Statistics Test

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Statistics Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

The standard deviation $$\sigma$$ of the first $$N$$ natural numbers can be obtained using which one of the following formula?

A
$$\sigma =\cfrac { { N }^{ 2 }-1 }{ 12 } $$

B
$$\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 12 } } $$

C
$$\sigma =\sqrt { \cfrac { N-1 }{ 12 } } $$

D
$$\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 6N } } $$

Solution

$$\sigma^2=\dfrac{1}{n}\sum_{i=1}^{n}i^2-\left(\dfrac{1}{n}\sum_{i=1}^{n}i\right)^2$$

$$=\dfrac{1}{n}(1^2+2^2+...+n^2)-\left(\dfrac{1}{n}(1+2+...+n)\right)^2$$

$$=\dfrac{1}{n} \times \dfrac{n(n+1)(2n+1)}{6}-(\dfrac{n+1}{2})^2=\dfrac{n^2-1}{12}$$

$$\therefore \sigma=\sqrt{\dfrac{n^2-1}{12}}$$
Question 2
1 / -0

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

A
$$\cfrac{(n+1)d}{(2n+1)}$$

B
$$\cfrac{nd}{2n+1}$$

C
$$\cfrac{(2n+1)d}{n(n+1)}$$

D
$$\cfrac{n(n+1)d}{2n+1}$$
Question 3
1 / -0

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

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

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

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

D
None of these

Solution
Question 4
1 / -0

Standard deviation of the first $$2$$n + $$1$$ natural numbers is

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

B
$$\sqrt{\dfrac {n(n+1)(2n + 1)}{3}}$$

C
$$\sqrt{\dfrac {n(n+1)}{3}}$$

D
$$\sqrt{\dfrac {n^2(n+1)}{2}}$$
Question 5
1 / -0

Variance of $$^{10}C_{0}, ^{10}C_{1}, ^{10}C_{2},^{10}C_{10}$$ is:

A
$$\dfrac{10.^{20}C_{10}-2^{10}}{100}$$

B
$$\dfrac{11.^{20}C_{10}-2^{10}}{11}$$

C
$$\dfrac{10.^{20}C_{10}-2^{20}}{100}$$

D
$$\dfrac{10.^{20}C_{10}-2^{10}}{121}$$
Question 6
1 / -0

If the sum of mean and variance of $$B.D$$ for $$5$$ trials is $$1.8$$, the binomial distribution is

A
$$ (0.8+0.2)^5$$

B
$$ (0.2+1.8)^5$$

C
$${\left( {0.8 + 0.2} \right)^{10}}$$

D
$${\left( {0.2 + 1.8} \right)^{10}}$$

Solution

a binomial distribution is characterized by $$2$$ parameters:
$$n$$ - number of trials
$$p$$ - probability of success in one trail.
We there by know n=5, we need to find $$p$$ .
$$Mean$$ of the binomial distribution is given by,
$$mean = np$$
variance is given by

$$variance\quad =\quad np(1-p)\\ \therefore \quad mean\quad +\quad variance\quad =\quad 1.8\\ =>\quad 5p\quad +\quad 5p(1-p)\quad =\quad 1.8\\ =>\quad 5{ p }^{ 2 }-10p+1.8=0\\ =>\quad p=\quad 0.2\quad or\quad 1.8$$
$$ =>\quad p=\quad 0.2\ [\because 1\ge p\ge 0]$$
$$q=0.8$$
Binomial distribution is $$(0.2+0.8)^n$$.
Hence the answer is A
Question 7
1 / -0

What is standard deviation of the set of observations $$32,28,29,30,31$$?

A
$$1.6$$

B
$$\sqrt { 2 }$$

C
$$2$$

D
$$none\ of\ these$$

Solution

$$\sigma =\displaystyle \sum _{ i=1 }^{ n }{ \sqrt { \dfrac { \left( { x }_{ i }-{ x }_{ av } \right) }{ m-1 } } } $$
$${ x }_{ au }=\dfrac { 32+28+29+30+31 }{ 5 } =30$$
$$\therefore =\sqrt { \dfrac { { \left( 32-30 \right) }^{ 2 }+{ \left( 28-30 \right) }^{ 2 }+{ \left( 29-30 \right) }^{ 2 }+{ \left( 30-30 \right) }^{ 2 }+{ \left( 31-30 \right) }^{ 2 } }{ 5-1 } } $$
$$=\sqrt { \dfrac { 4+4+1+1 }{ 4 } } =\sqrt { \dfrac { 10 }{ 4 } } =\sqrt { \dfrac { 10 }{ 2 } } =1.58$$
Question 8
1 / -0

The scores of a batsman in ten innings are: $$38, 70, 48, 34, 42, 55, 63, 46, 54, 44$$. Find the mean deviation about median.

