Question 12

NUMERICALMEDIUM

Consider a data consisting of 10 observations x1,x2,…,x10x_1, x_2, \dots, x_{10}, whose mean is 5 and variance is 7. If the mean and the variance of the first 8 observations x1,x2,…,x8x_1, x_2, \dots, x_8 are 4 and 3.5, respectively, and x9<x10x_9 < x_{10}, then the value of 3x9+2x103x_9 + 2x_{10} is ___________.

Correct Answer: 44

Detailed Solution

For 10 observations: ∑xi=10×5=50\sum x_i = 10 \times 5 = 50. Variance: ∑xi210−52=7⇒∑i=110xi2=320\frac{\sum x_i^2}{10} - 5^2 = 7 \Rightarrow \sum_{i=1}^{10} x_i^2 = 320. For first 8 observations: ∑i=18xi=8×4=32\sum_{i=1}^8 x_i = 8 \times 4 = 32. Variance: ∑i=18xi28−42=3.5⇒∑i=18xi2=8(19.5)=156\frac{\sum_{i=1}^8 x_i^2}{8} - 4^2 = 3.5 \Rightarrow \sum_{i=1}^8 x_i^2 = 8(19.5) = 156. Then x9+x10=50−32=18x_9 + x_{10} = 50 - 32 = 18 and x92+x102=320−156=164x_9^2 + x_{10}^2 = 320 - 156 = 164. (x9+x10)2=x92+x102+2x9x10⇒182=164+2x9x10⇒324−164=2x9x10⇒x9x10=80(x_9 + x_{10})^2 = x_9^2 + x_{10}^2 + 2x_9x_{10} \Rightarrow 18^2 = 164 + 2x_9x_{10} \Rightarrow 324 - 164 = 2x_9x_{10} \Rightarrow x_9x_{10} = 80. Solving t2−18t+80=0t^2 - 18t + 80 = 0 gives t=8,10t = 8, 10. Since x9<x10x_9 < x_{10}, x9=8x_9 = 8 and x10=10x_{10} = 10. 3x9+2x10=3(8)+2(10)=24+20=443x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44. Final Answer: 44.

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