Descriptive Statistics Test 23

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Descriptive Statistics Test 23

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Question 1
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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 2
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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 3
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The quartile deviation for the data
x :
2
3
4
5
6
f :
3
4
8
4
1
is

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

B
$$1$$

C
$$\cfrac { 1 }{ 4 } $$

D
$$0$$

Solution

$$x$$ $$f$$ Cumulative Frequency
2 3 3
3 4 7
4 8 15
5 4 19
6 1 20
Total 20
Here, $$n=20$$
$$Q_1=(\dfrac{20+1}{4})^{th}observation$$
$$=\dfrac{21}{4}=5.25^{th}observation$$
So, $$Q_1=3$$

$$Q_3=\dfrac{3(n+1)}{4})^{th} observation$$
$$=\dfrac{63}{4}=15.75^{th} observation$$
So, $$Q_3=5$$

Quartile deviation $$=\dfrac{Q_3-Q_1}{2}$$
$$=\dfrac{5-3}{2}=1$$
Question 4
1 / -0

The variance of the data:
x:
$$1$$
$$a$$
$${ a }^{ 2 }$$
......
$${ a }^{ n }$$
f:
$${ _{ }^{ n }{ C } }_{ 0 }$$
$${ _{ }^{ n }{ C } }_{ 1 }$$
$${ _{ }^{ n }{ C } }_{ 2 }$$
......
$${ _{ }^{ n }{ C } }_{ n }$$
is

A
$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$$

B
$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$$

C
$${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$$

D
none of these

Solution

$$ \displaystyle \bar { X } =\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \\ \displaystyle =\frac { ^{ n }C_{ 0 }+a.^{ n }C_{ 1 }+{ a }^{ 2 }.^{ n }C_{ 2 }+....{ a }^{ n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } \\ \displaystyle =\frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } } $$
.
$$ \displaystyle S.D.=\sqrt { \frac { \sum { { f }_{ i }{ { x }_{ i } }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } \right) }^{ 2 } } \\ \displaystyle =\sqrt { \frac { ^{ n }C_{ 0 }+{ a }^{ 2 }.^{ n }C_{ 1 }+{ a }^{ 4 }.^{ n }C_{ 2 }+....{ a }^{ 2n }.^{ n }C_{ n } }{ ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }+....^{ n }C_{ n } } -{ \left[ \frac { { \left( 1+a \right) }^{ n } }{ { 2 }^{ n } } \right] }^{ 2 } } \\ \displaystyle \sqrt { \frac { { \left( 1+{ a }^{ 2 } \right) }^{ n } }{ { 2 }^{ n } } -{ \left( \frac { 1+a }{ 2 } \right) }^{ 2n } } $$

$$Variance = (S.D)^2$$
Question 5
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Directions For Questions

The data given below shows how many letters were collected from a post box on each of the days of a certain week.

Monday $$\Box\Box\Box\Box\Box\Box\Box$$
Tuesday $$\Box\Box\Box\Box\Box$$
Wednesday $$\Box\Box\Box\Box$$
Thursday $$\Box\Box\Box\Box\Box\Box$$
Friday $$\Box\Box\Box\Box\Box\Box\Box\Box$$
Saturday $$\Box\Box\Box$$
$$\Box = 10$$ letters

...view full instructions

How many less letters were collected on Wednesday than on Tuesday?

A
$$40$$

B
$$30$$

C
$$20$$

D
$$10$$

Solution

Given that one square box represents $$10$$ letters.
On Tuesday there are 5 boxes.
Therefore, number of letters collected on Tuesday are $$10\times5=50$$
On Wednesday there are 4 boxes.
Therefore, number of letters collected on Wednesday are $$10\times4=40$$

There are $$50-40=10$$ letters less on Wednesday than on Tuesday.
Question 6
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The S.D. of the following frequency distribution is
Class 0-10 10-20 20-30 30-40
$$\displaystyle f_{i}$$ 1 3 4 2

A
$$7.8$$

B
$$9$$

C
$$8.1$$

D
$$0.9$$

Solution

Class $$x_i$$ $$f_i$$ $$u_i=\dfrac{x_i-a}{h}$$ $$f_iu_i$$ $$f_iu_i^2$$
0-10 5 1 -2 -2 4
10-20 15 3 -1 -3 3
20-30 25 4 0 0 0
30-40 35 2 1 2 2
$$N=10$$ $$\sum f_iu_i=-3$$ $$\sum f_iu_i^2=9$$
$$\displaystyle \sigma =h\sqrt{\frac{ \Sigma f_{1}u_{1}^{2}}{N}-\left ( \frac{\Sigma f_{1}u_{1}}{N} \right )^{2}}$$

$$\displaystyle =10\sqrt{\frac{9}{10}-\left ( \frac{-3}{10} \right )^{2}}=9$$
Question 7
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The adjoining diagram illustrates the information collected about the number
of brothers of the boys in a certain group. What is the total number of boys in
this group?

