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CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 4 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Dispersion
Quiz 4
Mean is a measure of _______.
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location (central value)
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dispersion
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correlation
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none of the above
The median of the following observations is
$$10, 8, -9, -12, 15, 0, 23, -3, -2, 13, -24, 28, 35, 42$$
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$$4$$
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$$-2$$
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$$7$$
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$$-9.75$$
What is the standard deviation of $$5, 5, 9, 9, 9, 10, 5, 10, 10$$?
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$$2.29$$
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$$6.48$$
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$$4.50$$
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$$8.00$$
If a constant $$25$$ is added to each observation of a set, the mean is _____.
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increased by $$25$$
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decreased by $$25$$
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$$25$$ times the original mean
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not affected
Which measure of dispersion is the quickest to compute?
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Standard deviation
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Quartile deviation
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Mean deviation
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Range
If all the observations are increased by $$10$$, then
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SD would be increased by $$10$$
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Mean deviation would be increased by $$10$$
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Both (A) and (B)
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None of above
The standard deviation of $$10, 16, 10, 16, 10, 10, 16, 16$$ is?
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$$4$$
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$$6$$
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$$3$$
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$$0$$
Explanation
Assume $$A=13;$$ hence $$d=A-13$$
x
d
$$d^2$$
$$10$$
$$-3$$
$$9$$
$$16$$
$$3$$
$$9$$
$$10$$
$$-3$$
$$9$$
$$16$$
$$3$$
$$9$$
$$10$$
$$-3$$
$$9$$
$$10$$
$$-3$$
$$9$$
$$16$$
$$-3$$
$$9$$
$$16$$
$$3$$
$$9$$
$$\displaystyle\sum d=0$$
$$\displaystyle\sum d^2=72$$
Use following formula:
$$\sqrt{\displaystyle\frac{\displaystyle\sum d^2}{n}-\left[\displaystyle\frac{\displaystyle\sum d}{n}\right]^2}$$
$$\sqrt{\displaystyle\frac{72}{8}-\left[\displaystyle\frac{0}{8}\right]^2}$$
$$\sqrt{9}$$
$$=3$$.
Class Intervals
$$30-40$$
$$40-50$$
$$50-60$$
$$60-70$$
$$70-80$$
Frequency
$$120$$
?
$$200$$
?
$$185$$
Cumulative Frequency
$$120$$
?
?
?
$$900$$
Cumulative frequency for class $$50-60=$$?
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$$715$$
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$$465$$
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$$145$$
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$$265$$
Explanation
Class Intervals Wages
Frequency(f)
Cumulative Frequency
$$30-40$$
$$120$$
$$120$$
$$40-50$$
$$f_1$$
$$120+f_1$$
$$50-60$$
$$200$$
$$320+f_1$$
$$60-70$$
$$f_2$$
$$320+f_1+f_2$$
$$70-80$$
$$185$$
$$900$$
Since median is given to be $$59.25$$, the median class is $$50-60$$. Thus, we can write, solving this for $$f_1$$, we get, $$f_1=145$$; Further, $$f_2=900-(120+145+200+185)=250$$.
The geometric mean of $$3, 6, 24 and 48$$ is ______.
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$$32$$
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$$12$$
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$$14$$
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$$24$$
Sum of the deviations about mean is ______.
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zero
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minimum
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maximum
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one
Harmonic mean gives more weight age to _______.
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small values
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large values
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positive values
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negative values
In a frequency distribution with open ends, one cannot find out ______.
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mean
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median
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mode
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all of the above
The middle value of an ordered series is called _______.
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$$2nd$$ quartile
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$$5th$$ decile
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$$50th$$ percentile
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All the above
Class Intervals
$$30-40$$
$$40-50$$
$$50-60$$
$$60-70$$
$$70-80$$
Frequency
$$120$$
?
$$200$$
?
$$185$$
Cumulative Frequency
$$120$$
?
?
?
$$900$$
Cumulative Frequency for class $$60-70=$$?
