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CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 11 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Dispersion
Quiz 11
If the data $$x_1, x_2, .., x_{10}$$ is such that the mean of first four of these is $$11$$, the mean of the remaining six is $$16$$ and the sum of square of all of these is $$2,000$$; then the standard deviation of this data is?
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$$4$$
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$$2$$
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$$\sqrt{2}$$
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$$2\sqrt{2}$$
If the mean deviation about the median of the numbers $$a,2a,....,50a$$ is $$50$$, then $$|a|$$ equals:-
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$$4$$
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$$5$$
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$$2$$
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$$3$$
Explanation
Given series:- $$a, 2a, 3a, ....., 50a$$
Median $$= \cfrac{25a + 26a}{2} = 25.5 a$$
Mean deviation about median $$= 50 \quad \left( \text{Given} \right)$$
Mean deviation $$= \cfrac{\sum_{i = 1}^{50}{\left| {x}_{i} - 25.5a \right|}}{50}$$
$$\therefore \cfrac{24.5a + 23.5a + 22.5a + ..... + 23.5a + 24.5a}{50} = 50$$
$$\Rightarrow a + 3a + 5a + ..... + 47a + 49a = 2500$$
$$\Rightarrow \cfrac{25}{2} \left( 2a + \left( 25 - 1 \right) 2a \right) = 2500$$
$$\Rightarrow 2a \times 25 = 200$$
$$\Rightarrow a = \cfrac{200}{50} = 4$$
Standard deviation of first $$n$$ odd natural numbers is
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$$\sqrt{n}$$
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$$\sqrt{\dfrac{(n+2)(n+1)}{3}}$$
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$$\sqrt{\dfrac{n^2-1}{3}}$$
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$$n$$
Explanation
Stndard deviation, $$\sigma = \sqrt{\dfrac{\sum x_i^2}{N}-(\bar{x})^2}$$
$$\therefore \bar{x} = \dfrac{\sum x_i}{N}$$
$$= \dfrac{1 + 3 +5 + ...(2n-1)}{n}$$
$$= \dfrac{\dfrac{n}{2}[1+2n-1]}{n}$$
$$=\dfrac{n^2}{n} = n$$
Again, $$\sum x_i^2 = 1^2 + 3^2 + 5^2 + ...(2n - 1)^2$$
$$=\sum (2n-1)^2$$
$$=\sum(4n^2 - 4n+1)$$
$$= 4\sum n^2 - 4\sum n + \sum 1$$
$$= \dfrac{4n(n+1)(2n+1)}{6} - \dfrac{4n(n+1)}{2}+n$$
$$= n \left[ \dfrac{2}{3} (n + 1)(2n + 1) - 2(n+1)+1\right]$$
$$= \dfrac{n}{3} [2(2n^2+3n+1) - 6(n+1) + 3]$$
$$= \dfrac{n}{3} [ 4n^2 + 6n + 2 - 6n - 6 + 3]$$
$$= \dfrac{n}{3} [4n^2 - 1]$$
$$\therefore \delta = \sqrt{\dfrac{n(4n^2 - 1)}{3n} - n^2}$$
$$= \sqrt{\dfrac{4n^2-1}{3}-n^2}$$
$$= \sqrt{\dfrac{4n^2-1-3n^2}{3}}$$
$$= \sqrt{\dfrac{n^2 - 1}{3}}$$
The standard deviation of the data $$6,5,9, 13, 12, 8, 10$$ is
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$$\dfrac{\sqrt{52}}{7}$$
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$$\dfrac{52}{7}$$
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$$\dfrac{\sqrt{53}}{7}$$
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$$\dfrac{53}{7}$$
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$$6$$
Explanation
Given data $$6,5,9,13,12,8,10$$ Mean of the given data $$(\bar{x})$$
$$=\dfrac{6+5+9+13+12+8+10}{7}$$
$$=\dfrac{63}{7}=9$$
The deviation of the respective data from the mean i.e. $$(x_i-\bar{x})$$ are
$$6-9,5-9,9-9,13-9,12-9,8-9,10-9$$
$$(x_i-\bar{x})=-3,-4,0,4,3,-1,1$$
$$(x_1-\bar{x})^2=9,16,0,16,9,1,1$$
$$\displaystyle \sum^7_{i=i} (x_1-\bar{x})^2=9+16+0+16+9+1+1=52$$
$$\therefore$$ standard deviation $$(\sigma)$$
$$\displaystyle =\sqrt{\dfrac{1}{n}\sum^7_{i=1} (x_1-\bar{x})^2}=\sqrt{\dfrac{52}{7}}$$
If $$ N = 10 \sum x = 120 $$ and $$ \sigma _{x} = 60 $$ , the variation coefficient is -
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$$5$$
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$$50$$
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$$500$$
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$$0.5$$
Explanation
Variation coefficient
$$= \dfrac{\sigma _{x}}{Mean} \times 100$$
$$ = \dfrac{60 \times 10}{120} \times = 500$$
Mean of variable series $$\bar{x} = 773$$ and mean deviation 64.4 , then coefficient of mean deviation is
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$$0.065$$
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$$12.003$$
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$$0.083$$
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$$0.073$$
Explanation
$$\dfrac{Mean deviation}{Mean} = \dfrac{64.4}{773} = 0.083 $$
Thus (C) is correct
Which one of the following is a false description?
