Question:
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to:
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to:
Updated On: Apr 13, 2026
- \(5.96 \)
- \(6.14 \)
- \(6.04 \)
- \(6.24 \)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
To find the variance of combined observations, we use the formula for combined variance, which accounts for the individual variances, the individual means, and the combined mean of the data sets.
Step 2: Key Formula or Approach:
1. Combined Mean: \(\bar{x}_{comb} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}\)
2. Combined Variance: \(\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\), where \(d_1 = \bar{x}_1 - \bar{x}_{comb}\) and \(d_2 = \bar{x}_2 - \bar{x}_{comb}\).
Step 3: Detailed Explanation:
Given: \(n_1 = 4, \bar{x}_1 = 1, \sigma_1^2 = 13\) \(n_2 = 6, \bar{x}_2 = 2, \sigma_2^2 = 1\) First, calculate the combined mean: \[ \bar{x}_{comb} = \frac{4(1) + 6(2)}{10} = \frac{4 + 12}{10} = 1.6 \] Calculate deviations: \[ d_1 = 1 - 1.6 = -0.6 \implies d_1^2 = 0.36 \] \[ d_2 = 2 - 1.6 = 0.4 \implies d_2^2 = 0.16 \] Now, calculate the combined variance: \[ \sigma^2 = \frac{4(13 + 0.36) + 6(1 + 0.16)}{10} \] \[ \sigma^2 = \frac{4(13.36) + 6(1.16)}{10} \] \[ \sigma^2 = \frac{53.44 + 6.96}{10} = \frac{60.4}{10} = 6.04 \]
Step 4: Final Answer:
The combined variance of the 10 observations is 6.04.
To find the variance of combined observations, we use the formula for combined variance, which accounts for the individual variances, the individual means, and the combined mean of the data sets.
Step 2: Key Formula or Approach:
1. Combined Mean: \(\bar{x}_{comb} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}\)
2. Combined Variance: \(\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}\), where \(d_1 = \bar{x}_1 - \bar{x}_{comb}\) and \(d_2 = \bar{x}_2 - \bar{x}_{comb}\).
Step 3: Detailed Explanation:
Given: \(n_1 = 4, \bar{x}_1 = 1, \sigma_1^2 = 13\) \(n_2 = 6, \bar{x}_2 = 2, \sigma_2^2 = 1\) First, calculate the combined mean: \[ \bar{x}_{comb} = \frac{4(1) + 6(2)}{10} = \frac{4 + 12}{10} = 1.6 \] Calculate deviations: \[ d_1 = 1 - 1.6 = -0.6 \implies d_1^2 = 0.36 \] \[ d_2 = 2 - 1.6 = 0.4 \implies d_2^2 = 0.16 \] Now, calculate the combined variance: \[ \sigma^2 = \frac{4(13 + 0.36) + 6(1 + 0.16)}{10} \] \[ \sigma^2 = \frac{4(13.36) + 6(1.16)}{10} \] \[ \sigma^2 = \frac{53.44 + 6.96}{10} = \frac{60.4}{10} = 6.04 \]
Step 4: Final Answer:
The combined variance of the 10 observations is 6.04.
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