The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12. If μ and σ² denote that mean and variance of the correct observations respectively, then 15( μ + μ² + σ²)  is equal to ______

The correct answer is 2521.

We are given n = 15 observations with:

Incorrect Mean = 12, Incorrect Standard Deviation = 3

An observation was recorded as 10 instead of the correct value 12.

Step 1: Find the Correct Mean (μ)

The incorrect sum of all observations:

$\Sigma x_{\mathrm{incorrect}}=15\times 12=180$

Replacing the wrong observation (10) with the correct one (12):

$\Sigma x_{\mathrm{correct}}=180-10+12=182$

Correct Mean:

$\mu =\frac{182}{15}$

Step 2: Find the Correct Variance (σ²)

From the incorrect data: Variance = (Standard Deviation)�� = 3² = 9.

Using the identity $\mathrm{Variance}=\frac{\Sigma x^2}{n}-{\left(\mathrm{Mean}\right)}^2$, we get:

$9=\frac{\Sigma x_{\mathrm{incorrect}}^2}{15}-{12}^2$

$\frac{\Sigma x_{\mathrm{incorrect}}^2}{15}=9+144=153$

$\Sigma x_{\mathrm{incorrect}}^2=15\times 153=2295$

Now correct the sum of squares by removing the wrong entry and adding the right one:

$\Sigma x_{\mathrm{correct}}^2=2295-{10}^2+{12}^2=2295-100+144=2339$

Correct Variance:

${\sigma }^2=\frac{\Sigma x_{\mathrm{correct}}^2}{n}-{\mu }^2$

${\sigma }^2=\frac{2339}{15}-{\left(\frac{182}{15}\right)}^2$

${\sigma }^2=\frac{2339}{15}-\frac{33124}{225}=\frac{35085}{225}-\frac{33124}{225}=\frac{1961}{225}$

Step 3: Calculate 15(μ + μ² + σ²)

Now combine all terms over a common denominator of 225:

$\mu =\frac{182}{15}=\frac{2730}{225}$

${\mu }^2=\frac{33124}{225}$

${\sigma }^2=\frac{1961}{225}$

$\mu +{\mu }^2+{\sigma }^2=\frac{2730+33124+1961}{225}=\frac{37815}{225}$

Finally:

$15(\mu +{\mu }^2+{\sigma }^2)=15\times \frac{37815}{225}=\frac{37815}{15}=$2521