Question 12

Sum of 10 observations = 50, sum of squares = 10(7 + 25) = 320. Sum of first 8 = 32, sum of squares = 8(3.5 + 16) = 156. So x₉ + x₁₀ = 18 and x₉² + x₁₀² = 164. Then 2x₉x₁₀ = 324 − 164 = 160, so x₉x₁₀ = 80. Roots of t² − 18t + 80 = 0: t = 8, 10. With x₉ < x₁₀: x₉ = 8, x₁₀ = 10. 3x₉ + 2x₁₀ = 24 + 20 = 44.

**Calculating population variance** I’m focusing on computing the standard population variance. First, I need the sum of 50 samples, where the first 32 sum to 18 and the total of squares is 320. I'll take the calculations further, adjusting the products based on these results. I’ll confirm if the final outcome will just be the JSON format the user expects, so I’ll make sure to produce only that when ready!

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I find the last two values sum to 18, using the totals from the first eight and all ten. Working out the sum of squares for both groups, I get that x9²+x10² = 164, so now I can solve for x9 and x10 using their sum and sum of squares. Expanding (x9+x10)² gives me x9x10=80, so x9 and x10 satisfy t²-18t+80=0, giving roots 10 and 8. Since x9<x10, x9=8 and x10=10, so 3x9+2x10=44. **Solution:** Sum of first 8 observations = 8×4 = 32 Sum of all 10 = 10×5 = 50 So x₉ + x₁₀ = 18 Sum of squares of first 8: variance = (Σx²)/8 - 16 = 3.5 → Σx² = 156 Sum of squares of all 10: variance = (Σx²)/10 - 25 = 7 → Σx² = 320 So x₉² + x₁₀² = 320 - 156 = 164 Using (x₉+x₁₀)² = x₉² + x₁₀² + 2x₉x₁₀: 324 = 164 + 2x₉x₁₀ → x₉x₁₀ = 80 x₉, x₁₀ satisfy t² - 18t + 80 = 0 → t = 8, 10 Since x₉ < x₁₀: x₉ = 8, x₁₀ = 10 3x₉ + 2x₁₀ = 24 + 20 = 44 ```json