Let the mean and the variance of 5 observations x₁, x₂, x₃, x₄, x₅ be ${24 \over 5}$ and ${194 \over 25}$ respectively. If the mean and variance of the first 4 observation are ${7 \over 2}$ and a respectively, then (4a + x₅) is equal to:

<p>Mean $(\overline x ) = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$</p> <p>Given, ${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = {{24} \over 5}$</p> <p>$\Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 24$ ...... (1)</p> <p>Now, Mean of first 4 observation</p> <p>$= {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4}$</p> <p>Given, $= {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} = {7 \over 2}$</p> <p>$\Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = 14$ ...... (2)</p> <p>From equation (1) and (2), we get</p> <p>$14 + {x_5} = 24$</p> <p>$\Rightarrow {x_5} = 10$</p> <p>Now, variance of first 5 observation</p> <p>$= {{\sum {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$</p> <p>$$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2}$$</p> <p>Given,</p> <p>$${{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2} = {{194} \over {24}}$$</p> <p>$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {25}}} \right)$$</p> <p>$\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 154$</p> <p>$\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + {(10)^2} = 154$</p> <p>$\Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54$</p> <p>Now, variance of first 4 observation</p> <p>$= {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2}$</p> <p>Given,</p> <p>${{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2} = a$</p> <p>$\Rightarrow {{54} \over 4} - {{49} \over 4} = a$</p> <p>$\Rightarrow a = {5 \over 4}$</p> <p>$\therefore$ $4a + {x_5}$</p> <p>$= 4 \times {5 \over 4} + 10 = 15$</p>