Statistics Test

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Statistics Test - 25

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Question 1
1 / -0

If the mean deviation about mean $$1,1+d,1+2d,....1+100d$$ from their mean is $$255$$, then the $$d$$ is equal to

A
$$10$$

B
$$20$$

C
$$10.1$$

D
$$20.2$$

Solution

Clearly given observation is in A.P, and have $$101$$ terms.
$$\therefore$$ Mean $$\displaystyle =\frac{1}{2}\left \{ 1+1+100d\right \}=1+50d$$
And thus mean deviation about mean $$\displaystyle =\frac{1}{101}\sum \left | x-\bar{x} \right |$$ $$\displaystyle =\frac{1}{101}.2d\left ( 1+2+3\cdots +50 \right )$$ $$\displaystyle =\frac{50\left ( 51 \right )d}{101}=255$$ (given)
$$\Rightarrow d = 10.1$$
Question 2
1 / -0

The mean deviation of the series $$a,a+d,a+2d+......,a+(2n-1)d,a+2nd$$ about the mean is

A
$$\cfrac { n+1 }{ 2n+1 } $$

B
$$\cfrac { n(n+1)d }{ 2n+1 } $$

C
$$\cfrac { (n+2)d }{ 2n } $$

D
$$\cfrac { (n-1)d }{ 2n+1 } $$

Solution

Mean $$\displaystyle\overline{x}=\frac{(a)+(a+d)+(a+2d)+...+(a+2nd)}{2n+1}$$
where $$\displaystyle\overline{x}=\frac{\sum x_i}{n}$$
Then, $$\displaystyle\frac{\frac{2n+1}{2}(a+a+2nd)}{2n+1}$$

From an A.P. $$\displaystyle S_n = \frac{n}{2} (a+l)$$ which gives $$\overline{x} = a+nd$$
The series being a,a+d,a+2d,...,a+(n-1)d,a+nd,a+(n+1)d,...,a+2nd
Mean deviation from the mean
= $$\displaystyle\frac{1}{N} \sum f_i [x_i- \overline{x}]$$
= $$\displaystyle\frac{1}{2n+1} \sum [x_i -a-nd]$$
= $$\displaystyle\frac{1}{2n+1} [nd+(n-1)d+(n-2)d+...+d+0+d+...+nd]$$
= $$\displaystyle\frac{2d}{2n+1} [n+(n-1)+(n-2)+...1]$$
= $$\displaystyle\frac{2d}{2n+1} .\frac{n(n+1)}{2} = \frac{n(n+1)d}{2n+1}$$
Question 3
1 / -0

Consider the frequency distribution
Class interval:
0-6
6-12
12-18
Frequency:
2
4
6
The variance of the above frequency distribution, is

A
$$24$$

B
$$12$$

C
$$20$$

D
$$25$$

Solution

Class interval $$f_i$$ $$x_i$$ $$d_i={x_i-A}$$ $$d_i^2$$ $$f_id_i^2$$ $$f_id_i$$
0-6 2 3 -6 36 72 -12
6-12 4 9 0 0 0 0
12-18 6 15 6 36 216 36
Here,$$N=\sum {f_i }=12$$
$$\sum {f_id_i^2}=288$$
$$\sum {f_id_i}=24$$

Now, variance $$\displaystyle { \sigma }^{ 2 }=\frac { \sum { f_{ i }d_{ i }^{ 2 } } }{ N } -{ \left( \frac { \sum { f_{ i }d_{ i } } }{ N } \right) }^{ 2 }$$

$$\Rightarrow \displaystyle { \sigma }^{ 2 }=\frac { 288 }{ 12 } -{ \left( \frac { 24 }{ 12 } \right) }^{ 2 }$$

$$\Rightarrow { \sigma }^{ 2 }=20$$
Question 4
1 / -0

If a variable $$X$$ takes values $$0,1,2,....,n$$ with frequencies $${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }\quad $$ respectively, then S.D. is equal to :

