Question

Let the Standard deviation of \(x_1,x_2\) and \(x_3\) be \(9\) .Then ,the variance of \(3x_1+4,\)  \(3x_2+4 \) and \(3x_3+4\) is 

• A. \(243\)
• B. \(729\)
• C. \(81\)
• D. \(9\)
• E. \(733\)

Answer 1

The Correct Option is B

Let the standard deviation of \( x_1, x_2, \) and \( x_3 \) be 9. 

This means the variance of \( x_1, x_2, \) and \( x_3 \) is \( 9^2 = 81 \).

Let \( Y_1, Y_2, \) and \( Y_3 \) be the transformed variables:

\[ Y_1 = 3x_1 + 4, \quad Y_2 = 3x_2 + 4, \quad Y_3 = 3x_3 + 4 \]

The variance of a linear transformation \( aX + b \) is \( a^2 \text{Var}(X) \). Therefore:

\[ \text{Var}(Y_1) = \text{Var}(3x_1 + 4) = 3^2 \text{Var}(x_1) = 9 \cdot 81 = 729 \] \[ \text{Var}(Y_2) = \text{Var}(3x_2 + 4) = 3^2 \text{Var}(x_2) = 9 \cdot 81 = 729 \] \[ \text{Var}(Y_3) = \text{Var}(3x_3 + 4) = 3^2 \text{Var}(x_3) = 9 \cdot 81 = 729 \]

The variance of \( Y_1, Y_2, \) and \( Y_3 \) is 729.

Answer 2

Let \( X_1, X_2, X_3 \) be random variables with standard deviation 9. 

This means their variance is \( 9^2 = 81 \).

Let \( Y_i = 3X_i + 4 \) for \( i = 1, 2, 3 \). We want to find the variance of \( Y_i \).

Recall that if \( Y = aX + b \), then \( \text{Var}(Y) = a^2 \text{Var}(X) \).

Therefore, the variance of \( Y_i \) is:

\[ \text{Var}(Y_i) = \text{Var}(3X_i + 4) = 3^2 \text{Var}(X_i) = 9 \text{Var}(X_i) = 9(81) = 729 \]

The variance of \( 3X_1+4 \), \( 3X_2+4 \), and \( 3X_3+4 \) is 729.