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 $(4 a + x_{5})$ is equal to

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answer is B.

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Detailed Solution

$\bar{x} = \frac{\sum x_{i}^{}}{5} = \frac{24}{5} \Rightarrow \sum x_{i} = 24 σ^{2} = \frac{\sum x_{i}^{2}}{5} - {(\frac{24}{5})}^{2} \Rightarrow \frac{194}{25} \Rightarrow \sum x_{i}^{2} = 154 \Rightarrow x_{1} + x_{2} + x_{3} + x_{4} = 14 \Rightarrow x_{5} = 10 σ^{2} = \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}}{4} - \frac{48}{4} = a \Rightarrow x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = 4 a + 49 \Rightarrow x_{5}^{2} = 154 - 4 a - 49 \Rightarrow 100 = 105 - 4 a \Rightarrow 4 a = 5 \Rightarrow a = \frac{5}{4} 4 a + x_{5} = 5 + 10 = 15$