The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and ${9 \over 4}$ respectively. Then ${x^4} + {y^2}$ is equal to :
Solution
<p>Mean $= {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6$</p>
<p>$\therefore$ $x + y = 12$ ..... (i)</p>
<p>And variance</p>
<p>$$ = {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}$$</p>
<p>$= {9 \over 4}$</p>
<p>$\therefore$ ${(x - 6)^2} + {(y - 6)^2} = 8$ ..... (ii)</p>
<p>From (i) and (ii)</p>
<p>x = 4 and y = 8</p>
<p>$\therefore$ ${x^4} + {y^2} = 320$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Dispersion
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