Statistics Question 4
Question 4 - 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
$\begin{aligned} & \frac{x_1+x_2+\ldots .+x_7}{7}=8 \\ & \frac{x_1+x_2+x_3 \ldots .+x_6+14}{7}=8 \\ & \Rightarrow x_1+x_2+\ldots .+x_6=42 \\ & \therefore \frac{x_1+x_2 \ldots .+x_6}{6}=\frac{42}{6}=7=a \\ & \frac{\sum x _i^2}{7}-8^2=16 \\ & \Sigma xi^2=560 \\ & \Rightarrow x_1^2+x_2^2+\ldots+x_6^2=364 \\ & b=\frac{x_1^2+x_2^2+\ldots .+x_6^2}{6}-7^2 \\ & =\frac{364}{6}-49 \\ & b=\frac{70}{6} \\ & a+3 b-5=7+3 \times \frac{70}{6}-5 \\ & =37 \end{aligned}$