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0%
$$715$$
0%
$$465$$
0%
$$145$$
0%
$$265$$
Explanation
Class Intervals Wages
Frequency(f)
Cumulative Frequency
$$30-40$$
$$120$$
$$120$$
$$40-50$$
$$f_1$$
$$120+f_1$$
$$50-60$$
$$200$$
$$320+f_1$$
$$60-70$$
$$f_2$$
$$320+f_1+f_2$$
$$70-80$$
$$185$$
$$900$$
Since median is given to be $$59.25$$, the median class is $$50-60$$. Thus, we can write, Solving this for $$f_1$$ we get, $$f_1=145$$; Further $$f_2=900-(120+145+200+185)=250$$.
Extreme value have no effect on _______.
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average
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median
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geometric mean
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harmonic mean
Paasche Price Index Number = ?
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$$220.00$$
0%
$$216.30$$
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$$219.12$$
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$$221.98$$
Explanation
Paasche Price Index Number
$$P^{p}_{01} =\dfrac{\Sigma P_{1}Q_{1}}{\Sigma P_{0}Q_{1}} \times {100} = \dfrac{398}{184}\times {100} = 216.30$$.
Find the present value of Rs. $$10,000$$ to be required after $$5$$ years if the interest rate be $$9\%$$. Given that $$(1.09)^5=1.5386$$.
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$$6,994.42$$
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$$6,949.24$$
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$$6,449.24$$
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$$6,499.42$$
Explanation
Here, $$i=0.09=9\%$$
$$n=5$$
$$A_n=10,000$$
Required present value $$=\displaystyle\frac{A_n}{(1+i)^n}$$
$$=\displaystyle\frac{10,000}{(1+0.09)^5}$$
$$=Rs. 6499.42$$.
Calculate standard deviation of the following data.
X
$$0-10$$
$$10-20$$
$$20-30$$
$$30-40$$
$$40-50$$
f
$$10$$
$$8$$
$$15$$
$$8$$
$$4$$
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$$-12$$
0%
$$12$$
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$$12.36$$
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$$152.77$$
Explanation
Marks
Mid-values(X)
f
dxi$$=X-25/10$$
$$fdx_if$$
$$dx_i^2$$
$$0-10$$
$$5$$
$$10$$
$$-2$$
$$-20$$
$$40$$
$$10-20$$
$$15$$
$$8$$
$$-1$$
$$-8$$
$$8$$
$$20-30$$
$$25$$
$$15$$
$$0$$
$$0$$
$$0$$
$$30-40$$
$$35$$
$$8$$
$$1$$
$$8$$
$$8$$
$$40-50$$
$$45$$
$$4$$
$$2$$
$$8$$
$$16$$
Total
$$45$$
$$-12$$
$$72$$
$$\sqrt{\displaystyle\frac{72}{45}-\frac{144}{45\times 45}}\times 10$$
$$=12.36$$
Calculate standard deviation of the following data.
X
$$10$$
$$11$$
$$12$$
$$13$$
$$14$$
$$15$$
$$16$$
$$17$$
$$18$$
f
$$2$$
$$7$$
$$10$$
$$12$$
$$15$$
$$11$$
$$10$$
$$6$$
$$3$$
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$$3.95$$
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$$1.99$$
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$$14$$
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None of the above
Explanation
X
f
fx
X-Mean
$$f(X-x)^2$$
$$10$$
$$2$$
$$20$$
$$-4$$
$$32$$
$$11$$
$$7$$
$$77$$
$$-3$$
$$63$$
$$12$$
$$10$$
$$120$$
$$-2$$
$$40$$
$$13$$
$$12$$
$$156$$
$$-11$$
$$2$$
$$14$$
$$15$$
$$210$$
$$0$$
$$0$$
$$15$$
$$11$$
$$165$$
$$1$$
$$11$$
$$16$$
$$10$$
$$160$$
$$2$$
$$40$$
$$17$$
$$6$$
$$102$$
$$3$$
$$54$$
$$18$$
$$3$$
$$54$$
$$4$$
$$48$$
Total
$$76$$
$$fX=1064$$
$$f(X-X)^2=300$$
$$^-{X}=\displaystyle\frac{1064}{76}=14$$
$$\sigma x=\sqrt{\displaystyle\frac{300}{76}}$$
$$=\sqrt{3.95}$$
$$=1.99$$.
