Descriptive Statistics Test 33

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Descriptive Statistics Test 33

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Weekly Quiz Competition

Question 1
1 / -0

The variance of numbers $$x_1, x_2, x_3,..... x_n$$ is V. Consider the following statements :
1. If every $$x_1$$, is increased by 2, the variance of the new set of numbers is V.
2. If the numbers $$x_1$$ is squared, the variance of the new set is $$V^2$$.
Which of the following statements is/are correct ?

A
1 only

B
2 only

C
Both 1 and 2

D
Neither 1 nor 2

Solution

In case 1, a deterministic value of 2 units is being added to every $$x_i$$ and hence, the variance of the set remains unchanged as $$V$$. New variance would be $$\dfrac{1}{n}\sum\left(x_i+2-\left(\sum\dfrac{x_i+2}{n}\right)\right)=\dfrac{1}{n}\sum\left(x_i-\sum\dfrac{x_i}{n}\right)=V$$
In case 2, each $$x_i$$ is squared and hence, the mean becomes $$\dfrac{1}{n}\sum x_i^2 $$. Hence, the new variance is $$\dfrac{1}{n}\sum\left( x_i^2-\dfrac{1}{n}\sum x_i^2\right)^2$$
and not $$V^2=\dfrac{1}{n^2}\sum\left( x_i-\dfrac{1}{n}\sum x_i\right)^4$$.
Question 2
1 / -0

Calculate the standard deviation of the first $$13$$ natural numbers.

A
$$3.74$$.

B
$$2.74$$.

C
$$3.48$$.

D
None of these

Solution

$$\Rightarrow$$ $$1st$$ $$13$$ natural numbers are $$1,2,3,4,5,6,7,8,9,10,11,12,13.$$
$$\sum x=\dfrac{n(n+1)}{2}$$
$$=\dfrac{13(13+1)}{2}$$
$$=\dfrac{13\times 14}{2}$$
$$\therefore$$ $$\sum x=91$$
Now, $$\sum x^2=\dfrac{n(n+1)(2n+1)}{6}$$
$$=\dfrac{13\times 14\times 27}{6}$$
$$\therefore$$ $$\sum x^2=819$$
$$\Rightarrow$$ $$SD=\sqrt{\dfrac{\sum x^2}{n}-\left(\dfrac{\sum x}{n}\right)^2}$$
$$=\sqrt{\dfrac{819}{13}-\left(\dfrac{91}{13}\right)^2}$$
$$=\sqrt{63-49}$$
$$=\sqrt{14}$$
$$=3.74$$
Question 3
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 4

1 / -0

Calculate the standard deviation of the following data.
$$10, 20, 15, 8, 3, 4$$

$$5.97$$

$$59.7$$

$$4.97$$

None of these

Solution

$$\Rightarrow$$ The given numbers are $$3,4,8,10,15,20$$.

$$\Rightarrow$$ $$\overline{X}=\dfrac{3+4+8+10+15+20}{6}=10$$

$$X$$	$$X-\overline{X}$$	$$(X-\overline{X})^2$$
$$3$$	$$-7$$	$$49$$
$$4$$	$$-6$$	$$36$$
$$8$$	$$-2$$	$$4$$
$$10$$	$$0$$	$$0$$
$$15$$	$$5$$	$$25$$
$$20$$	$$10$$	$$100$$
		$$\sum (X-\overline{X})^2=214$$

$$\Rightarrow$$ $$S=\sqrt{\dfrac{(X-\overline{X})^2}{N}}$$

$$=\sqrt{\dfrac{214}{6}}$$

$$=\sqrt{35.66}$$

$$=5.97$$

Question 5
1 / -0

What is the ratio of the total sales of branch B2 for both years to the total sales of branch B4 for both years?

A
2:3

B
3:5

C
4:5

D
7:9

Solution

Total sale of $$B2=75+65=140$$
Total sale of $$B4=85+95=180$$
Ratio of sales of $$B2$$ and $$B4$$ $$=\dfrac{140}{180}=\dfrac{7}{9}$$
So option $$D$$ is correct.
Question 6
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 7
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 8
1 / -0

For the data set below, which value is in the 75th percentile?
$$1, 3, 3, 4, 6, 7, 7, 7, 8, 9, 9, 10, 12, 15, 16, 17$$

A
16

B
9

C
12

D
15

Solution

$$1, 3, 3, 4, 6, 7, 7, 7, 8, 9, 9, 10, 12, 15, 16, 17$$
Logic:- first quartile lies halfway between the lowest valve of the Median & the third Quartile lies halfway between the Median & the largest value.
$$N=16$$
Third quartile or $$75$$ percentile
$$=\dfrac{3(n+1)}{4}$$th term
$$=\dfrac{3\times 17}{4}=12.75^{th}$$ term
$$=12^{th}$$ term $$+0.75\times(13^{th}$$ term$$-12^{th}$$ term$$)$$
$$=10+0.75\times (12-10)$$
$$=10+0.75\times 2$$
$$=11.5\approx 12$$.
Question 9
1 / -0

Directions For Questions

The graph gives expenditure of a company in the years $$2003,2004$$ and $$2005$$ for the month January to July. Read the graph and answer the questions.

...view full instructions

What is the total expenditure (Rs. in lakhs) of the company during the period January to July in the year 2003?

A
$$3,800$$

B
$$3,950$$

C
$$4,600$$

D
$$5,350$$

Solution

Required expenditure of the company
$$=600+500+500+650+500+600+600$$
$$3950$$ lakh
Question 10
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)
$$\Rightarrow x^{2} = 120 + 1 = 121$$
$$\Rightarrow x = 11$$

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Descriptive Statistics Test 33

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Descriptive Statistics Test 33

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SHARING IS CARING

Question 1

1 only

2 only

Both 1 and 2

Neither 1 nor 2

Solution

Question 2

$$3.74$$.

$$2.74$$.

$$3.48$$.

None of these

Solution

Question 3

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

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

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

None of these

Solution

Question 4

$$5.97$$

$$59.7$$

$$4.97$$

None of these

Solution

Question 5

2:3

3:5

4:5

7:9

Solution

Question 6

$$6$$

$$7$$

$$12$$

$$14$$

$$25$$

Solution

Question 7

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

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

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

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

Solution

Question 8

16

9

12

15

Solution

Question 9

$$3,800$$

$$3,950$$

$$4,600$$

$$5,350$$

Solution

Question 10

$$9$$

$$13$$

$$12$$

$$10$$

$$11$$

Solution

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