The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :
Solution
Let 8, 16, x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub> be the observations.<br><br>Now, ${{{x_1} + {x_2} + .... + {x_5} + 14} \over 7} = 8$<br><br>$\Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 42}$ .... (1)<br><br>Also, ${{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16$<br><br>$\Rightarrow \sum\limits_{i = 1}^5 {x_i^2 = 560 - 100 = 460}$ ..... (2)<br><br>So variance of x<sub>1</sub>, x<sub>2</sub>, ......., x<sub>5</sub><br><br>$$ = {{460} \over 5} - {\left( {{{42} \over 5}} \right)^2} = {{2300 - 1764} \over {25}} = {{536} \over {25}}$$
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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