Y bought a CD Player costing Rs. $$13,000$$ by making a down payment of Rs. $$3,000$$ and agreeing to make equal annual payment for four years. How much would be each payment if the interest on unpaid amount be $$14\%$$ compounded annually?$$[P(4, 0.14)2.91371]$$
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Rs. $$3,432.05$$
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Rs. $$3,423.50$$
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Rs. $$3,342.05$$
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Rs. $$3,234.50$$
Explanation
V$$=$$A.P(n, i)
Here $$n=4$$ and $$i=0.14$$
$$=\displaystyle\frac{10,000}{P(4, 0.14)}$$
$$[P(20, 0.10)=8.51356$$ from table $$2$$(a)]
$$=\displaystyle\frac{10,000}{2.91371}$$
$$=$$Rs. $$3,432.05$$. Therefore each payment would be Rs. $$3,432.05$$.
The mean and standard deviation of $$10$$ observations are $$35$$ and $$2$$ respectively. Find the changed mean and standard deviation if each observation is increased by $$4$$.
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$$39$$ & $$6$$
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$$2$$ & $$39$$
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$$39$$ & $$2$$
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$$6$$ & $$39$$
Explanation
When each observation is increased by $$4$$, the mean of the changed observations will also increase and it will become equal to $$35+4=39$$.
Since increment in all observations by a constant does not change the standard deviation.
Thus, the mean and standard deviation when each observation is increased by $$4$$ would be $$39$$ and $$2$$ respectively.
Paasche Price Index = ?
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$$157.33$$
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$$153.14$$
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$$153.33$$
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$$157.14$$
Explanation
Paasche Price Index Number
$$P^{P}_{01} =\dfrac{\Sigma P_{1}Q_{1}}{\Sigma P_{0}Q_{1}} \times {100} = \dfrac{22}{14}\times{100} = 157.14$$.
Fisher Price Index Number = ?
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$$220.00$$
0%
$$216.30$$
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$$219.12$$
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$$221.98$$
Explanation
Fisher Price Index Number
$$P^{p}_{01} = \sqrt{P^{l}_{01}\times P^{p}_{01}} = \sqrt{(221.98\times216.3)\times{100}} = 219.12$$.
Find the mean and standard deviation of the following observations: X $$=2, 5, 7, 8, 13$$.
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$$7$$ & $$3.63$$
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$$3.63$$ & $$7$$
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$$7.63$$ & $$3$$
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$$3$$ & $$7.63$$
Explanation
$$\displaystyle\frac{2+5+7+8+13}{5}=7$$
$$\sqrt{\displaystyle\frac{4+25+49+64+169}{5}}-49=3.63$$.
Let $${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$$ be $$n$$ observations with mean $$m$$ and standard deviation $$s$$. Then the standard deviation of the observations $$a{ x }_{ 1 },a{ x }_{ 2 },.....a{ x }_{ n }$$, is
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$$a+s$$
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$$s/a$$
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$$\left| a \right| s$$
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$$as$$
Explanation
$$S.D(ax) = \sqrt{E[ax-a\bar{x}]^2}=\sqrt{E[a(x-\bar{x})^2]}=\sqrt{a^2.E[(x-\bar{x})^2]}=|a|.\sqrt{E[(x-\bar{x})^2]}=|a|.S.D(x)=|a| s$$
The Mean deviation about A.M. of the numbers $$3, 4, 5, 6, 7 $$ is :
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$$25$$
0%
$$5$$
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$$1.2$$
0%
$$0$$
Explanation
Mean deviation is defined as the average of the absolute deviation of the observations from their mean.
