Problems from Absolute deviation and standard deviation

Development:

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

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

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

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

The standard deviation is then $$D_{\overline{x}}=\dfrac{217,3}{31}=7,01$$.

Solution:

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

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Development:

The goals are: $$58, 58, 57, 69, 53, 60, 59$$ and $$57$$. The average is $$$\overline{x}=\dfrac{58+58+57+69+53+60+59+57}{8}\simeq58,9$$$ Next we calculate the deviations indicating the result in a table:

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

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 $$$\overline{x}=\dfrac{65+54+54+54+53+50+49+48+46+29+24+24+23+19+17}{15}=$$$ $$$=\dfrac{609}{15}=40,6$$$

The standard deviation is then: $$$D_{\overline{x}}=\dfrac{|65-40,6|+|54-40,6|+|54-40,6|+|54-40,6|+|53-40,6|+|50-40,6|}{15}+$$$ $$$+\dfrac{|49-40,6|+|48-40,6|+|46-40,6|+|29-40,6|+|24-40,6|+|24-40,6|}{15}+$$$ $$$+\dfrac{|23-40,6|+|19-40,6|+|17-40,6|}{15}=$$$ $$$=\dfrac{24,4+13,4+13,4+13,4+12,4+9,4+8,4+7,4+5,4+11,6+16,6+16,6}{15} $$$ $$$+\dfrac{17,6+21,6+23,6}{15}=\dfrac{215,2}{15}\simeq14,35$$$

Solution:

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

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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 $$\overline{x}=\dfrac{70}{20}=3,5$$, and find the two columns that simplify the calculation of the standard deviation:

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

The standard deviation is then: $$D_{\overline{x}}=\dfrac{30}{20}=1,5$$.

Solution:

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

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Development:

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

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

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

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 $$$\overline{x}=x_i-D_i=14-3,2=10,8$$$ the average is found.

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

Solution:

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

b) $$D_i=-7,8$$

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