Descriptive Statistics Test 18

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Descriptive Statistics Test 18

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following statements is untrue for tabulation?

A
Statistical analysis of data requires tabulation

B
It facilitates comparison between rows and not columns

C
Complicated data can be presented

D
Diagrammatic representation of data requires tabulation
Question 2
1 / -0

Consider the following statements:
I. In a bar graph not only height but also width of each rectangle matters.
II. In a bar graph height of each rectangle matters and not its width
III. In a histogram the height as well as the width of each rectangle matters
IV. A bar graph is two dimensional
Of these statements:

A
I alone is correct

B
III alone is correct

C
II and III are correct

D
I and IV are correct
Question 3
1 / -0

The mean and standard deviation of a random variable $$x$$ is given by $$5$$ and $$3$$ respectively. The standard deviation of $$2-3x$$ is _____

A
$$-7$$

B
$$81$$

C
$$34$$

D
$$9$$

Solution

$$\sigma(2-3x)=\sigma(-3x)=|-3|\sigma(x)=3\times 3=9$$
Question 4
1 / -0

Maximum number of teachers are in between which age group?

A
$$45-50$$

B
$$25-30$$

C
$$40-45$$

D
$$30-35$$

Solution

$$\Rightarrow$$ From given histogram we are going to form frequency distribution table:

$$Ages$$
$$(Years)$$ $$No.\,of\,teachers$$
$$20-25$$ $$4$$
$$25-30$$ $$5$$
$$30-35$$ $$6$$
$$35-40$$ $$3$$
$$40-45$$ $$2$$
$$45-50$$ $$5$$
$$\Rightarrow$$ From above table we can see, the maximum number of teachers are between age group $$30-35$$ which is $$6$$
Question 5
1 / -0

If $$\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9} $$ and $$\sum\limits_{i = 1}^{18} {{{({x_i} - 8)}^2} = 45} $$, then standard deviation of $${x_1},{x_2},...,{x_{18}}$$ is

A
$$\dfrac{4}{9}$$

B
$$\dfrac{9}{4}$$

C
$$\dfrac{3}{2}$$

D
$$\dfrac{3}{4}$$

Solution

Given: $$\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right) = 9} $$

$$\displaystyle\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2} = 45} $$

Let $$X$$ be a random variable taking values $${x_1},........{x_{18.}}$$

Then $$X-8$$ has the values $${x_1} - 8,....,{x_{18}} - 8.$$

Now, $$E\left( {X - 8} \right) =\displaystyle {{\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right)} } \over {18}}$$

$$ = \displaystyle{9 \over {18}} = {1 \over 2}$$

And $$E\left[ {{{\left( {X - 8} \right)}^2}} \right] = \displaystyle{{\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2}} } \over {18}}$$

$$ =\displaystyle {{45} \over {18}} = {5 \over 2}$$

Thus, $$Var\left( {X - 8} \right) = E\left[ {{{\left( {X - 8} \right)}^2}} \right] - \left[ {E{{\left( {X - 8} \right)}}} \right]^2$$

$$ =\displaystyle {5 \over 2} - {\left( {{1 \over 2}} \right)^2}$$

$$ = \displaystyle{5 \over 2} - {1 \over 4}$$

$$ = \displaystyle{9 \over 4}$$

We know $$Var\left( {1.X - 8} \right) = {1^2}Var\left( X \right)$$, thus,

$$Var\left( X \right) = \displaystyle{9 \over 4}$$

Standard deviation of $$X = \sqrt {Var\left( X \right)} $$

$$ = \displaystyle\sqrt {{9 \over 4}} = {3 \over 2}$$
Question 6
1 / -0

Solve:
$$\log_5 \dfrac{(25)^4}{\sqrt{625}}$$

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

We have,
$$\log_5 \dfrac{(25)^4}{\sqrt{625}}$$

$$\Rightarrow \log_5 \dfrac{(25)^4}{(25)}$$

$$\Rightarrow \log_5 (25)^3$$

$$\Rightarrow \log_5 5^6$$

$$\Rightarrow 6\log_5 5$$ $$\therefore \log m^n=n\log m$$

$$\Rightarrow 6\times 1$$ $$\therefore \log_a a=1$$

$$\Rightarrow 6$$

Hence, this is the answer.
Question 7
1 / -0

Coefficient of skewnwss for the value $$Median = 18.8,{Q_1} = 14.6,{Q_3} = 25.2$$ is

