Statistics Test

Result

Statistics Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

A graph drawn using vertical bars is called

A
Bar graph

B
Line graph

C
Pictograph

D
Pie graph

Solution

Bar graph represents the data in the form of vertical bars.
Question 2
1 / -0

The mean of first six multiples of $$4$$ is

A
$$13.5$$

B
$$14.5$$

C
$$14$$

D
$$16$$

Solution

$$\textbf{Step 1: Write all 6 multiples of 4}$$

$$\text{First six multiples of 4 are: }4, 8, 12, 16, 20, 24$$

$$\textbf{Step 2: Find Mean of all 6 observations}$$

$$\text{Mean }= \cfrac{4 + 8 + 12 + 16 + 20+ 24}{6}$$ $$\left[\because \boldsymbol{\textbf{Mean }= \cfrac{\textbf{Sum of observations}}{\textbf{Total observations}}}\right]$$

$$\text{Mean }= \cfrac{112}{6}$$

$$\text{Mean }= 14$$

$$\textbf{Hence, Option C is correct.}$$
Question 3
1 / -0

The arithmetic mean of first five natural numbers is

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$

Apply the above formula, we get
$$\displaystyle AM=\frac{1+2+3+4+5}{5}=3$$
Question 4
1 / -0

Let $$\displaystyle m$$ be the mid-point and $$\displaystyle l$$ the upper class limit of a class in a continuous frequency distribution. The lower limit of the class is

A
$$\displaystyle 2 m + l$$

B
$$\displaystyle 2 m - l$$

C
$$\displaystyle m - l$$

D
$$\displaystyle m - 2l$$

Solution

Given, $$m$$ be the mid-point and $$l$$ the upper class limit of a class in a continuous frequency distribution.
Since, Mid point $$= \displaystyle \frac {\text {Lower class limit} + \text {Upper class limit}}{2} $$
$$\Rightarrow m = \displaystyle \frac {\text {Lower class limit} + l}{2} $$
$$\Rightarrow 2m =$$ Lower class limit $$+ l$$
$$\Rightarrow $$ Lower class limit $$ = 2m - l$$
Hence, option B is correct.
Question 5
1 / -0

The upper class limit of inclusive type class interval 10 - 20 is

A
$$10.5$$

B
$$20$$

C
$$20.5$$

D
$$17.5$$

Solution

In inclusive form class limits are obtained by subtracting 0.5 from lower limit and adding 0.5 to the upper limit.
$$\therefore $$ class limits of 10 - 20 class
interval in inclusive form are 9.5 - 20.5.
so, upper class limit of 9.5 - 20.5 is 20.5
$$\therefore $$ Option C is correct.
Question 6
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 7
1 / -0

The statistical data are of two types. These types are

A
technical data and presentation data

B
primary data and secondary data

C
primary data and personal data

D
none of the above

Solution

The statistical data are two type .These are
1.Primary data-The primary data is the first hand information that collect and published by the organisation.
2.Secondary Data-The secondary data is the second hand information that is already collected by someone for some purpose.
Question 8
1 / -0

Directions For Questions

Study the following frequency table and choose the appropriate answer :
Class Intervals Frequency
$$0-10$$ $$6$$
$$10-20$$ $$2$$
$$20-30$$ $$13$$
$$30-40$$ $$17$$
$$40-50$$ $$11$$
$$50-60$$ $$4$$
$$60-70$$ $$8$$
$$70-80$$ $$7$$

...view full instructions

The size of the each class intervals is:

A
$$5$$

B
$$0$$

C
$$10$$

D
None of these

Solution

The size of the each class intervals is the difference between the upper and lower limits of any class $$= 10-0 = 10.$$
Question 9
1 / -0

The number of times a particular item occurs in a given data is called its

A
variation

B
frequency

C
cumulative frequency

D
None of these

Solution

The number of times a particular item occurs in a given data is called its frequency.
for ex: In given data $$10,20,5,20,3,10,20,5,14,11,16$$
$$20$$ is occur 3 times.
$$\therefore $$ frequency of $$20$$ is = $$ 3$$
Option B is correct.
Question 10
1 / -0

The classmark of a class interval is

A
Lower limit + Upper limit

B
Upper limit - Lower limit

C
$$\displaystyle \frac{1}{2}$$ (Lower limit + Upper limit)

D
$$\displaystyle \frac{1}{4}$$ (Lower limit + Upper limit)

Solution

Class mark is the average of lower limit and upper limit.
Thus, Class mark $$= \displaystyle \frac{1}{2}$$ (Lower limit + Upper limit)

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Class Intervals	Frequency
$$0-10$$	$$6$$
$$10-20$$	$$2$$
$$20-30$$	$$13$$
$$30-40$$	$$17$$
$$40-50$$	$$11$$
$$50-60$$	$$4$$
$$60-70$$	$$8$$
$$70-80$$	$$7$$

Statistics Test - 20

Result

Statistics Test - 20

Score

-

Rank

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SHARING IS CARING

Question 1

Bar graph

Line graph

Pictograph

Pie graph

Solution

Question 2

$$13.5$$

$$14.5$$

$$14$$

$$16$$

Solution

Question 3

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 4

$$\displaystyle 2 m + l$$

$$\displaystyle 2 m - l$$

$$\displaystyle m - l$$

$$\displaystyle m - 2l$$

Solution

Question 5

$$10.5$$

$$20$$

$$20.5$$

$$17.5$$

Solution

Question 6

6

7

5.5

11

Solution

Question 7

technical data and presentation data

primary data and secondary data

primary data and personal data

none of the above

Solution

Question 8

$$5$$

$$0$$

$$10$$

None of these

Solution

Question 9

variation

frequency

cumulative frequency

None of these

Solution

Question 10

Lower limit + Upper limit

Upper limit - Lower limit

$$\displaystyle \frac{1}{2}$$ (Lower limit + Upper limit)

$$\displaystyle \frac{1}{4}$$ (Lower limit + Upper limit)

Solution

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