Data Handling Test

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Data Handling Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

The average of $$33.5, 30.4, 25.6, 31.5\ and \ 29$$ is:

A
$$28.5$$

B
$$30$$

C
$$29.7$$

D
$$30.4$$

Solution

$$\displaystyle = \frac{33.5+30.4+25.6+31.5+29}{5}$$
$$\displaystyle = \frac{150}{5}=30$$
Question 2
1 / -0

Mean of a set of observations is the value which

A
Occurs most frequently

B
Divides observations into two equal parts

C
Is a representative of a whole group

D
Is the sum of observations

Solution

Mean refers to an average that describes the central tendency of data.

Hence, it represents the whole group.

$$ \therefore $$ Option C is correct.
Question 3
1 / -0

Mode of a set of observations is the value which

A
occurs most frequently

B
divides the observations into two equal parts

C
is the mean of the middle two observations

D
is the sum of the observations

Solution

The value which occurs most number of times i.e. most frequently is called the mode of the series.
Question 4
1 / -0

A cricket player scored the following runs in $$11$$ one-day international matches:
$$65$$, $$30$$, $$7$$, $$60$$, $$65$$, $$65$$, $$30$$, $$28$$, $$30$$, $$15$$, $$30$$ Find the modal runs.

A
$$20$$ runs

B
$$25$$ runs

C
$$30$$ runs

D
$$35$$ runs

Solution

From the given observations:
$$65$$, $$30$$, $$7$$, $$60$$, $$65$$, $$65$$, $$30$$, $$28$$, $$30$$, $$15$$, $$30$$
Here, $$30$$ is the most frequent observation. hence, $$30$$ is the mode of the given data.
Question 5
1 / -0

In a cricket test match the scores of ten players are: $$85$$, $$32$$, $$0$$, $$54$$, $$29$$, $$101$$, $$73$$, $$64$$, $$29$$ and $$36$$. Find the mean of the runs.

A
$$36.4$$ runs

B
$$45.5$$ runs

C
$$50.3$$ runs

D
$$63.7$$ runs

Solution

Runs scored by $$10$$ players $$= 85,32,0,54,29,101,73,64,29,36$$
Mean $$= \cfrac{\text{Sum}}{\text{Number of players}}$$
Mean $$= \cfrac{85 + 32 + 0 + 54 + 29 + 101 + 73 + 64 + 29 + 36}{10}$$
Mean $$= \cfrac{503}{10}$$
Mean $$= 50.3$$
Question 6
1 / -0

The average of $$0.3, 0.03$$ and $$0.003$$ is ____

A
$$0.3$$

B
$$0.1$$

C
$$0.111$$

D
none of these

Solution

$$\text{Average} = \displaystyle \frac{0.3+0.03+0.003}{3}$$
$$=\displaystyle \frac{0.333}{3}\\=0.111$$
Question 7
1 / -0

The average of $$5, 0, 6,$$$$\displaystyle \frac{1}{4}$$ and $$\displaystyle 8\frac{3}{4}$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$\displaystyle \frac{5+0+6+\frac{1}{4}+8\frac{3}{4}}{5}$$
= $$\displaystyle \frac{20}{5}=4$$
Question 8
1 / -0

The range of observations 2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3 is

A
6

B
7

C
5.5

D
11

Solution

Given, observations: 2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3

Largest term $$=x_l = 9$$

Smallest term $$=x_s = 2$$

Hence Range of the given distribution $$=x_l-x_s=7$$
Question 9
1 / -0

Find the mean of first ten odd natural numbers.

A
$$7$$

B
$$10$$

C
$$13$$

D
$$18$$

Solution

$${\textbf{Step - 1: Listing first 10 odd natural numbers}}$$
$${\text{First 10 odd natural numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$$
$${\textbf{Step - 2: Calculating mean}}$$
  $${\text{We know that, mean of n numbers = }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$$
  $$\therefore {\text{ Mean of first 10 odd natural numbers,}}$$
$${\text{M = }}\dfrac{{{\text{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}}}}{{{\text{10}}}}$$
  $$ \Rightarrow {\text{ M = }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{ = 10}}$$
$${\textbf{Thus, the mean of first 10 odd natural numbers is 10}}$$
Question 10
1 / -0

In a monthly test the marks obtained in mathematics by $$16$$ students of a class are as follows
$$0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8 $$
The arithmetic mean of the marks obtained is

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Since, Mean $$= \cfrac {0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8}{16} $$
      $$\Rightarrow $$ Mean $$= \cfrac {64}{16} $$
   $$\Rightarrow $$ Mean $$= 4 $$
   $$\therefore $$ Option $$B$$ is correct.

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Re-Attempt Weekly Quiz Competition

Data Handling Test - 13

Result

Data Handling Test - 13

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SHARING IS CARING

Question 1

$$28.5$$

$$30$$

$$29.7$$

$$30.4$$

Solution

Question 2

Occurs most frequently

Divides observations into two equal parts

Is a representative of a whole group

Is the sum of observations

Solution

Question 3

occurs most frequently

divides the observations into two equal parts

is the mean of the middle two observations

is the sum of the observations

Solution

Question 4

$$20$$ runs

$$25$$ runs

$$30$$ runs

$$35$$ runs

Solution

Question 5

$$36.4$$ runs

$$45.5$$ runs

$$50.3$$ runs

$$63.7$$ runs

Solution

Question 6

$$0.3$$

$$0.1$$

$$0.111$$

none of these

Solution

Question 7

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 8

6

7

5.5

11

Solution

Question 9

$$7$$

$$10$$

$$13$$

$$18$$

Solution

Question 10

$$3$$

$$4$$

$$5$$

$$6$$

Solution

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