If the mean of 15 observations is 12 and standard deviation is 3. If 12 is replaced by 10 (in data) then the new mean is μ and variance is σ² then what is the value of 15 (μ + μ² + σ²)

The correct answer is 2429.

Let the 15 observations be $x_1,x_2,...,x_{15}$.

Given that the number of observations is:

$n=15$

The mean of these observations is 12, which gives:

$\overline{x}=\frac{\sum x_i}{15}=12$

So, the sum of the original observations is:

$\sum x_i=12\times 15=180$

The standard deviation is given as 3, which means the variance is:

$s^2=3^2=9$

Using the formula for variance:

$s^2=\frac{\sum x_i^2}{15}-{\overline{x}}^2$

Substitute the known values into the equation:

$9=\frac{\sum x_i^2}{15}-{12}^2$

$9=\frac{\sum x_i^2}{15}-144$

$\frac{\sum x_i^2}{15}=153$

Thus, the sum of squares of the original observations is:

$\sum x_i^2=153\times 15=2295$

Now, the observation 12 is replaced by 10. Let the new observations be denoted by $x_i^{\text{'}}$.

The new sum of observations is:

$\sum x_i^{\text{'}}=180-12+10=178$

So, the new mean $\mu$ is:

$\mu =\frac{178}{15}$

This also implies:

$15\mu =178$

The new sum of squares of the observations is:

$\sum {\left(x_i^{\text{'}}\right)}^2=2295-{12}^2+{10}^2=2295-144+100=2251$

The new variance ${\sigma }^2$ is given by:

${\sigma }^2=\frac{\sum {\left(x_i^{\text{'}}\right)}^2}{15}-{\mu }^2=\frac{2251}{15}-{\mu }^2$

We are asked to find the value of $15(\mu +{\mu }^2+{\sigma }^2)$. Substituting the expression for ${\sigma }^2$ into it:

$15(\mu +{\mu }^2+{\sigma }^2)=15(\mu +{\mu }^2+\frac{2251}{15}-{\mu }^2)$

Simplify inside the parentheses:

$=15(\mu +\frac{2251}{15})$

Distribute the 15:

$=15\mu +2251$

Substitute $15\mu =178$:

$=178+2251=2429$