The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is :
Solution
Let the two remaining observations be x and y.<br><br>$\because$ $\bar x = 10 = {{5 + 7 + 10 + 12 + 14 + 15 + x + y} \over 8}$<br><br>$\Rightarrow x + y = 17$ ....(1)<br><br>$\because$ $${\mathop {\rm var}} (x) = 13.5 = {{25 + 49 + 100 + 144 + 196 + 225 + {x^2} + {y^2}} \over 8} - {(10)^2}$$<br><br>$\Rightarrow {x^2} + {y^2} = 169$ ....(2)<br><br>From (1) and (2)<br><br>(x, y) = (12, 5) or (5, 12)<br><br>So $\left| {x - y} \right| = 7$
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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