A
$$\cfrac { n }{ 4 } $$

B
$$\cfrac { n }{ 2 } $$

C
$$\cfrac { \sqrt { n } }{ 2 } $$

D
$$\cfrac { \sqrt { n } }{ 4 } $$

Solution

Here, $$\displaystyle \mu1'$$ $$=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$$ $$\displaystyle $$
and
$$\displaystyle \mu 2' $$$$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$$

$$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1
\right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$$

$$\therefore$$ Variance $$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4} \mbox{or} S.D = \sigma = \sqrt{\sigma^2} = \frac{\sqrt{n}}{2}$$
Question 5
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Mean deviation for $$n$$ observations $${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$$ from their mean $$\bar { X } $$ is given by:

A
$$\sum _{ i=1 }^{ n }{ ( { x }_{ i }-\bar { X }) }$$

B
$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\bar { X } \right| } $$

C
$$\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$$

D
$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } } $$

Solution

It is fundamental concept that mean deviation of $$n$$ observations $$x_1, x_2, x_3, ......x_n$$ about mean $$\bar{X}$$ is given by, $$\displaystyle \frac{1}{n} \sum_{i=1}^{i=n} |x_i-\bar{X}|$$
Question 6
1 / -0

If the mean deviation about the median of the numbers $$a,2a,3a,.....50a$$ is $$50$$, then $$\left| a \right| $$ equals

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

Given data is
$$a,2a, 3a,....49a,50a$$
Here, $$n=50 (even)$$
So, median $$M= \dfrac{\text{value of }25^{th}\text{observation}+\text{value of }26^{th}\text{observation}}{2}$$
$$M=\dfrac{25a+26a}{2}=25.5a$$

Mean deviation about median $$M.D=\dfrac{\sum |x_i-25.5a|}{50}$$
$$\Rightarrow 50=\dfrac{|a-25.5a|+|2a-25.5a|+.......+|24a-25.5a|+|25a-25.5a|+|26a-25.5a|+.....|50a-25.5a|}{50}$$

$$\Rightarrow 2500=|a|+2(1.5+2.5+.....24.5)|a|$$

$$\Rightarrow 2500=|a|+2\times 12 (3+23)|a|$$
$$\Rightarrow |a|=\dfrac{2500}{625}=4$$
Question 7
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Directions For Questions

The standard deviation of variate $$x$$ is the square root of the A.M. of the squares of all deviations of $$x$$ from the A.M. of observations and we denote it by sigma $$ \displaystyle \left ( \sigma \right ) $$ if $$ \displaystyle x/f_{i}\left ( i = 1,2,3,...,n \right ) $$ is a frequency distribution, then
$$ \displaystyle \sigma =\sqrt{\frac{1}{N}\sum f_{i}\left ( x_{i}-\bar{x} \right )^{2}} $$           ........(i)
where $$ \displaystyle \bar{x} $$ is the A.M.of the distribution & $$ \displaystyle N=\sum_{i=1}^{n}f_{i} $$
The square of the standard deviation is called the variance & given by
$$ \displaystyle \sigma ^{2}=\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2} $$     ........(ii)
The calculation of coefficient of dispersion & coefficient of variation are count down by $$ \displaystyle \frac{\sigma }{\bar{x}} $$ & $$ \displaystyle \frac{\sigma }{\bar{x}}\times 100 $$ respectively.
If deviation of $$x$$ are measured from an assumed mean $$A$$ then root mean square deviation of $$x$$ is denoted by $$S$$ and given by
$$ \displaystyle S= \sqrt{\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i} -A\right )^{2}} $$
which set up the relationship between S.D. and Root mean square deviation given by $$ \displaystyle S^{2}=\sigma ^{2}+d^{2}\left ( Where, d=\sum_{i=1}^{n} x_i-A \right ) $$
Consider the frequency distribution
Size 4 6 8 10 12 14 16
Frequency 1 2 3 5 3 2 1
On the basis of above information answer the following questions.