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In a moderately asymmetrical distribution, the empirical relationship between Mean, Mode and Median suggested by Karl Pearson is Mean Mode = 3 (Mean Median)
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Coeffcient of variation is an absolute measure of dispersion
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Measure of skewness in the distribution of numerical values in the data set
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Kurtosis refers to the degree of flatness or peakedness in the region around the mode of a frequency curve
Mean and standard deviation of a data are $$48$$ and $$12$$ respectively. The coefficient of variation is.
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$$42$$
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$$25$$
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$$28$$
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$$48$$
Given $$\sum (x-\bar{x})^2=48, \bar{x}=20$$ and $$n=12$$. The coefficient of variation is.
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$$25$$
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$$20$$
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$$30$$
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$$10$$
The sum of the squares of deviations of 10 items about mean 50 is 250 .The coefficient of variation is
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10%
0%
50%
0%
30%
0%
none of these
Explanation
$$\textbf{Step -1: Find the standard deviation.}$$
$$\text{It is given, }$$
$$\text{Mean, }\overline{x}=50,$$
$$n=10\text{ and }$$
$$\sum(x-\overline{x})^2=250$$
$$\therefore\text{Standard deviation}=\sqrt{\dfrac{\sum(x-\overline{x})^2}{n}}$$
$$=\sqrt{\dfrac{250}{10}}$$
$$=5$$
$$\textbf{Step -2: Find the coefficient of variance.}$$
$$\text{Coefficient of variation}=\dfrac{\text{Standard deviation}}{\text{Mean}}\times100\%.$$
$$=\dfrac{5}{50}\times100\%$$
$$=10\%$$
$$\textbf{Hence, option A is correct.}$$
The probability distribution of a random variable $$X$$ is given below:
$$X=x$$
0
1
2
3
$$P(X=x)$$
$$\frac{1}{10}$$
$$\frac{2}{10}$$
$$\frac{3}{10}$$
$$\frac{4}{10}$$
Then the variance of $$X$$ is
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1
0%
2
0%
3
0%
4
Explanation
$$E[x^2]=0+1^2\cdot \displaystyle \frac{2}{10}+2^2\cdot \frac{3}{10}+3^2\cdot \frac{4}{10}=\frac{[2+12+36]}{10}=5.0$$
$$E[x]=0+\displaystyle \frac{2}{10}+2\cdot \frac{3}{10}+3\cdot \frac{4}{10}=\frac{[2+6+12]}{10}=2$$
$$\therefore$$ Var (x) $$= E[x^2] - E[x]^2 = 5-2^2=1$$
For two data sets, each of size $$ 5$$, the variances are given to be $$4$$ and $$5$$ and the corresponding means are given to be $$2$$ and $$4,$$ respectively. The variance of the combined data set is
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$$\displaystyle \frac{11}{2}$$
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$$6$$
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$$\displaystyle \frac{13}{2}$$
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$$\displaystyle \frac{5}{2}$$
Explanation
$$\sigma_{\mathrm{x}}^{2}=4$$
$$\sigma_{\mathrm{y}}^{2}=5$$
$$\overline{\mathrm{x}}=2$$
$$\overline{\mathrm{y}}=4$$
$$\displaystyle \frac{\Sigma \mathrm{x}_{\mathrm{i}}}{5}=2, \Sigma \mathrm{x}_{\mathrm{i}}=10;\Sigma \mathrm{y}_{\mathrm{i}}=20$$
$$
\sigma_{\mathrm{x}}^{2}=(\frac{1}{5}\Sigma \mathrm{x}_{\mathrm{i}}^{2})-(\overline{x})^{2}=\frac{1}{5}(\Sigma \mathrm{x}_{1}^{2})-4
$$
$$\sigma_{\mathrm{y}}^{2}=(\frac{1}{5}\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\overline{y})^{2}=\frac{1}{5}(\Sigma \mathrm{y}_{1}^{2})-4$$
$$
\Sigma \mathrm{x}_{\mathrm{i}}^{2}=40
$$
$$
\Sigma \mathrm{y}_{\mathrm{i}}^{2}=105
$$
$$\sigma_{\mathrm{z}}^{2}=\displaystyle \frac{1}{10}(\Sigma \mathrm{x}_{\mathrm{i}}^{2}+\Sigma \mathrm{y}_{\mathrm{i}}^{2})-(\frac{\overline{\mathrm{x}}+\overline{\mathrm{y}}}{2})=\frac{1}{10}(40+105)-9=\frac{145-90}{10}=\frac{55}{10}=\frac{11}{2}$$
Which of the following are limitations of range?
