Question

The mean & variance of \(x_1, x_2, x_3, x_4\) is 1 and 13 respectively and the mean and variance of \(y_1, y_2, \dots, y_6\) be 2 and 1 respectively, the variance of \(x_1, x_2, \dots, x_4, y_1, y_2, \dots, y_6\) will be

• A. 6.04
• B. 6.00
• C. 5.85
• D. 5.99

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The combined variance formula for two groups with sizes $n_1$, $n_2$, means $\bar{x}_1$, $\bar{x}_2$, and variances $\sigma_1^2$, $\sigma_2^2$ is: \[ \sigma^2 = \frac{n_1\sigma_1^2 + n_2\sigma_2^2}{n_1+n_2} + \frac{n_1 n_2}{(n_1+n_2)^2}(\bar{x}_1 - \bar{x}_2)^2 \] Step 2: Listing Given Values:
\[ n_1 = 4,\quad \bar{x}_1 = 1,\quad \sigma_1^2 = 13 \] \[ n_2 = 6,\quad \bar{x}_2 = 2,\quad \sigma_2^2 = 1 \] \[ n_1 + n_2 = 10 \] Step 3: Computing the Combined Mean:
\[ \bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} = \frac{4(1)+6(2)}{10} = \frac{4+12}{10} = \frac{16}{10} = 1.6 \] Step 4: Applying the Combined Variance Formula:
\[ \sigma^2 = \frac{4(13) + 6(1)}{10} + \frac{4 \times 6}{10^2}(1-2)^2 \] \[ = \frac{52 + 6}{10} + \frac{24}{100}(1) \] \[ = \frac{58}{10} + \frac{24}{100} \] \[ = 5.8 + 0.24 = 6.04 \] Step 5: Final Answer:
The combined variance is $\mathbf{6.04}$.
The answer is Option (1).