The variance of the first $$n$$ natural number is

$$\dfrac{1}{{12}}\left( {{n^2} - 1} \right)$$

$$\dfrac{1}{{6}}\left( {{n^2} - 1} \right)$$

$$\dfrac{1}{{6}}\left( {{n^2} +1} \right)$$

$$\dfrac{1}{{12}}\left( {{n^2} +1} \right)$$

MCQ Questions for 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 ________________.

0%
46,66
0%
96.32
0%
209,97
0%
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

0%
24
0%
22
0%
20
0%
25

In a series of

100

$100$ observations, half of them is equal to

p

$p$ and remaining half equal to

- p

$-p$ if the standard deviation of these observations is

10

$10$ then

| p | =

$|p|=$ __

0%
$10$
0%
$\dfrac {1}{10}$
0%
$\dfrac {1}{\sqrt {10}}$
0%
$\sqrt {10}$

\sum_{i = 1}^{10} {(X_{i} - 10)}^{2} = 2000 \sum_{i = 1}^{10} X_{i} - 10 = 100

$\sum _{ i=1 }^{ 10 }{ { \left( { X }_{ i }-10 \right) }^{ 2 }=2000 } \sum _{ i=1 }^{ 10 }{ { X }_{ i }-10 } =100$ then the standard deviation of data set

X_{i}

${X}_{i}$ .

0%
$1$
0%
$5$
0%
$10$
0%
$2$

If the standard deviation of

x_{1}, x_{2}, . . ., x_{n}

$x_1,x_2,...,x_n$ is 3.5 then the standard deviation of

2 x_{1} - 3, 2 x_{2} - 3, . . ., 2 x_{n} - 3

$2x_1-3,2x_2-3,...,2x_n-3$ is

0%
-7
0%
10.5
0%
7
0%
-10

Find the standard deviation of 10 observation 111, 211, 311,.......,1011.

0%
100 $\sqrt { 3 }$
0%
250
0%
300
0%
50 $\sqrt { 33 }$

The variance of the first

n

$n$ natural number is

0%
$\dfrac{1}{{12}}\left( {{n^2} - 1} \right)$
0%
$\dfrac{1}{{6}}\left( {{n^2} - 1} \right)$
0%
$\dfrac{1}{{6}}\left( {{n^2} +1} \right)$
0%
$\dfrac{1}{{12}}\left( {{n^2} +1} \right)$

Explanation

Variance

(σ^{2}) = \frac{1}{n} \sum_{i = 1}^{n} (x_{i} - \bar{x})^{2}

$(\sigma ^{2})=\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}$

= \frac{1}{n} \sum_{i = 1}^{n} {[x_{i} - (\frac{n + 1}{2})]}^{2}

$=\dfrac{1}{n}\sum_{i=1}^{n}\left[x_{i}-(\dfrac{n+1}{2})\right]^{2}$

= \frac{1}{n} \sum_{i = 1}^{n} x_{i}^{2} - \frac{1}{n} \sum_{i = 1}^{n} 2 (\frac{n + 1}{2}) x_{i} + \frac{1}{n} \sum_{i = 1}^{n} (\frac{n + 1}{2})^{2}

$=\dfrac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\dfrac{1}{n}\sum_{i=1}^{n}2(\dfrac{n+1}{2})x_{i}+\dfrac{1}{n}\sum_{i=1}^{n}(\dfrac{n+1}{2})^{2}$

= \frac{1}{n} \frac{n (n + 1) (2 n + 1)}{6} - (\frac{n + 1}{n}) [\frac{n (n + 1)}{2}] + \frac{(n + 1)^{2}}{4 n} \times n

$=\dfrac{1}{n}\dfrac{n(n+1)(2n+1)}{6}-(\dfrac{n+1}{n})[\dfrac{n(n+1)}{2}]+\frac{(n+1)^{2}}{4n}\times n$

= \frac{(n + 1) (2 n + 1)}{6} - \frac{(n + 1)^{2}}{2} + \frac{(n + 1)^{2}}{4}

$=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^{2}}{2}+\dfrac{(n+1)^{2}}{4}$

= \frac{(n + 1) (2 n + 1)}{6} - \frac{(n + 1)^{2}}{4}

$=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^{2}}{4}$

= (n + 1) [\frac{4 n + 2 - 3 n - 3}{12}]

$=(n+1)\left[\dfrac{4n+2-3n-3}{12}\right]$

= \frac{(n + 1) (n - 1)}{12}

$=\dfrac{(n+1)(n-1)}{12}$

= \frac{n^{2} - 1}{12}

$=\dfrac{n^{2}-1}{12}$

The standard deviation of the data 6, 5, 9, 13, 12, 8, 10 is

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

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 12 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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