A
$$81$$

B
$$22$$

C
$$73$$

D
$$34$$

Solution

From the figure,
there are $$2$$ boys who have $$1$$ brother, thus total $$2$$ boys
there are $$5$$ boys who have $$2$$ brothers, thus total $$10$$ boys
there are $$4$$ boys who have $$3$$ brothers, thus total $$12$$ boys
there are $$7$$ boys who have $$4$$ brothers, thus total $$28$$ boys
there are $$3$$ boys who have $$5$$ brothers, thus total $$15$$ boys
there are $$1$$ boys who have $$6$$ brothers, thus total $$6$$ boys
Hence, total number of boys $$= 2 + 10 + 12 + 28 + 15 + 6 = 73$$
Question 8
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The horizontal line in a bar graph is called:

A
$$x$$-axis

B
$$y$$-axis

C
Imaginary axis

D
None of these

Solution

In the Cartesian coordinate system, the horizontal reference line is called $$x$$-axis.
Therefore, the horizontal line in a bar graph is called $$x$$-axis.
Hence, option $$A$$ is correct.
Question 9
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The incomplete pictograph shows the number of stamps(1 ticket = 4 stamps) collected by four pupils:
The missing tickets represents the number of stamps collected by Anjali.
Then how many tickets collected by Anjali?

A
$$5$$

B
$$7$$

C
$$6$$

D
$$8$$

Solution

From the given pictograph, the total tickets coolected by Lata, Kalyan, Vaishnavi are $$13$$.
Given that one ticket symbol represents $$4$$ stamps.
Therefore, total number of stamps collected by Lata, Kalyan, Vaishnavi are $$13\times4=52$$.
Given that, total number of stamps collected by four pupils is $$80$$.

Therefore the number of stamps collected by Anjali = Total number of stamps collected by four pupil - number of stamps collected by three pupils (excluding Anjali) $$=80-52=28$$

Hence, the number of tickets collected by Anjali are $$ \dfrac{28}4=7$$
Question 10
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Answer the following question by reading the given pictograph.
Number of TV sets sold in the year 2005 is

A
$$1000$$

B
$$500$$

C
$$50$$

D
$$2000$$

Solution

From the given pictograph, $$2005$$ has one tv symbol.
Given that one symbol represents $$500$$ tv sets.
Therefore the $$2005$$ has $$1\times500=500$$ tv sets.

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x:	$$1$$	$$a$$	$${ a }^{ 2 }$$	......	$${ a }^{ n }$$
f:	$${ _{ }^{ n }{ C } }_{ 0 }$$	$${ _{ }^{ n }{ C } }_{ 1 }$$	$${ _{ }^{ n }{ C } }_{ 2 }$$	......	$${ _{ }^{ n }{ C } }_{ n }$$

Class	$$x_i$$	$$f_i$$	$$u_i=\dfrac{x_i-a}{h}$$	$$f_iu_i$$	$$f_iu_i^2$$
0-10	5	1	-2	-2	4
10-20	15	3	-1	-3	3
20-30	25	4	0	0	0
30-40	35	2	1	2	2
		$$N=10$$		$$\sum f_iu_i=-3$$	$$\sum f_iu_i^2=9$$

$$x$$	$$f$$	Cumulative Frequency
2	3	3
3	4	7
4	8	15
5	4	19
6	1	20
Total	20

Descriptive Statistics Test 23

Result

Descriptive Statistics Test 23

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Question 1

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

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

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

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

Solution

Question 2

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

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

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

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

Solution

Question 3

$$\cfrac { 1 }{ 2 } $$

$$1$$

$$\cfrac { 1 }{ 4 } $$

$$0$$

Solution

Question 4

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{ n }$$

$${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{n }-{ \left( \cfrac { 1+{ a } }{ 2 } \right) }^{2n }$$

$${ \left( \cfrac { 1+a }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ { a }^{ 2 } } }{ 2 } \right) }^{ n }$$

none of these

Solution

Question 5

$$40$$

$$30$$

$$20$$

$$10$$

Solution

Question 6

$$7.8$$

$$9$$

$$8.1$$

$$0.9$$

Solution

Question 7

$$81$$

$$22$$

$$73$$

$$34$$

Solution

Question 8

$$x$$-axis

$$y$$-axis

Imaginary axis

None of these

Solution

Question 9

$$5$$

$$7$$

$$6$$

$$8$$

Solution

Question 10

$$1000$$

$$500$$

$$50$$

$$2000$$

Solution

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