JEE Mathematics 2024 Question with Solution

Given: The incorrect mean is $12$, the incorrect standard deviation is $3$, and one observation $10$ should be replaced by $12$. The number of observations is $15$.

Find: The value of $15(\mu + \mu^2 + \sigma^2)$ for the corrected data.

From the mean,

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

After correcting one observation,

$$\sum x = 180 - 10 + 12 = 182$$

Hence the corrected mean is

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

Now use the relation between variance, mean, and sum of squares:

$$\sigma^2 = \frac{\sum x^2}{15} - \mu^2$$

For the incorrect data, the variance is

$$3^2 = 9$$

So,

$$9 = \frac{\sum x^2}{15} - 12^2$$ $$\frac{\sum x^2}{15} = 9 + 144 = 153$$ $$\sum x^2 = 15 \times 153 = 2295$$

Correcting the mistaken entry gives

$$\sum x^2 = 2295 - 10^2 + 12^2 = 2295 - 100 + 144 = 2339$$

Therefore,

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

Now compute the required expression:

$$15(\mu + \mu^2 + \sigma^2)$$

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

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

The $\mu^2$ terms cancel, so

$$15\left(\frac{182}{15} + \frac{2339}{15}\right)$$ $$= 182 + 2339 = 2521$$

Therefore, the required value is $2521$.

Given: Incorrect mean $= 12$, incorrect standard deviation $= 3$, one value $10$ is corrected to $12$, and the number of observations is $15$.

Find: $15(\mu + \mu^2 + \sigma^2)$.

Use the identity

$$\mu^2 + \sigma^2 = \frac{\sum x^2}{15}$$

Hence,

$$15(\mu + \mu^2 + \sigma^2) = 15\mu + \sum x^2$$

Now,

$$15\mu = \sum x = 180 - 10 + 12 = 182$$

Also, from the incorrect data,

$$\sum x^2 = 15(\sigma^2 + \mu^2) = 15(9 + 12^2) = 15(153) = 2295$$

After correction,

$$\sum x^2 = 2295 - 100 + 144 = 2339$$

Therefore,

$$15(\mu + \mu^2 + \sigma^2) = 182 + 2339 = 2521$$

Therefore, the correct answer is $2521$. This method works because $\mu^2 + \sigma^2$ immediately converts to the average of the squares, avoiding separate variance computation.