Statistics Test

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Statistics Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The coefficient of range of the following distribution $$10, 14, 11, 9, 8, 12, 6$$

A
$$0.4$$

B
$$2.5$$

C
$$8$$

D
$$0.9$$

Solution

Here greatest term in the given observation is, $$x_m =14$$
and least term is, $$x_l = 6$$
Hence coefficient of range is $$=\cfrac{x_m-x_l}{x_m+x_l}=\cfrac{8}{20}=0.4$$
Question 2
1 / -0

If the coefficient of variation and standard deviation of a distribution are 50% and 20 respectively, the its mean is

A
40

B
30

C
20

D
None of these

Solution

We know if a distribution having mean $$\bar{x}$$ and standard deviation $$\sigma$$
then coefficient of variation $$=\cfrac{\sigma}{\bar{x}}\times 100$$
$$\therefore \cfrac{20}{\bar{x}}\times 100=50\Rightarrow \bar{x} = 40$$
Hence required mean is $$=40$$
Question 3
1 / -0

Find the median of the following data:
$$5, 7, 9, 11, 15, 17, 2, 23$$ and $$19$$

A
$$7$$

B
$$9$$

C
$$11$$

D
$$15$$

Solution

Median is middle score in the ordered data.
Here given data is $$5,7,9,11,15,17,2,23,19$$.
Arranging them in order, we have
$$2, 5, 7, 9, 11, 15,17,19,23$$
Middle score is $$11$$.
Hence, median is $$11$$.
Question 4
1 / -0

Find the mode of:
$$7, 7, 8, 10, 10, 11, 10, 13, 14$$

A
$$7$$

B
$$8$$

C
$$10$$

D
$$13$$

Solution

Mode is defined as the maximum number of times an observation appears in the whole set.
Here, by observation, we get that $$10$$ is appearing the maximum (i.e. $$3$$) times and hence this is the mode.
Question 5
1 / -0

Mean deviations of the series $$a,a+d,a+2d,...,a+2nd$$ from its mean is

A
$$\displaystyle \frac{n\left ( n+1 \right )d}{\left ( 2n+1 \right )}$$

B
$$\displaystyle \frac{nd}{2n+1}$$

C
$$\displaystyle \frac{\left ( n+1 \right )d}{2n+1}$$

D
$$\displaystyle \frac{\left ( 2n+1 \right )d}{n\left ( n+1 \right )}$$

Solution

Clearly given observation is in A.P, and have $$2n+1$$ terms.
$$\therefore$$ Mean $$\displaystyle =\frac{1}{2}\left \{ a+{a+2nd}\right \}=a+nd$$
Use for $$x = a$$, $$|x-\bar{x}| =|a+nd-a|=nd$$
$$x = a+d$$, $$|x-\bar{x}| =|a+nd-a-d|=(n-1)d$$
Similarly, write till $$x=a+2nd$$, to get sum of $$\sum \left | x-\bar{x} \right |$$
And thus mean deviation about mean $$\displaystyle =\frac{1}{2n+1}\sum \left | x-\bar{x} \right |$$ $$\displaystyle =\frac{2}{2n+1}.d\left ( 1+2+3\cdots +n \right )$$ $$\displaystyle =\frac{n\left ( n+1 \right )d}{(2n+1)}$$
Question 6
1 / -0

The A.M. of the first ten odd numbers is

A
$$10$$

B
$$100$$

C
$$1000$$

D
$$1$$

Solution

First ten odd numbers are $$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$$ respectively.
$$A.M. (\overline {x}) = \dfrac{\text {sum of all the terms}}{\text {total number of terms}}$$

$$A.M. \left(\overline { x } \right) = \displaystyle\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$$

$$= \displaystyle\frac{100}{10}$$

$$=10$$

$$\therefore$$ the $$A.M$$ of first ten odd numbers is $$10$$
Question 7
1 / -0

The mean of a distribution is 4. If its coefficient of variation is 58%. Then the S.D. of the distribution is

A
2.23

B
3.23

C
2.32

D
none of these

Solution

Given, mean $$\bar{x} = 4,$$ and coefficient of variation $$=58$$ %
If S.D of the given distribution is $$\sigma$$ then we know that,
Coefficient of variation $$=\cfrac{\sigma}{\bar{x}}\times 100$$ %
$$\Rightarrow 58 = \cfrac{\sigma}{4}\times 100\Rightarrow \sigma = \cfrac{58\times 4}{100}=2.32$$
Question 8
1 / -0