Here, the arithmetic mean of the given observations is $$\dfrac{3 + 4 + 5 + 6 + 7}{5} = 5$$
Thus, mean deviation = $$\dfrac{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}{5} = \dfrac{6}{5} = 1.2$$
Hence, option 'C' is correct.
Mean deviation for $$n$$ observations $${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$$ from their mean $$\bar { X } $$ is given by:
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$$\sum _{ i=1 }^{ n }{ ( { x }_{ i }-\bar { X }) }$$
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$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\bar { X } \right| } $$
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$$\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$$
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$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } } $$
Explanation
It is fundamental concept that mean deviation of $$n$$ observations $$x_1, x_2, x_3, ......x_n$$ about mean $$\bar{X}$$ is given by, $$\displaystyle \frac{1}{n} \sum_{i=1}^{i=n} |x_i-\bar{X}|$$
If a variable $$X$$ takes values $$0,1,2,....,n$$ with frequencies $${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }\quad $$ respectively, then S.D. is equal to :
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$$\cfrac { n }{ 4 } $$
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$$\cfrac { n }{ 2 } $$
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$$\cfrac { \sqrt { n } }{ 2 } $$
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$$\cfrac { \sqrt { n } }{ 4 } $$
Explanation
Here, $$\displaystyle \mu1'$$ $$=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$$ $$\displaystyle $$
and
$$\displaystyle \mu 2' $$$$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$$
$$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1
\right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$$
$$\therefore$$ Variance $$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4} \mbox{or} S.D = \sigma = \sqrt{\sigma^2} = \frac{\sqrt{n}}{2}$$
If the mean deviation about mean $$1,1+d,1+2d,....1+100d$$ from their mean is $$255$$, then the $$d$$ is equal to
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$$10$$
0%
$$20$$
0%
$$10.1$$
0%
$$20.2$$
Explanation
Clearly given observation is in A.P, and have $$101$$ terms.
$$\therefore$$ Mean $$\displaystyle =\frac{1}{2}\left \{ 1+1+100d\right \}=1+50d$$
And thus mean deviation about mean $$\displaystyle =\frac{1}{101}\sum \left | x-\bar{x} \right |$$ $$\displaystyle =\frac{1}{101}.2d\left ( 1+2+3\cdots +50 \right )$$ $$\displaystyle =\frac{50\left ( 51 \right )d}{101}=255$$ (given)
$$\Rightarrow d = 10.1$$
The mean deviation of the series $$a,a+d,a+2d+......,a+(2n-1)d,a+2nd$$ about the mean is
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$$\cfrac { n+1 }{ 2n+1 } $$
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$$\cfrac { n(n+1)d }{ 2n+1 } $$
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$$\cfrac { (n+2)d }{ 2n } $$
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$$\cfrac { (n-1)d }{ 2n+1 } $$
Explanation
Mean $$\displaystyle\overline{x}=\frac{(a)+(a+d)+(a+2d)+...+(a+2nd)}{2n+1}$$
where $$\displaystyle\overline{x}=\frac{\sum x_i}{n}$$
Then, $$\displaystyle\frac{\frac{2n+1}{2}(a+a+2nd)}{2n+1}$$
From an A.P. $$\displaystyle S_n = \frac{n}{2} (a+l)$$ which gives $$\overline{x} = a+nd$$
The series being a,a+d,a+2d,...,a+(n-1)d,a+nd,a+(n+1)d,...,a+2nd
Mean deviation from the mean
= $$\displaystyle\frac{1}{N} \sum f_i [x_i- \overline{x}]$$
= $$\displaystyle\frac{1}{2n+1} \sum [x_i -a-nd]$$
= $$\displaystyle\frac{1}{2n+1} [nd+(n-1)d+(n-2)d+...+d+0+d+...+nd]$$
= $$\displaystyle\frac{2d}{2n+1} [n+(n-1)+(n-2)+...1]$$
= $$\displaystyle\frac{2d}{2n+1} .\frac{n(n+1)}{2} = \frac{n(n+1)d}{2n+1}$$
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