Statistics Test

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Statistics Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The coefficient of mean deviation from median of observations $$40, 62, 54, 90, 68, 76$$ is

A
$$2.16$$

B
$$0.2$$

C
$$5$$

D
None of these

Solution

Arrange the given observations in ascending order
$$40,54,62,68,76,90$$
Here, number of terms $$n=6 (even) $$
$$\displaystyle \therefore $$ Median (M) $$\displaystyle =\frac{\left ( \frac{n}{2} \right )th\:term+\left ( \frac{n}{2}+1 \right )th\:term}{2}=\frac{62+68}{2}=65$$

$$\Sigma \left | x_{i}-M \right |=25+11+3+3+11+25=78$$
Mean deviation from median $$\displaystyle =\frac{\Sigma \left | x_{i}-M \right |}{n}=\frac{78}{6}=13 $$
$$\therefore $$ Coefficient of M.D.=$$\displaystyle =\frac{M.D.}{median}=\frac{13}{65}=0.2$$
Question 2
1 / -0

The arithmetic mean of first ten natural numbers is

A
$$5.5$$

B
$$6$$

C
$$7.5$$

D
$$10$$

Solution

Formula used:
$$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$

So, by using the above formula we get,
$$A.M=\dfrac{1+2+3+...+10}{10}$$
$$=\dfrac{55}{10}$$
$$=5.5$$

Hence, option $$A$$ is correct.
Question 3
1 / -0

The sum of the squares of deviation of 10 observations from their mean 50 is 250, then coefficient of variation is

A
10%

B
40%

C
50%

D
none of these

Solution

Given, $$\sum (x-\bar{x})^2 = 250, n = 10, \bar{x} =50$$
Thus standard deviation $$ = \sqrt{\cfrac{\sum (x-\bar{x})^2}{n}}=\sqrt{25}=5$$
$$\therefore$$ Coefficient of variation $$=\cfrac{\sigma}{\bar{x}}\times 100 =\cfrac{5}{50}\times 100$$ % $$= 10$$%
Question 4
1 / -0

The S.D. of the following freq. dist.
Class 0-10 10-20 20-30 30-40
$$f_i$$ 1 3 4 2

A
$$7.8$$

B
$$9$$

C
$$8.1$$

D
$$0.9$$

Solution

Class $$x_i$$ $$f_i$$ $$u_i=\dfrac{x_i-A}{h}$$ $$u_i^2$$ $$f_iu_i$$ $$f_iu_i^2$$
0-10 5 1 -2 4 -2 4
10-20 15 3 -1 1 -3 3
20-30 25 4 0 0 0 0
30-40 35 2 1 1 2 2
Total 10 -3 9
$$S.D.$$$$=h\sqrt { \cfrac { \sum { { f }_{ i }{ u }_{ i }^{ 2 } } }{ \sum { { f }_{ i } } } -{ \left( \cfrac { \sum { f } _{ i }{ u }_{ i } }{ \sum { { f }_{ i } } } \right) }^{ 2 } } $$
$$S.D.=10\sqrt { \cfrac { 9 }{ 10 } -{ \left( \cfrac { -3 }{ 10 } \right) }^{ 2 } } =9$$
Question 5
1 / -0

The mean of a dist. is $$4$$. if its coefficient of variation is $$58\%$$. Then the S.D. of the dist. is

A
$$2.23$$

B
$$3.23$$

C
$$2.32$$

D
None of these

Solution

We know if a distribution having mean $$\bar{x}$$ and standard deviation $$\sigma$$
then coefficient of variation $$=\cfrac{\sigma}{\bar{x}}\times 100$$
$$\therefore \cfrac{\sigma}{4}\times 100=58\Rightarrow \sigma = \cfrac{58}{25}=2.32$$
Hence required standard deviation is $$=2.32$$
Question 6
1 / -0

The mean deviation from mean of observations 5, 10, 15, 20, ......85 is

A
$$43.71$$

B
$$21.17$$

C
$$38.7$$

D
None of these

Solution

Given series is $$5,10,15,....,85$$, which is an AP.
Last term $$85=a+(n-1)d$$
$$\Rightarrow n=17$$
Sum of $$n$$ terms of AP $$=\dfrac{17}{2}(5+85)$$
$$\sum x_i=765$$

$$\bar x=\dfrac{\sum x_i}{n}$$

$$\bar x=\dfrac{765}{17}=45$$

So, deviations from mean are $$40,35,25,20,15,10,5,0,5,10,15,20,25,30,35,40$$
Sum of deviations $$=360$$

Mean deviation from mean $$=\dfrac{360}{17}=21.17$$
Question 7
1 / -0

The mean deviation of $$\displaystyle \frac{a+b}{2}\: \: and\: \: \frac{a-b}{2} $$ (where a and b >0) is?