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It is not based on all the values.
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As long as the minimum and maximum values remain unaltered, any change in other values does not affect range.
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It can not be calculated for open-ended frequency distribution.
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All of these
Which one of the following is not a measure of dispersion?
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Quartile
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Range
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Mean Deviation
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Standard Deviation
The formula of students t-distribution is_____________.
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$$ t = \dfrac {s}{\sqrt{n}}$$
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$$ t = \dfrac {| \bar{X} - \mu |}{s}$$
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$$ t = \dfrac {{\dfrac{| \bar{X} - \mu |}{s}}}{\sqrt{n}}$$
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$$ t = \dfrac {{\dfrac {\sqrt{n}}{| \bar{X} - \mu |}}}{s}$$
Explanation
The correct formula for calculating students t-distribution is given in option C, where s is the standard deviation of the sample and n is the sample size. The t distribution is used to estimate population parameters when the sample size is small.
While constructing Lorenz Curve we take into account _________.
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cumulative percentages of the variable on Y axis and cumulative percentages of frequencies on X-axis
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cumulative percentages of the variable on X axis and cumulative percentages of frequencies on Y-axis
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cumulative variable on X axis and cumulative frequencies on Y-axis
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None of these
Coefficient of Quartile Deviation =
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[ Q(3) + Q(1) ] / [ Q(3) - Q(1) ]
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Q(3) - Q(1)
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[ Q(3) - Q(1) ] / [ Q(3) + Q(1) ]
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[ Q(3) - Q(1) ] / 2
Lorenz curve is based on which of the following?
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Cumulative frequencies as a percentage
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Cumulative totals of class mid points as a percentage
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Both A and B
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None of these
Inter-Quartile Range is based upon ________.
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First 50% of the values
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middle 50% of the values
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Last 50 % of the values
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25% of the values
In a frequency distribution, Range is the difference between _________ and __________
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upper limit of the highest class, the lower limit of the lowest class
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lower limit of the highest class, the upper limit of the lowest class
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A or B
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None of the above
A graphical
measure called ________ is
available for estimating dispersion.
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Lorenz Curve
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Bar diagram
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Histogram
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None of these
Which of the following is another term for quartile division?
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Inter quartile range
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Semi Inter Quartile Range
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Inter Quartile Deviation
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None of these
The median for the following distribution is
X
2
3
4
5
6
7
8
9
10
11
Y
3
6
9
18
20
14
10
10
7
2
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$$6$$
0%
$$5$$
0%
$$8$$
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$$9$$
The standard deviation of $$5$$ items is found to be $$15$$. What will be the standard deviation if the values of all the items are increased by $$5$$?
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$$15$$
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$$20$$
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$$10$$
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None of the above
The $$P_82$$ for the following distribution is
X
2
3
4
5
6
7
8
9
10
11
Y
3
6
9
18
20
14
10
10
7
2
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$$6$$
0%
$$5$$
0%
$$8$$
0%
$$9$$
In a group, $$2$$ students spend Rs$$8$$ daily, $$3$$ students spend Rs$$10$$ daily and $$5$$ students spend Rs$$6$$ daily. The average spending of all $$10$$ students is _______.
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$$7.6$$
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$$5.8$$
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$$8.5$$
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$$6.7$$
Given mean = $$70.2$$ and mode = $$70.5$$, find median using empirical relationship among them.
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$$70.3$$
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$$70.5$$
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$$70.6$$
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$$70.4$$
The mean of $$100$$ observations is $$18.4$$ and sum of sqares of deviations from mean is $$1444$$, the Co-efficient of variation is ______.
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$$30.6$$
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$$35.6$$
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$$20.6$$
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$$10.6$$
For a grouped date, the formula for median is based on _______.
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interpolation method
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extrapolation method
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trial and error
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iterative method
The mean of $$25$$ observations is $$73.408$$. If one observation $$64$$ is removed, the revised mean is ______.
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$$72.8$$
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$$73.8$$
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$$80.8$$
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$$76.8$$
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