JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Statistics

The mean of 6 distinct observations is 6.5 and their variance is 10.25 . If 4 out of 6 observations are

2, 4, 5

and 7 , then the remaining two observations are :

Answer: (A)

Solution

Six observations with mean

6.5

have

\sum x_{i} = 39

, and variance

10.25

gives

\sum x_{i}^{2} = 6 (10.25 + 6. 5^{2}) = 6 (52.5) = 315

The four known values contribute

2 + 4 + 5 + 7 = 18

to the sum and

4 + 16 + 25 + 49 = 94

to the sum of squares, so the remaining two values

p

and

q

satisfy

p + q = 21

and

p^{2} + q^{2} = 221

. Then

2 pq = 2 1^{2} - 221 = 220

, so

p

and

q

are the roots of

t^{2} - 21 t + 110 = 0

, which factorises as

(t - 10) (t - 11) = 0

. The two remaining observations are

10

and

11

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option