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CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Descriptive Statistics
Quiz 12
If the variance of the data 2,4,5,6,8,17 is 23,33, then variance of 5,11,14,17,23,50 is ________________.
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46,66
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96.32
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209,97
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213,97
Ina a class of 20 students, each student can score 10 or 0 marks in a certain examination. The maximum possible variance in the marks of the students in the class is
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24
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22
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20
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25
In a series of $$100$$ observations, half of them is equal to $$p$$ and remaining half equal to $$-p$$ if the standard deviation of these observations is $$10$$ then $$|p|=$$__
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$$10$$
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$$\dfrac {1}{10}$$
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$$\dfrac {1}{\sqrt {10}}$$
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$$\sqrt {10}$$
$$\sum _{ i=1 }^{ 10 }{ { \left( { X }_{ i }-10 \right) }^{ 2 }=2000 } \sum _{ i=1 }^{ 10 }{ { X }_{ i }-10 } =100$$ then the standard deviation of data set $${X}_{i}$$.
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$$1$$
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$$5$$
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$$10$$
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$$2$$
If the standard deviation of $$x_1,x_2,...,x_n$$ is 3.5 then the standard deviation of $$2x_1-3,2x_2-3,...,2x_n-3$$ is
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-7
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10.5
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7
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-10
Find the standard deviation of 10 observation 111, 211, 311,.......,1011.
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100$$\sqrt { 3 } $$
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250
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300
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50$$\sqrt { 33 } $$
The variance of the first $$n$$ natural number is
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$$\dfrac{1}{{12}}\left( {{n^2} - 1} \right)$$
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$$\dfrac{1}{{6}}\left( {{n^2} - 1} \right)$$
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$$\dfrac{1}{{6}}\left( {{n^2} +1} \right)$$
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$$\dfrac{1}{{12}}\left( {{n^2} +1} \right)$$
Explanation
Variance $$(\sigma ^{2})=\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$$
$$=\dfrac{1}{n}\sum_{i=1}^{n}\left[x_{i}-(\dfrac{n+1}{2})\right]^{2}$$
$$=\dfrac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\dfrac{1}{n}\sum_{i=1}^{n}2(\dfrac{n+1}{2})x_{i}+\dfrac{1}{n}\sum_{i=1}^{n}(\dfrac{n+1}{2})^{2}$$
$$=\dfrac{1}{n}\dfrac{n(n+1)(2n+1)}{6}-(\dfrac{n+1}{n})[\dfrac{n(n+1)}{2}]+\frac{(n+1)^{2}}{4n}\times n$$
$$=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^{2}}{2}+\dfrac{(n+1)^{2}}{4}$$
$$=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^{2}}{4}$$
$$=(n+1)\left[\dfrac{4n+2-3n-3}{12}\right]$$
$$=\dfrac{(n+1)(n-1)}{12}$$
$$=\dfrac{n^{2}-1}{12}$$
The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is
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$$\sqrt { \dfrac { 52 }{ 7 } } $$
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$$\sqrt { 6 } $$
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$$\dfrac { 52 }{ 7 } $$
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6
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