Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be [24 5] and [194 25] …

Question

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:

Accepted Answer

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

$$ = {{\sum {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$$

$$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2}$$

Given,

$${{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2} = {{194} \over {24}}$$

$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {25}}} \right)$$

$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 154$$

$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + {(10)^2} = 154$$

$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54$$

Now, variance of first 4 observation

$$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2}$$

Given,

$${{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2} = a$$

$$ \Rightarrow {{54} \over 4} - {{49} \over 4} = a$$

$$ \Rightarrow a = {5 \over 4}$$

$$\therefore$$ $4a + {x_5}$

$$ = 4 \times {5 \over 4} + 10 = 15$$

Solutions · 1