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:

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Answer: B. 15 Mean (\overline x ) = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} Given, {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = {{24} \over 5} \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 24 ...... (1) Now, Mean of first 4 observation = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} Given, = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} = {7 \over 2} \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = 14 ...... (2) From equation (1) and (2), we get 14 + {x_5} = 24 \Rightarrow {x_5} = 10 Now, variance of first 5 observation = {{\sum {x_i^2} } \over n} - {\left( {\overline x } ight)^2} = {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} ight)^2} Given, {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} ight)^2} = {{194} \over {24}} \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {25}}} ight) \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 154 \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + {(10)^2} = 154 \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54 Now, variance of first 4 observation = {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} ight)^2} Given, {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} ight)^2} = a \Rightarrow {{54} \over 4} - {{49} \over 4} = a \Rightarrow a = {5 \over 4} 	herefore 4a + {x_5} = 4 	imes {5 \over 4} + 10 = 15

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:

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