Question

Mean and variance of an ungrouped data of 15 numbers are \(12\) and \(14\) respectively. For another ungrouped data of 15 numbers the mean and variance are \(14\) and \(10\) respectively. If the two data are combined, then the variance of the combined data of 30 numbers is

• A. \(12\)
• B. \(16\)
• C. \(13\)
• D. \(15\)

Hint:

For combining two data sets, \[ \boxed{ \sigma^2 = \frac{ n_1(\sigma_1^2+d_1^2) + n_2(\sigma_2^2+d_2^2) } {n_1+n_2}, } \] where \[ d_i=\bar{x}_i-\bar{x}. \]

Answer 1

The Correct Option is C

Step 1: Find the combined mean. Given, \[ n_1=n_2=15,\qquad \bar{x}_1=12,\qquad \bar{x}_2=14. \] The combined mean is \[ \bar{x} = \frac{15(12)+15(14)}{30} =13. \]

Step 2: Use the combined variance formula. The combined variance is \[ \sigma^2 = \frac{ n_1\left(\sigma_1^2+(\bar{x}_1-\bar{x})^2\right) + n_2\left(\sigma_2^2+(\bar{x}_2-\bar{x})^2\right) } {n_1+n_2}. \] Substituting the given values, \[ \sigma^2 = \frac{ 15(14+1)+15(10+1) }{30}. \]

Step 3: Calculate the variance. Thus, \[ \sigma^2 = \frac{225+165}{30} = \frac{390}{30} = 13. \] Therefore, \[ \boxed{13}. \] Thus, \[ \boxed{(C)} \] is the correct answer.