...view full instructions

The coefficient of dispersion for the given distribution is

A
3.06

B
.306

C
30.6

D
None of these

Solution

First of all find the mean of the given frequency distribution after getting mean, then find deviation for each size.
$$\begin{matrix}
size\left

( x \right ) & frequency\left ( f \right ) & f(x) &

Deviation   x=X-\bar{x}\left ( say\:\bar{x}=10 \right ) & x^{2} &

fx^{2}\\
4 & 1 & 4 & -6 & 36 & 36\\
6 & 2 & 12 & -4 & 16 & 32\\
8 & 3 & 24 & -2 & 4 & 12\\
10 & 5 & 50 & 0 & 00 & 0\\
12 & 3 & 36 & +2 & 4 & 12\\
14 & 2 & 28 & +4 & 16 & 32\\
16 & 1 & 16 & +6 & 36 & 36\\
& \sum f=N=17 & \sum f\left ( x \right )=170 & & & \sum f(x^{2})=160
\end{matrix}$$

$$\therefore $$   $$\displaystyle \bar{x}=\frac{170}{17}=10$$
Hence coefficient of dispersion (variation) $$\displaystyle =\frac{\sigma }{\bar{X}}\times 100=0.306$$

Question 8

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Directions For Questions

The standard deviation of variate $$x$$ is the square root of the A.M. of the squares of all deviations of $$x$$ from the A.M. of observations and we denote it by sigma $$ \displaystyle \left ( \sigma \right ) $$ if $$ \displaystyle x/f_{i}\left ( i = 1,2,3,...,n \right ) $$ is a frequency distribution, then

$$ \displaystyle \sigma =\sqrt{\frac{1}{N}\sum f_{i}\left ( x_{i}-\bar{x} \right )^{2}} $$ ........(i)

where $$ \displaystyle \bar{x} $$ is the A.M.of the distribution & $$ \displaystyle N=\sum_{i=1}^{n}f_{i} $$

The square of the standard deviation is called the variance & given by

$$ \displaystyle \sigma ^{2}=\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i}-\bar{x} \right )^{2} $$ ........(ii)

The calculation of coefficient of dispersion & coefficient of variation are count down by $$ \displaystyle \frac{\sigma }{\bar{x}} $$ & $$ \displaystyle \frac{\sigma }{\bar{x}}\times 100 $$ respectively.

If deviation of $$x$$ are measured from an assumed mean $$A$$ then root mean square deviation of $$x$$ is denoted by $$S$$ and given by

$$ \displaystyle S= \sqrt{\frac{1}{N}\sum_{i=1}^{n}f_{i}\left ( x_{i} -A\right )^{2}} $$

which set up the relationship between S.D. and Root mean square deviation given by $$ \displaystyle S^{2}=\sigma ^{2}+d^{2}\left ( Where, d=\sum_{i=1}^{n} x_i-A \right ) $$

Consider the frequency distribution

Size	4	6	8	10	12	14	16
Frequency	1	2	3	5	3	2	1

On the basis of above information answer the following questions.

...view full instructions

The coefficient of variation for the given distribution is

$$28.6$$

$$0.306$$

$$30.6$$

$$306$$

Solution

Lett $$A=10$$

$$Size$$ $$x_i$$	$$Frequency$$ $$f_i$$	$$f_ix_i$$	$$(x_i-A)^2$$	$$f_i(x_i-A)^2$$
$$4$$	$$1$$	$$4$$	$$36$$	$$36$$
$$6$$	$$2$$	$$12$$	$$16$$	$$32$$
$$8$$	$$3$$	$$24$$	$$4$$	$$12$$
$$10$$	$$5$$	$$50$$	$$0$$	$$0$$
$$12$$	$$3$$	$$36$$	$$4$$	$$12$$
$$14$$	$$2$$	$$28$$	$$16$$	$$32$$
$$16$$	$$1$$	$$16$$	$$36$$	$$16$$
	$$\sum f_i=17$$	$$\sum f_ix_i=170$$		$$\sum f_i(x_i-A)^2=140$$

$$\Rightarrow$$ $$\overline{x}=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{170}{17}=10$$

