Verified Solution • 2025–26 Session moderate

QJEEM21Ma6506: JEE Main 2021 Mathematics - Class 11 JEEM21Ma6506 Solution: The mean of 6 distinct observations is 6.5 and their variance…

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 :

Multiple Choice Options

  • A 10, 11
  • B 3, 18
  • C 8, 13
  • D 1, 20

Step-by-Step Verified Solution

Let other two numbers be a, (21 −- a)

Now,

10.25=(4+16+25+49+a2+(21−a)26−(6.5)210.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}

(Using formula for variance)

⇒6(10.25)+6(6.5)2=94+a2+(21−a)2\Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}

⇒a2+(21−a)2=221\Rightarrow {a^2} + {(21 - a)^2} = 221

∴\therefore a = 10 and (21 −- a) = 21 −- 10 = 11

So, remaining two observations are 10, 11.

⇒\Rightarrow Option (1) is correct.

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