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
Select the correct option:
A
$\frac{4}{5}$
B
$\frac{77}{12}$
C
$\frac{5}{4}$
D
$\frac{105}{4}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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, $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$...
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
$$
x_5=10
$$
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
$$
\begin{aligned}
\operatorname{Var} & =\frac{\sum_{i=1}^4 x_i^2}{4}-\left(\frac{\sum_{i=1}^4 x_i}{4}\right)^2 \\
& =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}
\end{aligned}
$$
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