Measures of Dispersion Test

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Measures of Dispersion Test - 16

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Question 1
1 / -0

Fisher Price Index Number = ?

A
$$220.00$$

B
$$216.30$$

C
$$219.12$$

D
$$221.98$$

Solution

Fisher Price Index Number
$$P^{p}_{01} = \sqrt{P^{l}_{01}\times P^{p}_{01}} = \sqrt{(221.98\times216.3)\times{100}} = 219.12$$.
Question 2
1 / -0

Find the mean and standard deviation of the following observations: X $$=2, 5, 7, 8, 13$$.

A
$$7$$ & $$3.63$$

B
$$3.63$$ & $$7$$

C
$$7.63$$ & $$3$$

D
$$3$$ & $$7.63$$

Solution

$$\displaystyle\frac{2+5+7+8+13}{5}=7$$
$$\sqrt{\displaystyle\frac{4+25+49+64+169}{5}}-49=3.63$$.
Question 3
1 / -0

Let $${ x }_{ 1 },{ x }_{ 2 },.....{ x }_{ n }$$ be $$n$$ observations with mean $$m$$ and standard deviation $$s$$. Then the standard deviation of the observations $$a{ x }_{ 1 },a{ x }_{ 2 },.....a{ x }_{ n }$$, is

A
$$a+s$$

B
$$s/a$$

C
$$\left| a \right| s$$

D
$$as$$

Solution

$$S.D(ax) = \sqrt{E[ax-a\bar{x}]^2}=\sqrt{E[a(x-\bar{x})^2]}=\sqrt{a^2.E[(x-\bar{x})^2]}=|a|.\sqrt{E[(x-\bar{x})^2]}=|a|.S.D(x)=|a| s$$
Question 4
1 / -0

If a variable $$x$$ takes values $$0,1,2,....n$$ with frequencies proportional to the binomial coefficients $${ _{ }^{ n }{ C } }_{ 0 },{ _{ }^{ n }{ C } }_{ 1 },{ _{ }^{ n }{ C } }_{ 2 },......{ _{ }^{ n }{ C } }_{ n }$$, then mean of distribution is

A
$$\dfrac {n(n+1)}{2}$$

B
$$\dfrac { n }{ 2 } $$

C
$$\dfrac { 2 }{ n } $$

D
$$\dfrac {n(n-1)}{2}$$

Solution

Here, $$\displaystyle \mu1 '=\frac{\sum r\frac{n}{r}^{n-1}C_{r-1}}{\sum ^{n}C_{r}}$$
$$\displaystyle =\frac{n.2^{n-1}}{2^{n}}=\frac{n}{2}$$
and
$$\displaystyle \mu 2'=\frac{1}{2^{n}}\sum_{0}^{n}\left\{ \left ( r-1 \right )+r \right
\}^{n}C_{r}$$
$$\displaystyle=\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )^{n}C_{r}+\frac{n}{2}$$
$$\displaystyle =\frac{1}{2^{n}}\sum_{0}^{n}r\left ( r-1 \right )\frac{n\left ( n-1 \right )}{r\left ( r-1 \right )}^{n-2}C_{r-2}+\frac{n}{2}$$
$$\displaystyle =\frac{n\left ( n-1 \right)}{2^{n}}.2^{n-2}+\frac{n}{2}$$
$$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}$$
$$\therefore$$ Variance $$\displaystyle \sigma^2=\mu 2'-\left (\mu 1' \right )^{2}$$ $$\displaystyle =\frac{n\left ( n-1 \right )}{4}+\frac{n}{2}-\left ( \frac{n}{2} \right )^{2}=\frac{n}{4}$$
Question 5
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 6
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 7
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 8
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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 9
1 / -0

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 10
1 / -0

The Mean deviation about A.M. of the numbers $$3, 4, 5, 6, 7 $$ is :

A
$$25$$

B
$$5$$

C
$$1.2$$

D
$$0$$

Solution

Mean deviation is defined as the average of the absolute deviation of the observations from their mean.
Here, the arithmetic mean of the given observations is $$\dfrac{3 + 4 + 5 + 6 + 7}{5} = 5$$
Thus, mean deviation = $$\dfrac{|3 - 5| + |4 - 5| + |5 - 5| + |6 - 5| + |7 - 5|}{5} = \dfrac{6}{5} = 1.2$$
Hence, option 'C' is correct.

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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

Measures of Dispersion Test - 16

Result

Measures of Dispersion Test - 16

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SHARING IS CARING

Question 1

$$220.00$$

$$216.30$$

$$219.12$$

$$221.98$$

Solution

Question 2

$$7$$ & $$3.63$$

$$3.63$$ & $$7$$

$$7.63$$ & $$3$$

$$3$$ & $$7.63$$

Solution

Question 3

$$a+s$$

$$s/a$$

$$\left| a \right| s$$

$$as$$

Solution

Question 4

$$\dfrac {n(n+1)}{2}$$

$$\dfrac { n }{ 2 } $$

$$\dfrac { 2 }{ n } $$

$$\dfrac {n(n-1)}{2}$$

Solution

Question 5

$$10$$

$$20$$

$$10.1$$

$$20.2$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$24$$

$$12$$

$$20$$

$$25$$

Solution

Question 8

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

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

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

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

Solution

Question 9

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

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

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

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

Solution

Question 10

$$25$$

$$5$$

$$1.2$$

$$0$$

Solution

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