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CBSE Questions for Class 11 Commerce Economics Measures Of Dispersion Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Dispersion
Quiz 12
In a class of 100, the mean on a certain exam was 50, the standard deviation,This means..........................
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half the class had scores less than 50
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there was a high correlation between ability and grade
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everyone had a score of exactly 50
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half the class had 0's and half had 50's
Explanation
Standard deviation is the square root of the arithmetic mean of the squares of the deviations measured from the arithmetic mean of the data. If standard deviation of a certain series is zero than it denotes that all the values of that series is equal to the mean of the series which made all the deviations zero and therefore standard deviation also zero.
The following data show the number of hours worked by 200 statistics students:
Number of Hours
Frequency
0-9
40
10-19
50
20-29
70
30-39
40
The class width for this distribution is.........
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9
0%
10
0%
11
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Varies from class to class
A student obtained the mean and the standard deviation of 100 observations as 40 and 5.It was later found that one observation was wrongly copied as 50, the correct figure beingFind the correct mean and the S. D.
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Mean = 38.8, S. D. = 5
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Mean = 39.9, S. D. = 5
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Mean = 39.9, S. D. = 4
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None
The standard deviation of the data:
$$x:$$ $$1$$ $$a$$ $${a}^{2}$$........$${a}^{n}$$
$$f:$$ $${ _{ }^{ n }{ C } }_{ 0 }$$ $${ _{ }^{ n }{ C } }_{ 1 }$$.....$${ _{ }^{ n }{ C } }_{ n }$$ is
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$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a }^{ } }{ 2 } \right) }^{ n }$$
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$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ a }^{ } }{ 2 } \right) }^{ n }$$
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$${ \left( \cfrac { 1+{ a }^{ 2 } }{2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }$$
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None of these
A researcher has collected the following sample data. The mean of the sample is 5.
3 5 12 3 2
The standard deviation is...............
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8.944
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4.062
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13.2
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16.5
Variance of $$^{10}C_{0}, ^{10}C_{1}, ^{10}C_{2},^{10}C_{10}$$ is:
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$$\dfrac{10.^{20}C_{10}-2^{10}}{100}$$
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$$\dfrac{11.^{20}C_{10}-2^{10}}{11}$$
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$$\dfrac{10.^{20}C_{10}-2^{20}}{100}$$
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$$\dfrac{10.^{20}C_{10}-2^{10}}{121}$$
When a line connects points that are the cumulative percent of observation below the upper limit of each interval in a cumulative frequency distribution is known as ______________.
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0%
Mode
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Histogram
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Frequency polygon
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Ogive
The scores of a batsman in ten innings are: $$38, 70, 48, 34, 42, 55, 63, 46, 54, 44$$. Find the mean deviation about median.
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$$\dfrac{43}{5}$$
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$$\dfrac{44}{5}$$
0%
$$\dfrac{41}{5}$$
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$$\dfrac{42}{5}$$
Explanation
$$\begin{array}{l} 38,70,48,42,55,63,46,54,44 \\ Arranging\, :34,38,42,44,48,55,63,70 \\ Median=\frac { { \frac { N }{ 2 } ,\frac { n }{ 2 } +1 } }{ 2 } =\frac { { 5,6 } }{ 2 } = \\ =\frac { { 46+48 } }{ 2 } =47 \\ Mean\, \, deviation \\ =\left| { 34-47 } \right| +\left| { 38-47 } \right| +\left| { 42-47 } \right| +\left| { 44-47 } \right| \\ +\left| { 46-47 } \right| +\left| { 48-47 } \right| +\left| { 54-47 } \right| +\left| { 55-47 } \right| \\ \left| { 63-47 } \right| +\left| { 70-47 } \right| \\ \overline { 10 } \\ =13+9+5+3+1+1+7+8+16+23 \\ =\frac { { 86 } }{ { 10 } } =86 \end{array}$$
Whenever we group data into classes it is recommended that we have _________.
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Less than 5 classes
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Between 5 and 20 classes
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At least 2 classes
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Between 2 and 3 classes
Which of the following show direct variation?
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Height of a child $$(x)$$ and his weight $$(y)$$ as he grows
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Distance travelled $$(x)$$ and time taken$$(y)$$, if a car not moving at a uniform
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Wages of a worker $$(x)$$ and the number of hours of work $$(y)$$
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The number of students $$(x)$$ and the fees paid by them $$(y)$$
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Number of rainy days $$(x)$$ and the amount of rainfall on those days $$(y)$$
If mean deviation through median is 15 and median is 450, then coefficient of mean deviation is
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$$\dfrac{1}{30}$$
0%
30
0%
15
0%
45
Explanation
$$\textbf{Step 1: Apply the Mean Deviation formula.}$$
$$\text{Given: Mean deviation through median}=15$$
$$\text{Median}=450$$
$$\text{Coefficient of Mean Deviation }$$
$$= \dfrac{Mean\ deviation\ through\ median}{Median}$$
$$=\dfrac{15}{450}$$
$$=\dfrac{1}{30}$$
$$\textbf{Hence, the coefficient of Mean Deviation is }\dfrac{1}{30}.$$
If $$\sum _{ i=1 }^{ 18 }{ } $$ $$(X_{i}-8)=9$$ and $$\sum _{ i=1 }^{ 18 }{ } $$$$(x_{i}-8)^{2}=45$$ , then the standard deviation of $$x_{1},x_{2},.....,x_{18}$$ is
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4/9
0%
9/4
0%
3/2
0%
2/3
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