \( x = np \). Then: \[ x(2
- p) = 5, \] \[ x^2 (1
- p) = 6. \] From the first equation: \[ x = \frac{5}{2
- p}. \] Substitute into the second equation: \[ \left( \frac{5}{2
- p} \right)^2 (1
- p) = 6. \] Simplify: \[ \frac{25 (1
- p)}{(2
- p)^2} = 6. \] Multiply through by \( (2
- p)^2 \): \[ 25 (1
- p) = 6 (2
- p)^2. \] Expand \( (2
- p)^2 \): \[ 25 (1
- p) = 6 (4
- 4p + p^2). \] Simplify: \[ 25
- 25 p = 24
- 24 p + 6 p^2. \] Rearrange: \[ 6 p^2 + p
- 1 = 0. \] Solve the quadratic equation: \[ p = \frac{
-1 \pm \sqrt{1 + 24}}{12} = \frac{
-1 \pm 5}{12}. \] Thus: \[ p = \frac{4}{12} = \frac{1}{3} \quad \text{or} \quad p = \frac{
-6}{12} =
-\frac{1}{2}. \] Since \( p \) must be between 0 and 1, \( p = \frac{1}{3} \). Substitute \( p = \frac{1}{3} \) into \( x = \frac{5}{2
- p} \): \[ x = \frac{5}{2
- \frac{1}{3}} = \frac{5}{\frac{5}{3}} = 3. \] Thus: \[ np = 3 \quad \Rightarrow \quad n = \frac{3}{p} = \frac{3}{\frac{1}{3}} = 9. \] Now, compute \( 6(n + p
- q) \): \[ n + p
- q = 9 + \frac{1}{3}
- \frac{2}{3} = 9
- \frac{1}{3} = \frac{26}{3}. \] Multiply by 6: \[ 6(n + p
- q) = 6 \cdot \frac{26}{3} = 52. \] Final Answer: \[ \boxed{52} \]