A
$$\dfrac{43}{5}$$

B
$$\dfrac{44}{5}$$

C
$$\dfrac{41}{5}$$

D
$$\dfrac{42}{5}$$

Solution

$$\begin{array}{l} 38,70,48,42,55,63,46,54,44 \\ Arranging\, :34,38,42,44,48,55,63,70 \\ Median=\frac { { \frac { N }{ 2 } ,\frac { n }{ 2 } +1 } }{ 2 } =\frac { { 5,6 } }{ 2 } = \\ =\frac { { 46+48 } }{ 2 } =47 \\ Mean\, \, deviation \\ =\left| { 34-47 } \right| +\left| { 38-47 } \right| +\left| { 42-47 } \right| +\left| { 44-47 } \right| \\ +\left| { 46-47 } \right| +\left| { 48-47 } \right| +\left| { 54-47 } \right| +\left| { 55-47 } \right| \\ \left| { 63-47 } \right| +\left| { 70-47 } \right| \\ \overline { 10 } \\ =13+9+5+3+1+1+7+8+16+23 \\ =\frac { { 86 } }{ { 10 } } =86 \end{array}$$
Question 9
1 / -0

Which of the following show direct variation?

A
Height of a child $$(x)$$ and his weight $$(y)$$ as he grows

B
Distance travelled $$(x)$$ and time taken$$(y)$$, if a car not moving at a uniform

C
Wages of a worker $$(x)$$ and the number of hours of work $$(y)$$

D
The number of students $$(x)$$ and the fees paid by them $$(y)$$

E
Number of rainy days $$(x)$$ and the amount of rainfall on those days $$(y)$$

Solution
Question 10
1 / -0

Find the range in the following frequency distribution of economics marks.
No of students Marks
5 10-20
10 20-30
25 30-40
20 40-50
10 50-60
5 60-70
10 70-80
10 80-90
5 90-100

A
90 marks

B
70 marks

C
50 marks

D
60 marks

Solution

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Re-Attempt Weekly Quiz Competition

No of students	Marks
5	10-20
10	20-30
25	30-40
20	40-50
10	50-60
5	60-70
10	70-80
10	80-90
5	90-100

Statistics Test - 48

Result

Statistics Test - 48

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SHARING IS CARING

Question 1

$$\sigma =\cfrac { { N }^{ 2 }-1 }{ 12 } $$

$$\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 12 } } $$

$$\sigma =\sqrt { \cfrac { N-1 }{ 12 } } $$

$$\sigma =\sqrt { \cfrac { { N }^{ 2 }-1 }{ 6N } } $$

Solution

Question 2

$$\cfrac{(n+1)d}{(2n+1)}$$

$$\cfrac{nd}{2n+1}$$

$$\cfrac{(2n+1)d}{n(n+1)}$$

$$\cfrac{n(n+1)d}{2n+1}$$

Question 3

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

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

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

None of these

Solution

Question 4

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

$$\sqrt{\dfrac {n(n+1)(2n + 1)}{3}}$$

$$\sqrt{\dfrac {n(n+1)}{3}}$$

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

Question 5

$$\dfrac{10.^{20}C_{10}-2^{10}}{100}$$

$$\dfrac{11.^{20}C_{10}-2^{10}}{11}$$

$$\dfrac{10.^{20}C_{10}-2^{20}}{100}$$

$$\dfrac{10.^{20}C_{10}-2^{10}}{121}$$

Question 6

$$ (0.8+0.2)^5$$

$$ (0.2+1.8)^5$$

$${\left( {0.8 + 0.2} \right)^{10}}$$

$${\left( {0.2 + 1.8} \right)^{10}}$$

Solution

Question 7

$$1.6$$

$$\sqrt { 2 }$$

$$2$$

$$none\ of\ these$$

Solution

Question 8

$$\dfrac{43}{5}$$

$$\dfrac{44}{5}$$

$$\dfrac{41}{5}$$

$$\dfrac{42}{5}$$

Solution

Question 9

Height of a child $$(x)$$ and his weight $$(y)$$ as he grows

Distance travelled $$(x)$$ and time taken$$(y)$$, if a car not moving at a uniform

Wages of a worker $$(x)$$ and the number of hours of work $$(y)$$

The number of students $$(x)$$ and the fees paid by them $$(y)$$

Number of rainy days $$(x)$$ and the amount of rainfall on those days $$(y)$$

Solution

Question 10

90 marks

70 marks

50 marks

60 marks

Solution

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