Statistics Test

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Statistics Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

Consider the data: $$2, x, 3, 4, 5, 2, 4, 6, 4$$ where $$x >2$$
The mode of the data is _____

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

The value which occurs the maximum number of times in a given set of data is know as mode.
In the given data $$4$$ occurs 3 times and hence it is the mode. But $$x$$ is unknown.
We know that $$x>2$$
Hence even if we consider $$x$$ as $$3,4,5\ or\ 6$$ still $$4$$ will remain as the mode.
Hence $$4$$ is the ans.
Question 2
1 / -0

For the given data, SD $$= 10$$, AM $$= 20$$ the coefficient of variation is ...........

A
$$47$$

B
$$24$$

C
$$44$$

D
$$50$$

Solution

Coefficient of variation is the ratio of standard deviation to the mean.

Given that $$SD=10$$ and $$AM=20$$

Therefore of coefficient of variation is $$\dfrac{SD}{AM}\times100=\dfrac{10}{20}\times100=50\%$$
Question 3
1 / -0

Find the standard deviation of $$210, 240, 250, 260, 220, 230$$ and $$270$$.

A
$$18$$

B
$$20$$

C
$$22$$

D
$$25$$

Solution

Mean of the given numbers is $$\bar x =\dfrac{210+240+250+260+220+230+270}{7}=\dfrac{1680}{7}=240$$

total number of values $$n=7$$

Standard deviation is given by $$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$$

$$\implies SD=\sqrt{\dfrac{(210-240)^2+(240-240)^2+.....+(270-240)^2}{7}}$$

$$\implies SD=\sqrt{\dfrac{900+0+100+400+400+100+900}{7}}$$

$$\implies SD=\sqrt{\dfrac{2800}{7}}$$

$$\implies SD=\sqrt{400}=20$$

Therefore the standard deviation for the given values is $$20$$
Question 4
1 / -0

The ages (in years) of a family of 6 members are 1, 5, 12, 15, 38 and 40. The standard deviation is found to be 15.9. After 10 years the standard deviation is

A
increased

B
decreased

C
remains same

D
none of these

Solution

Then mean of six members=$$\frac{1+5+12+15+38+40}{6}=\frac{111}{6}=18.5$$
Then Variance $$\alpha ^{2}=\frac{(18.5-1)^{2}+(18.5-5)^{2}+(18.5-12)^{2}+(18.5-15)^{2}+(18.5-38)^{2}+(18.5-40)^{2}}{5}$$
$$=\frac{(17.5)^{2}+(13.5)^{2}+(6.5)^{2}+(3.5)^{2}+(-19.5)^{2}+(21.5)^{2}}{6}$$
$$=\frac{306.25+182.25+42.25+12.25+380.25++462.25}{6}=\frac{1385.50}{6}=230.91$$
So standard deviation $$\alpha =15.19$$
Given after 10 year The standard deviation is 15.9
Then standard deviation remain same
Question 5
1 / -0

Find the mode of 0,0,2,2,3,3,3,4,5,5,5,5,6,6,7,8

A
8

B
0

C
2

D
5

Solution

$$Number$$ $$Frequency$$
0 2
2 2
3 3
4 1
5 4
6 2
7 1
8 1

Frequency of 5 is maximum. Hence 5 is the mode.
Question 6
1 / -0

Find the mean for the following data using step deviation method.

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 2

$$X_i$$ $$F_i$$ $$d=\cfrac{X-A}{i}$$ $$F_i d_i$$
2 10 -1 -10
4=A 20 0 0
6 30 1 30
8 40 2 80
$$\Sigma f = 100$$ $$\Sigma fd = 100$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times w=4+\cfrac{100}{100} \times 2 ={4+2}=6$$
C)6
Question 7
1 / -0

Branch
A
Branch
B Branch
C Branch
D
Q1 4.1 7.4 8.0 5.4
Q2 3.6 5.2 3.7 6.2
Q3 5.0 4.5 4.9 4.8
Q4 4.9 6.3 5.9 5.6
$$\overline { x } $$ 4.4 5.85 5.625 5.5
$$s$$ 0.67 1.27 1.82 0.85

A company wants to close one of its branch due to recession. The table given shows each branch's quarterly profit data with mean $$(\bar{x})$$ and standard deviation $$(s)$$. Company's finance department recommendations were taken into final conclusion. Finance department recommends to close either the store with the lowest average profits or the store that performs the least consistently. According to the data, which branches will the finance department recommend for closure to the board?