A
$$\displaystyle \frac{b}{2}$$

B
$$\displaystyle \frac{a}{2}$$

C
$$a$$

D
$$b$$

Solution

The given observations are: $$\dfrac{a + b}{2}$$ and $$\dfrac{a - b}{2}$$
Mean of the observations = $$\dfrac{\dfrac{a + b}{2} + \dfrac{a - b}{2}}{2} = \dfrac{a}{2}$$

Now, mean deviation = $$\dfrac{|\dfrac{a + b}{2} - \dfrac{a}{2}| + |\dfrac{a - b}{2} - \dfrac{a}{2}|}{2}$$ = $$\dfrac{b}{2}$$
Question 8
1 / -0

Standard deviation of the distribution $$1, 3, 5,....., 13$$ will be

A
$$2$$

B
$$4$$

C
$$7$$

D
None

Solution

Mean of the given numbers is $$\bar x =\dfrac{1+3+5+7+9+11+13}{7}=\dfrac{49}{7}=7$$

total number of values $$n=7$$

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

$$\implies SD=\sqrt{\dfrac{(1-7)^2+(3-7)^2+(5-7)^2+(7-7)^2+(9-7)^2+(11-7)^2+(13-7)^2}{7-1}}$$

$$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{6}}$$

$$\implies SD=\sqrt{\dfrac{112}{6}}$$

$$\implies SD=\sqrt{18.6667}=4.32$$

Therefore the standard deviation for the given values is $$4.32$$
Question 9
1 / -0

In the following table pass percentage of three schools from the year $$2001$$ to the year $$2006$$ are given. Which school students performance is more consistent?

$$2001$$ $$2002$$ $$2003$$ $$2004$$ $$2005$$ $$2006$$
School (X) $$80$$ $$89$$ $$79$$ $$83$$ $$84$$ $$65$$
School (Y) $$92$$ $$94$$ $$76$$ $$75$$ $$80$$ $$63$$
School (Z) $$93$$ $$97$$ $$67$$ $$63$$ $$70$$ $$85$$

A
X

B
Y

C
Z

D
X and Y

Solution

Average of $$x=\cfrac { 80+89+79+83+84+65 }{ 6 } =\cfrac { 480 }{ 6 } =80$$
Average of $$y=\cfrac { 92+94+76+75+80+63 }{ 6 } =\cfrac { 480 }{ 6 } =80$$
Average of $$z=\cfrac { 93+97+67+63+70+85 }{ 6 } =79.16$$
Average of x and y are equal and more than z.
So,their performance is consistent.
Question 10
1 / -0

The mean deviation of first 8 composite numbers is

A
$$3.0$$

B
$$4.83$$

C
$$5.315$$

D
$$3.5625$$

Solution

Mean deviation is the mean of the absolute differences between each value in the data set and the mean of the data set.

Mean is the sum of the values in the data set divided by the total number of values.

A composite number is an integer which is not a prime.
The first $$8$$ composite numbers are $$4,6,8,9,10,12,14,15$$

Mean of the composite numbers is $$\dfrac{4+6+8+9+10+12+14+15}{8}$$

$$\implies mean =\dfrac{78}{8}=9.75$$

The absolute difference between the data values and mean is $$|9.75-4|,|9.75-6|,|9.75-8|,|9.75-9|,|9.75-10|,|9.75-12|,|9.75-14|,|9.75-15|$$

Therefore, the data values after finding the absolute differences are $$5.75,3.75,1.75,0.75,0.25,2.25,4.25,5.25$$

Therefore, the mean deviation is $$\dfrac{5.75+3.75+1.75+0.75+0.25+2.25+4.25+5.25}{8}=\dfrac{24}{8}=3$$

Therefore, the mean deviation of first eight composite numbers is $$3$$

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Re-Attempt Weekly Quiz Competition

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

	$$2001$$	$$2002$$	$$2003$$	$$2004$$	$$2005$$	$$2006$$
School (X)	$$80$$	$$89$$	$$79$$	$$83$$	$$84$$	$$65$$
School (Y)	$$92$$	$$94$$	$$76$$	$$75$$	$$80$$	$$63$$
School (Z)	$$93$$	$$97$$	$$67$$	$$63$$	$$70$$	$$85$$

Statistics Test - 28

Result

Statistics Test - 28

Score

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Rank

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SHARING IS CARING

Question 1

$$2.16$$

$$0.2$$

$$5$$

None of these

Solution

Question 2

$$5.5$$

$$6$$

$$7.5$$

$$10$$

Solution

Question 3

10%

40%

50%

none of these

Solution

Question 4

$$7.8$$

$$9$$

$$8.1$$

$$0.9$$

Solution

Question 5

$$2.23$$

$$3.23$$

$$2.32$$

None of these

Solution

Question 6

$$43.71$$

$$21.17$$

$$38.7$$

None of these

Solution

Question 7

$$\displaystyle \frac{b}{2}$$

$$\displaystyle \frac{a}{2}$$

$$a$$

$$b$$

Solution

Question 8

$$2$$

$$4$$

$$7$$

None

Solution

Question 9

X

Y

Z

X and Y

Solution

Question 10

$$3.0$$

$$4.83$$

$$5.315$$

$$3.5625$$

Solution

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