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

A
10%

B
40%

C
50%

D
None of these

Solution

Given $$\displaystyle \Sigma \left ( x_{i}-\overline{x} \right )^{2}=250$$,$$n=10,\overline{x}=50$$

Now, $$\sigma=\sqrt{\dfrac{1}{n}\Sigma \left ( x_{i}-\overline{x} \right )^{2}}$$

$$= \sqrt{\dfrac{1}{10}\times 250}=5$$
Hence coefficient of variation $$\displaystyle =\dfrac{\sigma }{\overline{x}}\times 100=\dfrac{5}{50}\times 100=10$$%
Question 9
1 / -0

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

A
2.16

B
1.2

C
5

D
none of these

Solution

Arranging the given data in ascending order
40,54,62,68,76,90
Here, $$n=6 (even)$$
$$M= \dfrac{\text{value of }3^{rd}\text{observation}+\text{value of }4^{th}\text{observation}}{2}$$
Median $$M=\dfrac{62+68}{2}=65$$

Mean deviation about median $$M.D=\dfrac{|40-65|+|54-65|+|62-65|+|68-65|+|76-65|+|90-65|}{65}$$

$$=\dfrac{25+11+3+3+11+25}{65}=1.2$$
Question 10
1 / -0

Mean deviation of the observations 70, 42, 63,34, 44, 54, 55, 46, 38, 48 from median is

A
$$7.8$$

B
$$8.6$$

C
$$7.6$$

D
$$8.8$$

Solution

Arranging the given data in ascending order-
$$34, 38, 42, 44, 46, 48, 54, 55, 63, 70$$
$$\because n = 10$$
Therefore,
Median $$\left( M \right) = \cfrac{{\left( \cfrac{n}{2} \right)}^{th} \text{ term } + {\left( \cfrac{n}{2} + 1 \right)}^{th} \text{ term}}{2}$$
$$\Rightarrow M = \cfrac{46 + 48}{2} = 47$$
Therefore,
$${x}_{i}$$ $$\left| {x}_{i} - M \right|$$
$$34$$ 13
$$38$$ 9
$$42$$ 5
$$44$$ 3
$$46$$ 1
$$48$$ 1
$$54$$ 7
$$55$$ 8
$$63$$ 16
$$70$$ 23

$$\sum{\left| {x}_{i} - M \right|} = 86$$
Therefore,
Mean deviation $$= \cfrac{\sum{\left| {x}_{i} - M \right|}}{n} \\ = \cfrac{86}{10} = 8.6$$
Hence the mean deviation of the observations is $$8.6$$.

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$${x}_{i}$$	$$\left\| {x}_{i} - M \right\|$$
$$34$$	13
$$38$$	9
$$42$$	5
$$44$$	3
$$46$$	1
$$48$$	1
$$54$$	7
$$55$$	8
$$63$$	16
$$70$$	23
	$$\sum{\left\| {x}_{i} - M \right\|} = 86$$

Statistics Test - 27

Result

Statistics Test - 27

Score

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Rank

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SHARING IS CARING

Question 1

$$0.4$$

$$2.5$$

$$8$$

$$0.9$$

Solution

Question 2

40

30

20

None of these

Solution

Question 3

$$7$$

$$9$$

$$11$$

$$15$$

Solution

Question 4

$$7$$

$$8$$

$$10$$

$$13$$

Solution

Question 5

$$\displaystyle \frac{n\left ( n+1 \right )d}{\left ( 2n+1 \right )}$$

$$\displaystyle \frac{nd}{2n+1}$$

$$\displaystyle \frac{\left ( n+1 \right )d}{2n+1}$$

$$\displaystyle \frac{\left ( 2n+1 \right )d}{n\left ( n+1 \right )}$$

Solution

Question 6

$$10$$

$$100$$

$$1000$$

$$1$$

Solution

Question 7

2.23

3.23

2.32

none of these

Solution

Question 8

10%

40%

50%

None of these

Solution

Question 9

2.16

1.2

5

none of these

Solution

Question 10

$$7.8$$

$$8.6$$

$$7.6$$

$$8.8$$

Solution

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