Now, $$S.D (\sigma)=\sqrt{\dfrac{\sum f_i(x_i-A)^2}{\sum f_i}}$$

$$=\sqrt{\dfrac{140}{17}}$$

$$=2.86$$

$$\Rightarrow$$ Coefficient of variation $$=\dfrac{\sigma}{\overline{x}}\times 100$$

$$=\dfrac{2.86}{10}\times 100$$

$$=28.6$$

Question 9
1 / -0

If $$n> 1, x> -1, x\neq 0$$, then the statement $$\left ( 1+x \right )^{n}> 1+nx$$ is true for

A
$$ \;n\;\epsilon \;N$$

B
$$\forall \;n\;> 1$$

C
$$x> -1 \;and\; x\neq 0$$

D
None of these

Solution

$$P(1)$$ is not true

For $$n=2,P\left( 2 \right) :{ \left( 1+x \right) }^{ 2 }>1+2x$$ is true if $$x\neq 0$$

Let $$P(k):{ \left( 1+x \right) }^{ k }>1+kx$$ be two

$$\therefore{ \left( 1+x \right) }^{ k+1 }=\left( 1+x \right) { \left( 1+x \right) }^{ k}>\left( 1+x \right) \left( 1+kx \right)> 1+\left( k+1 \right) x+k{ x }^{ 2}>1+\left(k+1\right) x$$

$$\left( \because k{ x }^{ 2 }>0 \right) $$
$$\therefore$$ By PMI
Given statement is true for every $$n\in N$$.
Question 10
1 / -0

If $$25$$ is the arithmetic mean of $$ X$$ and $$46,$$ then find $$X.$$

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution

Since $$25$$ is the arithmetic mean of $$X$$ and $$46$$,

$$\therefore 25=\displaystyle \frac{(X+46)}{2}$$
$$\therefore X+46=50$$
$$\therefore X=4$$

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Re-Attempt Weekly Quiz Competition

Class interval	$$f_i$$	$$x_i$$	$$d_i={x_i-A}$$	$$d_i^2$$	$$f_id_i^2$$	$$f_id_i$$
0-6	2	3	-6	36	72	-12
6-12	4	9	0	0	0	0
12-18	6	15	6	36	216	36

Statistics Test - 25

Result

Statistics Test - 25

Score

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Rank

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SHARING IS CARING

Question 1

$$10$$

$$20$$

$$10.1$$

$$20.2$$

Solution

Question 2

$$\cfrac { n+1 }{ 2n+1 } $$

$$\cfrac { n(n+1)d }{ 2n+1 } $$

$$\cfrac { (n+2)d }{ 2n } $$

$$\cfrac { (n-1)d }{ 2n+1 } $$

Solution

Question 3

$$24$$

$$12$$

$$20$$

$$25$$

Solution

Question 4

$$\cfrac { n }{ 4 } $$

$$\cfrac { n }{ 2 } $$

$$\cfrac { \sqrt { n } }{ 2 } $$

$$\cfrac { \sqrt { n } }{ 4 } $$

Solution

Question 5

$$\sum _{ i=1 }^{ n }{ ( { x }_{ i }-\bar { X }) }$$

$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-\bar { X } \right| } $$

$$\sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } }$$

$$\cfrac { 1 }{ n } \sum _{ i=1 }^{ n }{ { { (x }_{ i }-\bar { X } ) }^{ 2 } } $$

Solution

Question 6

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 7

3.06

.306

30.6

None of these

Solution

Question 8

$$28.6$$

$$0.306$$

$$30.6$$

$$306$$

Solution

Question 9

$$ \;n\;\epsilon \;N$$

$$\forall \;n\;> 1$$

$$x> -1 \;and\; x\neq 0$$

None of these

Solution

Question 10

$$2$$

$$4$$

$$8$$

$$16$$

Solution

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