"Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=\mathrm{b}, a\mathrm{b}$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots, 5$ is"

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"

First, find the values of $a$ and $b$ using the given conditions:

Mean Condition:

$ 5 = \frac{1 + 3 + a + 7 + b}{5} $

This implies:

$ a + b = 14 $

Variance Condition:

$ \frac{1 + 9 + a^2 + 49 + b^2}{5} - (5)^2 = 10 $

Simplifying, we get:

$ a^2 + b^2 = 116 $

Using $a + b = 14$, we have:

$ b = 14 - a $

Substitute $b$ into the equation for $a^2 + b^2$:

$ a^2 + (14 - a)^2 = 116 $

Expand and simplify:

$ a^2 + 196 - 28a + a^2 = 116 $

$ 2a^2 - 28a + 80 = 0 $

Simplify further:

$ a^2 - 14a + 40 = 0 $

Solve the quadratic equation:

$ (a - 10)(a - 4) = 0 $

Thus, $a = 10$ and $b = 4$ (since $a > b$).

New Observations:

The transformed observations are:

$ 1+1=2, \quad 3+2=5, \quad 10+3=13, \quad 7+4=11, \quad 4+5=9 $

Calculate the Variance of the New Observations:

First, find the mean of the new data:

$ \text{Mean} = \frac{2 + 5 + 13 + 11 + 9}{5} = 8 $

Calculate the variance:

$ \text{Variance} = \frac{(2-8)^2 + (5-8)^2 + (13-8)^2 + (11-8)^2 + (9-8)^2}{5} $

$ = \frac{36 + 9 + 25 + 9 + 1}{5} $

$ = \frac{80}{5} = 16 $

Therefore, the variance of the transformed observations is 16.

"

Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=\mathrm{b}, a>\mathrm{b}$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots, 5$ is

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Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=\mathrm{b}, a>\mathrm{b}$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots, 5$ is

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