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 :
Solution
Let other two numbers be a, (21 $-$ a)<br><br>Now,<br><br>$10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}$<br><br>(Using formula for variance)<br><br>$\Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}$<br><br>$\Rightarrow {a^2} + {(21 - a)^2} = 221$<br><br>$\therefore$ a = 10 and (21 $-$ a) = 21 $-$ 10 = 11<br><br>So, remaining two observations are 10, 11.<br><br>$\Rightarrow$ Option (1) is correct.
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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