JEE PYQ: Statistics Question 3
Question 3 - 2021 (17 Mar Shift 2)
Consider a set of $3n$ numbers having variance 4. In this set, the mean of first $2n$ numbers is 6 and the mean of the remaining $n$ numbers is 3. A new set is constructed by adding 1 into each of first $2n$ numbers, and subtracting 1 from each of the remaining $n$ numbers. If the variance of the new set is $k$, then $9k$ is equal to ____.
Show Answer
Answer: 68
Solution
$\sum a^2 + \sum b^2 = 87n$. New variance $= \frac{108}{3} - \left(\frac{16}{3}\right)^2 = \frac{68}{9}$. So $9k = 68$.
BETA
