Statistics Ans 4
Q4 - 29 January - Shift 2
Let $X={11,12,13, \ldots . ., 40,41}$ and $Y={61,62 , 63, \ldots \ldots, 90,91}$ be the two sets of observations. If $\overline{x}$ and $\overline{y}$ are their respective means and $\sigma^{2}$ is the variance of all the observations in $X \cup Y$, then $|\overline{x}+\overline{y}-\sigma^{2}|$ is equal to
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Answer: (603)
Solution:
Formula: Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data), Combined Mean, Arithmetic Progression
$\begin{aligned} & \frac{\mathrm{x}_1+\mathrm{x}_2+\ldots .+\mathrm{x}_7}{7}=8 \\ & \frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3 \ldots .+\mathrm{x}_6+14}{7}=8 \\ & \Rightarrow \mathrm{x}_1+\mathrm{x}_2+\ldots .+\mathrm{x}_6=42 \\ & \therefore \frac{\mathrm{x}_1+\mathrm{x}_2 \ldots .+\mathrm{x}6}{6}=\frac{42}{6}=7=\mathrm{a} \\ & \frac{\sum \mathrm{x}{\mathrm{i}}^2}{7}-8^2=16 \\ & \Sigma \mathrm{xi}^2=560 \\ & \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\ldots+\mathrm{x}_6^2=364 \\ & \mathrm{~b}=\frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\ldots .+\mathrm{x}_6^2}{6}-7^2 \\ & =\frac{364}{6}-49 \\ & \mathrm{~b}=\frac{70}{6} \\ & \mathrm{a}+3 \mathrm{~b}-5=7+3 \times \frac{70}{6}-5 \\ & =37 \end{aligned}$
