Statistics Test

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Statistics Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

For a collection of $$11$$ items, sum of all items is $$132$$, then the arithmetic mean is.

A
$$11$$

B
$$12$$

C
$$14$$

D
$$13$$

Solution
Question 2
1 / -0

If the mean of the numbers $$a,b,8,5,10$$ is $$6$$ and their variance is $$68$$, then $$ab$$ is equal to

A
$$6$$

B
$$7$$

C
$$12$$

D
$$14$$

E
$$25$$

Solution

Given,
$$\cfrac { a+b+8+5+10 }{ 5 } =6$$
$$\Rightarrow a+b+c+23=30$$
$$\Rightarrow a+b=7....(i)$$
and $$\quad \cfrac { \sum { { \left( { x }_{ 1 }-\overline { x } \right) }^{ 2 } } }{ n } ={ \sigma }^{ 2 }\quad $$
$$\cfrac { { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+4+1+16 }{ 5 } =\cfrac { 34 }{ 5 } \quad $$
$$\Rightarrow { (a-6) }^{ 2 }+{ (b-6) }^{ 2 }+21=34\quad $$
$$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }-84+93=34\quad \left[ \because a+b=7 \right] $$
$$\quad \Rightarrow { a }^{ 2 }+{ b }^{ 2 }=25...(i)\quad $$
From Eq.(i)
$$\Rightarrow { (a+b) }^{ 2 }={ 7 }^{ 2 }$$
$$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+2ab=49$$
$$\Rightarrow 25+2ab=49$$ (From Eq.(ii))
$$\Rightarrow 2ab=24\Rightarrow ab=12\quad $$
Question 3
1 / -0

If the median of the data 6,7,x-2,x,18,21 written in ascending order is 16, then the variance of that data is

A
$$30\dfrac{1}{5}$$

B
$$31\dfrac{1}{3}$$

C
$$32\dfrac{1}{2}$$

D
$$33\dfrac{1}{3}$$

Solution

$$M=\dfrac{x-2+x}{2}=16$$

Variance = $$\dfrac{1}{n}\sum (x_i-\overline{x})^{2}$$
Required variance=$$31\dfrac{1}{3}$$
Question 4
1 / -0

The runs scored by Virat in 7 different matches are given as:
$$89,91,54,66,13,97,06$$
Calculate the mean deviation about the median for this data.

A
$$31.26$$

B
$$30.06$$

C
$$29.142$$

D
$$27.85$$

Solution

Arranging data in ascending order,
$$06,13,54,66,89,91,97$$
Here, number of observations $$(n)=7(odd)$$
$$\therefore$$ $$Median=\left(\dfrac{7+1}{2}\right)^{th}observation=4^{th}observation.$$
$$\therefore$$ $$Median(M) =66$$

$$x$$ $$|x-M|$$
$$06$$ $$60$$
$$13$$ $$53$$
$$54$$ $$12$$
$$66$$ $$0$$
$$89$$ $$23$$
$$91$$ $$25$$
$$97$$ $$31$$
$$\sum |x-M|=204$$
$$\Rightarrow$$ Mean deviation about median $$=\dfrac{\sum|x-M|}{7}=\dfrac{204}{7}=29.142$$
Question 5
1 / -0

The mean deviation from the mean for the set of observations $$-1,0,4$$ is

A
Less than $$3$$

B
Less than $$1$$

C
Greater than $$2.5$$

D
Greater than $$4.9$$

Solution

We have $$\overset { \_ }{ x } =\dfrac { -1+0+4 }{ 3 } =1$$
$${ x }_{ i }$$ $$\left| { x }_{ i }-\overset { \_ }{ x } \right| $$
$$-1$$ $$|-1-1|=2$$
$$0$$ $$|0-1|=1$$
$$4$$ $$|4-1|=3$$
Mean deviation from mean $$=\dfrac { \sum { \left| { x }_{ i }-\overset { \_ }{ x } \right| } }{ n } =\dfrac { 2+1+3 }{ 3 } =2$$
So, option A is correct.
Question 6
1 / -0

If the variance of $$1, 2, 3, 4, 5, ..., x$$ is $$10$$, then the value of $$x$$ is

A
$$9$$

B
$$13$$

C
$$12$$

D
$$10$$

E
$$11$$

Solution

We know that, the variance of first $$x$$ natural number is
$$\text{Var} (x) = \dfrac {x^{2} - 1}{12} = 10$$ ....(given)
$$\Rightarrow x^{2} = 120 + 1 = 121$$
$$\Rightarrow x = 11$$
Question 7
1 / -0

If $$\displaystyle \sum_{i = 1}^9 (x_i - 5) = 9$$ and $$\displaystyle \sum_{i = 1}^9 (x_i - 5)^2 = 45$$, then the standard deviation of the 9 times $$x_1, x_2, ....., x_9$$ is

