JEE Mathematics 2023 Question with Solution

Given: The mean of $10$ students was taken as $50$ and the standard deviation was taken as $12$ using wrongly read marks $45$ and $50$ in place of $20$ and $25$.

Find: The correct variance.

Let the total sum of the marks of the $10$ students be $\Sigma x_i$. From the given mean:

Now correct the sum by replacing the wrongly read values:

Using the variance relation with the given standard deviation $12$:

Since $\sigma = 12$, we have

So,

Now correct the sum of squares:

Therefore, the corrected variance is

Thus, the correct variance is $269$.

Given: Recorded data gave mean $50$ and standard deviation $12$ for $10$ students.

Find: The corrected variance after replacing $45, 50$ by $20, 25$.

First use the incorrect data to recover the original totals. From the mean,

From the standard deviation,

Using

we get

Now correct the total sum:

Hence the corrected mean is

Correct the sum of squares in the same way:

Now apply the variance formula again with corrected values:

Therefore, the correct variance is $269$.