Variance of Combined Observation Sets

To find the variance of the combined set of $10$ observations, we can use the parameters given for the two individual sets.

1. Given Data:

• First Set: Number of observations, $n_1 = 4$

  • Mean, $\bar{x}_1 = 1$
  • Variance, $\sigma_1^2 = 13$

• Second Set: Number of observations, $n_2 = 6$

  • Mean, $\bar{x}_2 = 2$
  • Variance, $\sigma_2^2 = 1$

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2. Calculate the Combined Mean ($\bar{x}$): The combined mean $\bar{x}$ of all $N = n_1 + n_2 = 10$ observations is given by: $\bar{x} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2}$ $\bar{x} = \frac{4(1) + 6(2)}{4 + 6} = \frac{4 + 12}{10} = \frac{16}{10} = 1.6$

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3. Calculate the Sum of Squares for Each Set: The formula for variance is: $\sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 \implies \sum x^2 = n(\sigma^2 + \bar{x}^2)$

• For the first set: $\sum x_1^2 = 4 \left(13 + 1^2\right) = 4(14) = 56$

• For the second set: $\sum x_2^2 = 6 \left(1 + 2^2\right) = 6(5) = 30$

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4. Calculate the Combined Variance ($\sigma^2$): The total sum of squares for all $10$ observations is: $\sum x^2 = \sum x_1^2 + \sum x_2^2 = 56 + 30 = 86$

The combined variance $\sigma^2$ is then: $\sigma^2 = \frac{\sum x^2}{N} - \bar{x}^2$ $\sigma^2 = \frac{86}{10} - (1.6)^2$ $\sigma^2 = 8.6 - 2.56 = 6.04$

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Conclusion: The variance of all $10$ observations is $6.04$.

Thus, the correct option is C.