A
9

B
4

C
3

D
2

E
1

Solution

Given, $$\displaystyle \sum_{i = 1}^n = (x_i - 5) = 9$$
and $$\displaystyle \sum_{i = 1}^9 (x_i - 5)^2 = 45$$
$$\therefore$$ Standard deviation
$$= \sqrt{\dfrac{\left [\displaystyle \sum_{i = 1}^9 (x_i - 5) \right ]^2 - \displaystyle \sum_{i = 1}^9 (x_i - 5)^2}{9}}$$
$$= \sqrt{\dfrac{(9)^2 - 45}{9}} = \sqrt{\dfrac{81 - 45}{9}}$$
$$= \sqrt{\dfrac{36}{9}} = \sqrt{4} = 2$$
Question 8
1 / -0

The arithmetic mean of the observations 10,8,5,a,b is 6 and their variance 6.8. Then ab=

A
6

B
4

C
3

D
12

Solution

$$\dfrac{10+8+5+a+b}{5}=6,$$

$$\dfrac{1}{n}\sum \left ( x_i-\overline{x} \right )^{2}=6.8$$
solving above two equations we get values of a and b.
Question 9
1 / -0

The variance of first '$$n$$' natural number is

A
$$\dfrac { { n }^{ 2 }+1 }{ 12 } $$

B
$$\dfrac { { n }^{ 2 }-1 }{ 12 } $$

C
$$\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

D
None of these

Solution

By using formula of variance, we have
$${ \sigma }^{ 2 }=\dfrac { 1 }{ n } \displaystyle\sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } -{ \left( \dfrac { 1 }{ n } \displaystyle\sum _{ i=1 }^{ n }{ { x }_{ i } } \right) }^{ 2 }$$
$$=\dfrac { 1 }{ n } \left( { 1 }^{ 2 }+{ 2 }^{ 2 }+\cdots +{ n }^{ 2 } \right) -{ \left( \dfrac { 1 }{ n } \left( 1+2+\cdots +n \right) \right) }^{ 2 }$$
$$=\dfrac { 1 }{ n } \cdot \dfrac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 } -\dfrac { 1 }{ { n }^{ 2 } } \cdot { \left[ \dfrac { n\left( n+1 \right) }{ 2 } \right] }^{ 2 }$$
$$=\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } -\dfrac { { \left( n+1 \right) }^{ 2 } }{ 4 }$$
$$ =\dfrac { { n }^{ 2 }-1 }{ 12 } $$
Question 10
1 / -0

Mean of $$10$$ observations is $$50$$ and their standard deviation is $$10$$. If each observation is subtracted by $$5$$ and then divided by $$4$$, then the new mean and standard deviation are

A
$$22.45, 2.5$$

B
$$11.25, 2.5$$

C
$$11.5, 2.5$$

D
$$11, 2.5$$

E
$$11.75, 2.5$$

Solution

Given,mean of $$10$$ observations $$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i}}{10} = 50$$
$$\Rightarrow \displaystyle \sum_{i = 1}^{10} x_{i} = 500$$ .... (i)
Therefore, new mean $$= \dfrac {\displaystyle \sum_{i = 1}^{10} x_{i} - 5\times 10}{4\times 10} = \dfrac {500 - 50}{4\times 10}$$ [using Eq. (i)]
$$= \dfrac {450}{4\times 10} = 11.25$$
and standard deviation $$= \dfrac {\text{old}(SD)}{4} = \dfrac {10}{4} = 2.5$$.

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Re-Attempt Weekly Quiz Competition

$$x$$	$$\|x-M\|$$
$$06$$	$$60$$
$$13$$	$$53$$
$$54$$	$$12$$
$$66$$	$$0$$
$$89$$	$$23$$
$$91$$	$$25$$
$$97$$	$$31$$
	$$\sum \|x-M\|=204$$

$${ x }_{ i }$$	$$\left\| { x }_{ i }-\overset { \_ }{ x } \right\| $$
$$-1$$	$$\|-1-1\|=2$$
$$0$$	$$\|0-1\|=1$$
$$4$$	$$\|4-1\|=3$$

Statistics Test - 36

Result

Statistics Test - 36

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Rank

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SHARING IS CARING

Question 1

$$11$$

$$12$$

$$14$$

$$13$$

Solution

Question 2

$$6$$

$$7$$

$$12$$

$$14$$

$$25$$

Solution

Question 3

$$30\dfrac{1}{5}$$

$$31\dfrac{1}{3}$$

$$32\dfrac{1}{2}$$

$$33\dfrac{1}{3}$$

Solution

Question 4

$$31.26$$

$$30.06$$

$$29.142$$

$$27.85$$

Solution

Question 5

Less than $$3$$

Less than $$1$$

Greater than $$2.5$$

Greater than $$4.9$$

Solution

Question 6

$$9$$

$$13$$

$$12$$

$$10$$

$$11$$

Solution

Question 7

9

4

3

2

1

Solution

Question 8

6

4

3

12

Solution

Question 9

$$\dfrac { { n }^{ 2 }+1 }{ 12 } $$

$$\dfrac { { n }^{ 2 }-1 }{ 12 } $$

$$\dfrac { \left( n+1 \right) \left( 2n+1 \right) }{ 6 } $$

None of these

Solution

Question 10

$$22.45, 2.5$$

$$11.25, 2.5$$

$$11.5, 2.5$$

$$11, 2.5$$

$$11.75, 2.5$$

Solution

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