Let the Mean and Variance of five observations x_1=1,x_2=3,x_3=a,x_4=7 and x_5=b,ab, be 5…

Question

Let the Mean and Variance of five observations ${x}_{1}=1,{x}_{2}=3,{x}_{3}=a,{x}_{4}=7$ and ${x}_{5}=b,a>b$, be 5 and 10 respectively. Then the Variance of the observations $n+{x}_{n},n=1,2,\ldots ,5$ is

Accepted Answer

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:

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

Calculate the variance:

$    	ext{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.

Solutions · 1