MCQ Questions for 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?

0%
$4$
0%
$2$
0%
$\sqrt{2}$
0%
$2\sqrt{2}$

If the mean deviation about the median of the numbers

$a,2a,....,50a$ is

$50$ , then

$|a|$ equals:-

0%
$4$
0%
$5$
0%
$2$
0%
$3$

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

0%
$\sqrt{n}$
0%
$\sqrt{\dfrac{(n+2)(n+1)}{3}}$
0%
$\sqrt{\dfrac{n^2-1}{3}}$
0%
$n$

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

0%
$\dfrac{\sqrt{52}}{7}$
0%
$\dfrac{52}{7}$
0%
$\dfrac{\sqrt{53}}{7}$
0%
$\dfrac{53}{7}$
0%
$6$

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

$(\sigma)$

$\displaystyle =\sqrt{\dfrac{1}{n}\sum^7_{i=1} (x_1-\bar{x})^2}=\sqrt{\dfrac{52}{7}}$

$N = 10 \sum x = 120$ and

$\sigma _{x} = 60$ , the variation coefficient is -

0%
$5$
0%
$50$
0%
$500$
0%
$0.5$

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

0%
$0.065$
0%
$12.003$
0%
$0.083$
0%
$0.073$

Explanation

$\dfrac{Mean deviation}{Mean} = \dfrac{64.4}{773} = 0.083$
Thus (C) is correct

Which one of the following is a false description?

0%
In a moderately asymmetrical distribution, the empirical relationship between Mean, Mode and Median suggested by Karl Pearson is Mean Mode = 3 (Mean Median)
0%
Coeffcient of variation is an absolute measure of dispersion
0%
Measure of skewness in the distribution of numerical values in the data set
0%
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.

0%
$42$
0%
$25$
0%
$28$
0%
$48$

Given

$\sum (x-\bar{x})^2=48, \bar{x}=20$ and

$n=12$ . The coefficient of variation is.

0%
$25$
0%
$20$
0%
$30$
0%
$10$

The sum of the squares of deviations of 10 items about mean 50 is 250 .The coefficient of variation is

0%
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

0%
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

0%
$\displaystyle \frac{11}{2}$
0%
$6$
0%
$\displaystyle \frac{13}{2}$
0%
$\displaystyle \frac{5}{2}$

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?

0%
It is not based on all the values.
0%
As long as the minimum and maximum values remain unaltered, any change in other values does not affect range.
0%
It can not be calculated for open-ended frequency distribution.
0%
All of these

Which one of the following is not a measure of dispersion?

0%
Quartile
0%
Range
0%
Mean Deviation
0%
Standard Deviation

The formula of students t-distribution is_____________.

0%
$t = \dfrac {s}{\sqrt{n}}$
0%
$t = \dfrac {| \bar{X} - \mu |}{s}$
0%
$t = \dfrac {{\dfrac{| \bar{X} - \mu |}{s}}}{\sqrt{n}}$
0%
$t = \dfrac {{\dfrac {\sqrt{n}}{| \bar{X} - \mu |}}}{s}$

While constructing Lorenz Curve we take into account _________.

0%
cumulative percentages of the variable on Y axis and cumulative percentages of frequencies on X-axis
0%
cumulative percentages of the variable on X axis and cumulative percentages of frequencies on Y-axis
0%
cumulative variable on X axis and cumulative frequencies on Y-axis
0%
None of these

Coefficient of Quartile Deviation =

0%
[ Q(3) + Q(1) ] / [ Q(3) - Q(1) ]
0%
Q(3) - Q(1)
0%
[ Q(3) - Q(1) ] / [ Q(3) + Q(1) ]
0%
[ Q(3) - Q(1) ] / 2

Lorenz curve is based on which of the following?

0%
Cumulative frequencies as a percentage
0%
Cumulative totals of class mid points as a percentage
0%
Both A and B
0%
None of these

Inter-Quartile Range is based upon ________.

0%
First 50% of the values
0%
middle 50% of the values
0%
Last 50 % of the values
0%
25% of the values

In a frequency distribution, Range is the difference between _________ and __________

0%
upper limit of the highest class, the lower limit of the lowest class
0%
lower limit of the highest class, the upper limit of the lowest class
0%
A or B
0%
None of the above

A graphical measure called ________ is available for estimating dispersion.

0%
Lorenz Curve
0%
Bar diagram
0%
Histogram
0%
None of these

Which of the following is another term for quartile division?

0%
Inter quartile range
0%
Semi Inter Quartile Range
0%
Inter Quartile Deviation
0%
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

0%
$6$
0%
$5$
0%
$8$
0%
$9$

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

$5$ ?

0%
$15$
0%
$20$
0%
$10$
0%
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

0%
$6$
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 _______.

0%
$7.6$
0%
$5.8$
0%
$8.5$
0%
$6.7$

Given mean =

$70.2$ and mode =

$70.5$ , find median using empirical relationship among them.

0%
$70.3$
0%
$70.5$
0%
$70.6$
0%
$70.4$

The mean of

$100$ observations is

$18.4$ and sum of sqares of deviations from mean is

$1444$ , the Co-efficient of variation is ______.

0%
$30.6$
0%
$35.6$
0%
$20.6$
0%
$10.6$

For a grouped date, the formula for median is based on _______.

0%
interpolation method
0%
extrapolation method
0%
trial and error
0%
iterative method

The mean of

$25$ observations is

$73.408$ . If one observation

$64$ is removed, the revised mean is ______.

0%
$72.8$
0%
$73.8$
0%
$80.8$
0%
$76.8$

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 11 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 11 - MCQExams.com

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Practice Class 11 Commerce Economics Quiz Questions and Answers

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