A
Branches A or C

B
Branches A or D

C
Branches B or C

D
Branches B or D

Solution

In the given problem finance department will recommend branch with least avg profit (i.e least mean profit) or branch with least consistency (i.e. least standard deviation)
therefore, Branch A has least mean value and Branch C has maximum std deviation .
Ans is A (Branches A or C)
Question 8
1 / -0

Choose the formula used for arithmetic mean of grouped data by step deviation method is

A
$$\bar { x } =A-\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

B
$$\bar { x } =A\times \dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

C
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

D
$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } +i$$

Solution

The formula used for arithmetic mean of groped data by step deviation method is $$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$
$$A =$$ Assumed mean of the given data
$$\sum { f } =$$ Summation of the frequencies given in the grouped data
$$\sum { fd\prime } =$$ Summation of the frequencies and deviation of a given mean data
$$d\prime =\dfrac { x-A }{ i } $$
$$i =$$ Class interval width
$$\bar { x } =$$ Arithmetic mean
Question 9
1 / -0

Find the mean for the following data using step deviation method.

A
$$9$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

Answer:- By shortcut Method
Class interval width (w) = 3

X F $$d=\cfrac{X-A}{i}$$ Fd
3 5 -1 -5
6=A 10 -0 -0
9 15 1 15
12 20 2 40
$$\Sigma f = 50$$ $$\Sigma fd = 50$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=6+\left(\cfrac{50}{50} \times 3\right) = 6+3 = 9$$
A) 9
Question 10
1 / -0

Find the mean for the following data using step deviation method.

A
$$43$$

B
$$44$$

C
$$45$$

D
$$46$$

Solution

Answer:- By shortcut Method
Class interval width (i)= 24-12 = 12

X F $$d=\cfrac{x-A}{i} $$ Fd
12=A 1 0 0
24 2 1 2
36 3 2 6
48 4 3 12
60 5 4 20
$$\Sigma f = 15$$ $$\Sigma fd = 40$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f} \times i=12+\left(\cfrac{40}{15} \times 12\right)={12+32}=44$$
B)44

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Re-Attempt Weekly Quiz Competition

$$X_i$$	$$F_i$$	$$d=\cfrac{X-A}{i}$$	$$F_i d_i$$
2	10	-1	-10
4=A	20	0	0
6	30	1	30
8	40	2	80
	$$\Sigma f = 100$$		$$\Sigma fd = 100$$

	Branch A	Branch B	Branch C	Branch D
Q1	4.1	7.4	8.0	5.4
Q2	3.6	5.2	3.7	6.2
Q3	5.0	4.5	4.9	4.8
Q4	4.9	6.3	5.9	5.6
$$\overline { x } $$	4.4	5.85	5.625	5.5
$$s$$	0.67	1.27	1.82	0.85

X	F	$$d=\cfrac{X-A}{i}$$	Fd
3	5	-1	-5
6=A	10	-0	-0
9	15	1	15
12	20	2	40
	$$\Sigma f = 50$$		$$\Sigma fd = 50$$

X	F	$$d=\cfrac{x-A}{i} $$	Fd
12=A	1	0	0
24	2	1	2
36	3	2	6
48	4	3	12
60	5	4	20
	$$\Sigma f = 15$$		$$\Sigma fd = 40$$

Statistics Test - 30

Result

Statistics Test - 30

Score

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Rank

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 2

$$47$$

$$24$$

$$44$$

$$50$$

Solution

Question 3

$$18$$

$$20$$

$$22$$

$$25$$

Solution

Question 4

increased

decreased

remains same

none of these

Solution

Question 5

8

0

2

5

Solution

Question 6

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 7

Branches A or C

Branches A or D

Branches B or C

Branches B or D

Solution

Question 8

$$\bar { x } =A-\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

$$\bar { x } =A\times \dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$$

$$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } +i$$

Solution

Question 9

$$9$$

$$10$$

$$11$$

$$12$$

Solution

Question 10

$$43$$

$$44$$

$$45$$

$$46$$

Solution

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