JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Statistics

Consider a set of

3 n

numbers having variance 4. In this set, the mean of first

2 n

numbers is 6 and the mean of the remaining n number is 3. A new set is constructed by adding 1 into each of first

2 n

numbers, and subtracting 1 from each of the remaining

n

numbers. If the variance of the new set is

k

, then

9 k

is equal to ____.

Answer: 68

Solution

The mean of all

3 n

numbers is

\frac{2 n ( 6 ) + n ( 3 )}{3 n} = 5

, so from the variance,

4 = \frac{\sum x _{i}^{2}}{3 n} - 25 ⟹ \sum x_{i}^{2} = 87 n

After the change the total becomes

(12 n + 2 n) + (3 n - n) = 16 n

, so the new mean is

\frac{16}{3}

. Also

\sum y_{i}^{2} = \sum x_{i}^{2} + 2 first 2 n \sum x_{i} - 2 last n \sum x_{i} + 3 n = 87 n + 24 n - 6 n + 3 n = 108 n

Hence

k = \frac{108 n}{3 n} - (\frac{16}{3})^{2} = 36 - \frac{256}{9} = \frac{68}{9}

and

9 k = 68

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer