Question

the variance of 15 observation is 4 if eacg observation is increased by 9 the variance of theresulting observation mcq​

Answer 1

Given that the variance of 15 observations is 4, i.e.,

$\longrightarrow\sigma^2=4$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2-\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2=4$

Note that the variance can be defined as square of mean of observations subtracted from mean of square of observations. $\displaystyle\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2$ is mean of squares and $\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$ is square of mean.

Now each observation is increased by 9, i.e.,

• $y_i=x_i+9,\quad i\in\mathbb{Z},\quad\!\!1\leq i\leq 15.$

Then, mean of squares,

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i+9)^2$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}\left((x_i)^2+18x_i+81)$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{1}{15}\sum_{i=1}^{15}18x_i+\dfrac{1}{15}\sum_{i=1}^{15}81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{18}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\cdot15\cdot81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{6}{5}\sum_{i=1}^{15}x_i+81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2=\dfrac{6}{5}\sum_{i=1}^{15}x_i+81\quad\quad\dots(1)$

And, square of mean,

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}(x_i+9)\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\sum_{i=1}^{15}9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\cdot15\cdot9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2+\dfrac{18}{15}\sum_{i=1}^{15}x_i+81$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2=\dfrac{6}{5}\sum_{i=1}^{15}x_i+81\quad\quad\dots(2)$

Comparing (1) and (2) where the RHS are same,

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2=\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$

Here the LHS is the variance of new observations and RHS is the variance is the old observations, i.e.,

$\displaystyle\longrightarrow\underline{\underline{(\sigma')^2=\sigma^2}}$

That is, the variance and standard deviation never change whenever each observation is increased or decreased by a constant.

Hence the answer to the question is 4.

Answer 2

$\huge\bf\fbox\red{Answer:-}$

Given that the variance of 15 observations is 4, i.e.,

$\longrightarrow\sigma^2=4$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2-\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2=4$

Note that the variance can be defined as square of mean of observations subtracted from mean of square of observations. $\displaystyle\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2$is mean of squares and $\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$

is square of mean.

Now each observation is increased by 9, i.e.,

$y_i=x_i+9,\quad i\in\mathbb{Z},\quad\!\!1\leq i\leq 15.$

Then, mean of squares,

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i+9)^2$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}\left((x_i)^2+18x_i+81)$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{1}{15}\sum_{i=1}^{15}18x_i+\dfrac{1}{15}\sum_{i=1}^{15}81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{18}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\cdot15\cdot81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2+\dfrac{6}{5}\sum_{i=1}^{15}x_i+81$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2=\dfrac{6}{5}\sum_{i=1}^{15}x_i+81\quad\quad\dots(1)$

And, square of mean,

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}(x_i+9)\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\sum_{i=1}^{15}9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+\dfrac{1}{15}\cdot15\cdot9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i+9\right)^2$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2+\dfrac{18}{15}\sum_{i=1}^{15}x_i+81⟶( ∑)$

$\displaystyle\longrightarrow\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2=\dfrac{6}{5}\sum_{i=1}^{15}x_i+81\quad\quad\dots(2)$

Comparing (1) and (2) where the RHS are same,

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2=\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$

$\displaystyle\longrightarrow\dfrac{1}{15}\sum_{i=1}^{15}(y_i)^2-\left(\dfrac{1}{15}\sum_{i=1}^{15}y_i\right)^2=\dfrac{1}{15}\sum_{i=1}^{15}(x_i)^2-\displaystyle\left(\dfrac{1}{15}\sum_{i=1}^{15}x_i\right)^2$

Here the LHS is the variance of new observations and RHS is the variance is the old observations, i.e.,

$\displaystyle\longrightarrow\underline{\underline{(\sigma')^2=\sigma^2}}$

That is, the variance and standard deviation never change whenever each observation is increased or decreased by a constant.

Hence the answer to the question is 4.