Let the mean and the variance of 5 observatio | vedclass

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(B) Given the mean of $5$ observations is $\bar{x} = \frac{\sum_{i=1}^{5} x_{i}}{5} = \frac{24}{5}$,so $\sum_{i=1}^{5} x_{i} = 24$.The variance is $\s

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$15$

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(B) Given the mean of $5$ observations is $\bar{x} = \frac{\sum_{i=1}^{5} x_{i}}{5} = \frac{24}{5}$,so $\sum_{i=1}^{5} x_{i} = 24$.The variance is $\sigma^{2} = \frac{\sum x_{i}^{2}}{5} - (\bar{x})^{2} = \frac{194}{25}$.Substituting $\bar{x} = \frac{24}{5}$,we get $\frac{\sum x_{i}^{2}}{5} - \frac{576}{25} = \frac{194}{25}$ $\Rightarrow \frac{\sum x_{i}^{2}}{5} = \frac{770}{25} = \frac{154}{5}$,so $\sum_{i=1}^{5} x_{i}^{2} = 154$.For the first $4$ observations,the mean is $\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} = \frac{7}{2} \Rightarrow x_{1}+x_{2}+x_{3}+x_{4} = 14$.Since $\sum_{i=1}^{5} x_{i} = 24$,we have $x_{5} = 24 - 14 = 10$.The variance of the first $4$ observations is $a = \frac{\sum_{i=1}^{4} x_{i}^{2}}{4} - (\frac{7}{2})^{2} = \frac{\sum_{i=1}^{4} x_{i}^{2}}{4} - \frac{49}{4}$.Thus,$\sum_{i=1}^{4} x_{i}^{2} = 4a + 49$.We know $\sum_{i=1}^{5} x_{i}^{2} = \sum_{i=1}^{4} x_{i}^{2} + x_{5}^{2} = 154$.Substituting the values: $(4a + 49) + 10^{2} = 154$.$4a + 49 + 100 = 154$ $\Rightarrow 4a + 149 = 154$ $\Rightarrow 4a = 5$.Finally,$4a + x_{5} = 5 + 10 = 15$.

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observations are $\frac{7}{2}$ and $a$ respectively,then $(4a + x_{5})$ is equal to

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