If the mean of the discrete distribution 8, 9 | vedclass

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(B) We have,$	ext{Mean} = 6$. $\frac{8+9+6+5+x+4+6+5}{8} = 6$ $\Rightarrow 43+x = 48$ $\Rightarrow x = 5$. So,the data set is $8, 9, 6, 5, 5, 4, 6, 5

Vedclass · Accepted Answer

$1.58$

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(B) We have,$	ext{Mean} = 6$. $\frac{8+9+6+5+x+4+6+5}{8} = 6$ $\Rightarrow 43+x = 48$ $\Rightarrow x = 5$. So,the data set is $8, 9, 6, 5, 5, 4, 6, 5$. The sum of observations $\Sigma x_i = 48$ and the sum of squares $\Sigma x_i^2 = 8^2 + 9^2 + 6^2 + 5^2 + 5^2 + 4^2 + 6^2 + 5^2 = 64 + 81 + 36 + 25 + 25 + 16 + 36 + 25 = 308$. The variance is given by $\sigma^2 = \frac{1}{N} \Sigma x_i^2 - (\bar{x})^2 = \frac{308}{8} - (6)^2 = 38.5 - 36 = 2.5$. The standard deviation is $\sigma = \sqrt{2.5} \approx 1.58$.

$x_{i}$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
$f_{i}$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$

If the mean of the discrete distribution $8, 9, 6, 5, x, 4, 6, 5$ is $6$,then its standard deviation (nearest to two decimal places) is

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