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

Answer 1

The Correct Option is B

Given that:

Standard deviation of \(x_1,x_2\) and \(x_3\) be .

The variance  of random variables \(3x_1 + 4, 3x_2 + 4\), and \(3x_3 + 4,\) to be found ;

For a random variable \(X\) with standard deviation \(σ\) , the variance of a linear transformation\((aX + b)\), where \('a'\) and \('b'\) are constants and can be calculated as follows:

\(Var(aX + b) = a^2 × Var(X)\)

applying this to all three given variables we get;

For, \(3x_1 + 4: Var(3x_1 + 4)\)

\(= 3^2 × Var(x1) = 9 × 9 = 81\)

For, \(3x_2 + 4: Var(3x_2 + 4)\)

\(= 3^2 × Var(x_2) = 9 × 9 = 81\)

For, \(3x_3 + 4: Var(3x_3 + 4)\)

\(= 3^2 × Var(x3) = 9 × 9 = 81\)

Hence, the variance for the random variable is \(81\) for each of them.(_Ans)