Statistics Question 6
Question 6 - 2024 (29 Jan Shift 2)
If the mean and variance of five observations are $\frac{24}{5}$ and $\frac{194}{25}$ respectively and the mean of first four observations is $\frac{7}{2}$, then the variance of the first four observations in equal to
(1) $\frac{4}{5}$
(2) $\frac{77}{12}$
(3) $\frac{5}{4}$
(4) $\frac{105}{4}$
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Answer (3)
Solution
$\bar{X}=\frac{24}{5} ; \sigma^{2}=\frac{194}{25}$
Let first four observation be $\mathrm{x}{1}, \mathrm{x}{2}, \mathrm{x}{3}, \mathrm{x}{4}$
Here, $\frac{\mathrm{x}{1}+\mathrm{x}{2}+\mathrm{x}{3}+\mathrm{x}{4}+\mathrm{x}_{5}}{5}=\frac{24}{5} \ldots$
Also, $\frac{\mathrm{x}{1}+\mathrm{x}{2}+\mathrm{x}{3}+\mathrm{x}{4}}{4}=\frac{7}{2}$
$\Rightarrow \mathrm{x}{1}+\mathrm{x}{2}+\mathrm{x}{3}+\mathrm{x}{4}=14$
Now from eqn -1
$\mathrm{x}_{5}=10$
Now, $\sigma^{2}=\frac{194}{25}$
$$ \begin{aligned} & \frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}}{5}-\frac{576}{25}=\frac{194}{25} \ & \Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=54 \end{aligned} $$
Now, variance of first 4 observations
$$ \begin{aligned} \operatorname{Var} & =\frac{\sum_{i=1}^{4} x_{i}^{2}}{4}-\left(\frac{\sum_{i=1}^{4} x_{i}}{4}\right)^{2} \ & =\frac{54}{4}-\frac{49}{4}=\frac{5}{4} \end{aligned} $$
