The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then $\mathrm{a+3 b-5}$ is equal to ___________.
Answer (integer)
37
Solution
<p>$\sum {{x_i} = 7 \times 8 = 56}$</p>
<p>${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} = 16$</p>
<p>${{\sum {x_i^2} } \over 7} - 64 = 16$</p>
<p>$\sum {x_i^2 = 560}$</p>
<p>when 14 is omitted</p>
<p>$\sum {{x_i} = 56 - 14 = 42}$</p>
<p>New mean $= a = {{\sum {{x_i}} } \over 6} = 7$</p>
<p>$\sum {x_i^2 = 560 - 196 = 364}$</p>
<p>new variance, $b = {{\sum {x_i^2} } \over 6} - {\left( {{{\sum {{x_i}} } \over 6}} \right)^2}$</p>
<p>$= {{364} \over 6} - 49 = {{35} \over 3}$</p>
<p>$3b = 35$</p>
<p>$a + 3b - 5 = 7 + 35 - 5 = 37$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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