A
$$0.2$$

B
$$0.5$$

C
$$0.7$$

D
None

Solution

$$\Rightarrow$$ Given data : $$M_d=18.8,Q_1=14.6,Q_3=25.2$$
The coefficient of skewness based on quartile is known as Bowley's coefficient of skewness is given by
$$S_k(B)=\dfrac{Q_3+Q_1-2M_d}{Q_3-Q_1}$$

$$=\dfrac{25.2+14.6-(2\times 18.8)}{25.2-14.6}$$

$$=\dfrac{39.8-37.6}{10.6}$$

$$=\dfrac{2.2}{10.6}$$

$$=0.2$$
Question 8
1 / -0

The variance of the data $$6,\ 8,\ 10,\ 12,\ 14,\ 16,\ 18,\ 20,\ 22,\ 24$$ is

A
$$15$$

B
$$20$$

C
$$30$$

D
$$33$$

Solution

$$\overline{x}=\dfrac{6+8+10+12+14+16+18+20+22+24}{10}=\dfrac{150}{10}=15$$
$$Variance=\dfrac{\sum(x-\overline{x})^2}{10}$$

$$=\dfrac{(6-15)^2+(8-15)^2+(10-15)^2+(12-15)^2+(14-15)^2+(16-15)^2+(18-15)^2+(20-15)^2+(22-15)^2+(24-15)^2}{10}$$

$$=\dfrac{(-9)^2+(-7)^2+(-5)^2+(-3)^2+(-1)^2+(1)^2+(3)^2+(5)^2+(7)^2+(9)^2}{10}$$

$$=\dfrac{81+49+25+9+1+1+9+25+49+81}{10}$$

$$=\dfrac{330}{10}$$
$$=33$$

$$\therefore$$ Variance $$=33$$
Question 9
1 / -0

Use the graph to answer the question.How many students have more than 100 rupees?

A
7

B
6

C
100

D
8

Solution

From the graph
Number of students have more than 100 rupees $$=4+3+1=8$$
Question 10
1 / -0

If $$V$$ is the variance and $$M$$ is the mean of first $$15$$ natural numbers, then what is $$V+M^2$$ equal to ?

A
$$\dfrac{124}{3}$$

B
$$\dfrac{148}{3}$$

C
$$\dfrac{248}{3}$$

D
$$\dfrac{124}{9}$$

Solution

first 15 natural numbers

$$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$$

$$N=5$$

mean,

$$\mu =\dfrac{1}{N}\sum_{i=1}^{N}x_i$$

$$\mu =\dfrac{1}{15}(1+2+3+4+5+6+7+8+9+10+11+12+13+14+15)$$

$$=\dfrac{120}{5}=8$$

variance,

$$\sigma ^2=\dfrac{1}{N}\sum_{i=1}^{N}(x_i-\mu )^2$$

$$=\dfrac{1}{15}\sum_{i=1}^{15}(x_i-8 )^2$$

$$=\dfrac{1}{15}(280)$$

Given,

$$V+M^2$$

$$=\dfrac{280}{15}+8^2$$

$$=\dfrac{1240}{15}$$

$$=\dfrac{248}{3}$$

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Re-Attempt Weekly Quiz Competition

$$Ages$$ $$(Years)$$	$$No.\,of\,teachers$$
$$20-25$$	$$4$$
$$25-30$$	$$5$$
$$30-35$$	$$6$$
$$35-40$$	$$3$$
$$40-45$$	$$2$$
$$45-50$$	$$5$$

Descriptive Statistics Test 18

Result

Descriptive Statistics Test 18

Score

-

Rank

-

SHARING IS CARING

Question 1

Statistical analysis of data requires tabulation

It facilitates comparison between rows and not columns

Complicated data can be presented

Diagrammatic representation of data requires tabulation

Question 2

I alone is correct

III alone is correct

II and III are correct

I and IV are correct

Question 3

$$-7$$

$$81$$

$$34$$

$$9$$

Solution

Question 4

$$45-50$$

$$25-30$$

$$40-45$$

$$30-35$$

Solution

Question 5

$$\dfrac{4}{9}$$

$$\dfrac{9}{4}$$

$$\dfrac{3}{2}$$

$$\dfrac{3}{4}$$

Solution

Question 6

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 7

$$0.2$$

$$0.5$$

$$0.7$$

None

Solution

Question 8

$$15$$

$$20$$

$$30$$

$$33$$

Solution

Question 9

7

6

100

8

Solution

Question 10

$$\dfrac{124}{3}$$

$$\dfrac{148}{3}$$

$$\dfrac{248}{3}$$

$$\dfrac{124}{9}$$

Solution

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