Problems from Absolute deviation and standard deviation

Development:

The table is filled in to make easier to calculate the average and the standard deviation:

	$x_i$	$f_i$	$x_i{f}_i$	$|x_i-\bar{x}|$	$|x_i-\bar{x}|\cdot f_i$
$[140,155)$	$147,5$	$3$	$442,5$
$[155,165)$	$160$	$6$	$960$
$[165,175)$	$170$	$17$	$2890$
$[175,190)$	$182,5$	$5$	$912,5$
			$5205$

To be able to fill in the last 2 columns the average is calculated $\bar{x}={\displaystyle \frac{5205}{31}}=167,9$

	$x_i$	$f_i$	$x_i{f}_i$	$|x_i-\bar{x}|$	$|x_i-\bar{x}|\cdot f_i$
$[140,155)$	$147,5$	$3$	$442,5$	$20,4$	$61,2$
$[155,165)$	$160$	$6$	$960$	$7,9$	$47,4$
$[165,175)$	$170$	$17$	$2890$	$2,1$	$35,7$
$[175,190)$	$182,5$	$5$	$912,5$	$14,6$	$73$
			$5205$		$217,3$

The standard deviation is then $D_{\bar{x}}={\displaystyle \frac{217,3}{31}}=7,01$.

Solution:

With these values, we have $D_{\bar{x}}=7,01$

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Development:

The goals are: $58,58,57,69,53,60,59$ and $57$. The average is $$\bar{x}={\displaystyle \frac{58+58+57+69+53+60+59+57}{8}}\simeq 58,9$$ Next we calculate the deviations indicating the result in a table:

Goals	$D_i=x_i-\bar{x}=x_i-58,9$
$53$	$-5,9$
$57$	$-1,9$
$57$	$-1,9$
$58$	$-0,9$
$58$	$-0,9$
$59$	$0,1$
$60$	$1,1$
$69$	$10,1$

Solution:

Note	$D_i$
$53$	$-5,9$
$57$	$-1,9$
$57$	$-1,9$
$58$	$-0,9$
$58$	$-0,9$
$59$	$0,1$
$60$	$1,1$
$69$	$10,1$

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Development:

The average is calculated first $$\bar{x}={\displaystyle \frac{65+54+54+54+53+50+49+48+46+29+24+24+23+19+17}{15}}=$$ $$={\displaystyle \frac{609}{15}}=40,6$$

The standard deviation is then: $$D_{\bar{x}}={\displaystyle \frac{|65-40,6|+|54-40,6|+|54-40,6|+|54-40,6|+|53-40,6|+|50-40,6|}{15}}+$$ $$+{\displaystyle \frac{|49-40,6|+|48-40,6|+|46-40,6|+|29-40,6|+|24-40,6|+|24-40,6|}{15}}+$$ $$+{\displaystyle \frac{|23-40,6|+|19-40,6|+|17-40,6|}{15}}=$$ $$={\displaystyle \frac{24,4+13,4+13,4+13,4+12,4+9,4+8,4+7,4+5,4+11,6+16,6+16,6}{15}}$$ $$+{\displaystyle \frac{17,6+21,6+23,6}{15}}={\displaystyle \frac{215,2}{15}}\simeq 14,35$$

Solution:

$D_{\bar{x}}\simeq 14,35$

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Development:

We add a column to help with the calculation of the average:

$x_i$	$f_i$	$x_i{f}_i$
$1$	$4$	$4$
$2$	$2$	$4$
$3$	$4$	$12$
$4$	$3$	$12$
$5$	$4$	$20$
$6$	$3$	$18$
		$70$

Find then the average $\bar{x}={\displaystyle \frac{70}{20}}=3,5$, and find the two columns that simplify the calculation of the standard deviation:

$x_i$	$f_i$	$x_i{f}_i$	$|x_i-\bar{x}|$	$|x_i-\bar{x}|\cdot f_i$
$1$	$4$	$4$	$2,5$	$10$
$2$	$2$	$4$	$1,5$	$3$
$3$	$4$	$12$	$0,5$	$2$
$4$	$3$	$12$	$0,5$	$1,5$
$5$	$4$	$20$	$1,5$	$6$
$6$	$3$	$18$	$2,5$	$7,5$
		$70$		$30$

The standard deviation is then: $D_{\bar{x}}={\displaystyle \frac{30}{20}}=1,5$.

Solution:

$D_{\bar{x}}=1,5$

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Development:

We must know the average before applying the formula $D_i=x_i-\bar{x}$.

To find the average we add up all the terms and divide by the number of throws: $$\bar{x}={\displaystyle \frac{1+1+1+3+3+4+4+5+6+6}{10}}={\displaystyle \frac{34}{10}}=3,4$$ Again the deviations are classified in a table, regardless of the frequencies:

Result of the throw	$D_i=x_i-\bar{x}=x_i-3,4$
$1$	$-2,4$
$3$	$-0,4$
$4$	$0,6$
$5$	$1,6$
$6$	$2,6$

Solution:

Taking $1,1,1,3,3,4,4,5,6,6$ as samples , we have:

Result of the throw	$D_i$
$1$	$-2,4$
$3$	$-0,4$
$4$	$0,6$
$5$	$1,6$
$6$	$2,6$

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Development:

a) Applying the formula $$\bar{x}=x_i-D_i=14-3,2=10,8$$ the average is found.

b) Next, we apply the equation $$D_i=x_i-\bar{x}=3-10,8=-7,8$$ to find the deviation of Marc's scoring with regard to the average.

Solution:

a) $\bar{x}=10,8$

b) $D_i=-7,8$

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