rectangles are of equal lengths

rectangles are drawn with unequal shapes between them

width of rectangles carry important meaning

MCQ Questions for Class 9 Maths Statistics Quiz 2

Find the class mark of the class

$11 - 19$

0%
$15$
0%
$11$
0%
$19$
0%
$14.5$

Explanation

Class interval :

$11-19$
Lower limit:

$11$
Upper limit:

$19$
Class Mark

$= \cfrac{\text{Lower} + \text{Upper}}{2}$
Class Mark

$= \cfrac{11 + 19}{2}$
Class Mark

$= 15$

The upper and lower limits of the fifth class.

0%
$52$ and $50$
0%
$55$ and $51$
0%
$54$ and $50$
0%
$58$ and $60$

Explanation

Class Interval	Frequency
$30 - 34$	$7$
$35 - 39$	$10$
$40 - 44$	$12$
$45 - 49$	$13$
$50 - 54$	$8$
$55 - 59$	$4$

The mean of first six multiples of

$4$ is

0%
$13.5$
0%
$14.5$
0%
$14$
0%
$16$

Explanation

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

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

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

$\text{Mean }= \cfrac{4 + 8 + 12 + 16 + 20+ 24}{6}$

$\left[\because \boldsymbol{\textbf{Mean }= \cfrac{\textbf{Sum of observations}}{\textbf{Total observations}}}\right]$

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

$\text{Mean }= 14$

$\textbf{Hence, Option C is correct.}$

Find the mean of first ten odd natural numbers.

0%
$10$
0%
$11$
0%
$12$
0%
None of these

Explanation

${\textbf{Step -1: Listing first 10 odd natural numbers .}}$

${\text{First 10 odd natural numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19}}$

${\textbf{Step -2: Calculating mean}}$

${\text{We know that, mean of n numbers = }}\dfrac{{{\text{Sum of n numbers}}}}{{\text{n}}}$

$\therefore {\text{ Mean of first 10 odd natural numbers,}}$

${\text{M = }}\dfrac{{{\text{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}}}}{{{\text{10}}}}$

$\Rightarrow {\text{ M = }}\dfrac{{{\text{100}}}}{{{\text{10}}}}{\text{ = 10}}$

${\textbf{Thus, the mean of first 10 odd natural numbers is 10 .}}$

The lower limit of class interval

$0-5$ is _____ and upper limit is _____.

0%
$0, 5$
0%
$0.1, 4.9$
0%
$2.5, 2.5$
0%
None of these

Explanation

Class interval is defined as "Lower Limit" - "Upper Limit"

The lower limit of class interval

$0-5$ is

$0$ land upper limit is

$5$

The lower limit of class having lowest frequency is:

0%
$0$
0%
$50$
0%
$10$
0%
None of these

Explanation

The class having lowest frequency is

$10-20$ .
The lower limit of class having lowest frequency is

$10$ .

The number of times a particular item occurs in a class interval is called its

0%
mean
0%
frequency
0%
cumulative frequency
0%
none of these

Explanation

The number of times a particular item occurs in a class interval is called its frequency.

For ex.

$1,2,3,4,3,5,3,6$ in the given data

$'3'$ appear

$3$ times so the frequency of

$3$ is

$3$ .

Mean of 10 values is 32.If another values is included the mean becomesThe included value is

0%
$16$
0%
$14$
0%
$15$
0%
$19$

Explanation

Included value

$\displaystyle =31\times 11-32.6\times 10=15$

The value of

$\pi$ upto

$50$ decimal places is :

$3.14159265358979323846264338327950288419716933993751$

What is the frequency of digit

$0$ .

0%
$0$
0%
$3$
0%
$2$
0%
$1$

Explanation

The given value of $\pi$ is $3.14159265358979323846264338327950288419716933993751$ Digits after decimal will be considered for counting the frequency Digits Frequency $0$ $1$ $1$ $5$ $2$ $5$ $3$ $9$ $4$ $4$ $5$ $5$ $6$ $4$ $7$ $4$ $8$ $5$ $9$ $8$ Hence, frequency of $0$ is $1$ .

In the class interval

$35 - 46$ ; the lower limit is

$35$ and upper limit is _____

0%
$36$
0%
$45$
0%
$46$
0%
$48$

Explanation

Class interval is defined by the lower limit- upper limit.
Hence, for the class interval

$35 - 46$
The lower limit is

$35$ and upper limit is

$46$

State true or false:

The class mark of class interval

$22 - 29$ is

$25.5$ .

0%
True
0%
False

Explanation

The class mark of

$22 - 29 =$ mean of lower limit and upper limit

$= \dfrac{22 + 29}{2} = 25.5$

Range of

$14, 12, 17, 18, 16$ and

$x$ is

$20$ . Find

$x$

$(x > 0)$

0%
$2$
0%
$28$
0%
$32$
0%
Cannot be determined

Explanation

Let

$x$ be the highest no. and here lowest no. is

$12$
Since, Range

$=$ Highest value

$-$ Lowest value

$\therefore 20 = x - 12$

$\therefore x =32$

$\therefore$ Option

$C$ is correct.

What is the average of squares of consecutive odd numbers between

$1$ and

$13$ ?

0%
$49$
0%
$51$
0%
$53$
0%
$57$

Explanation

The consecutive odd numbers from

$1$ to

$13 = 3, 5, 7, 9, 11$
Thus required average =

$\displaystyle \frac{3^{2}+5^{2}+7^{2}+9^{2}+11^{2}}{5}=\frac{9+25+49+81+121}{5}=\frac{285}{5}=57$

Mean of the first six even numbers is

0%
$7$
0%
$5$
0%
$3$
0%
$1$

Explanation

Formula used:

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

Apply the above formula, we get

Mean

$=\displaystyle \frac{2+4+6+8+10+12}{6}$
=

$\displaystyle \frac{42}{6}=7$

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

0%
variation
0%
frequency
0%
cumulative frequency
0%
None of these

Explanation

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.

A graph drawn using vertical bars is called

0%
Bar graph
0%
Line graph
0%
Pictograph
0%
Pie graph

The arithmetic mean of first five natural numbers is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Formula used:

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

Apply the above formula, we get

$\displaystyle AM=\frac{1+2+3+4+5}{5}=3$

The class interval having highest frequency is:

0%
$30-40$
0%
$50-60$
0%
$20-30$
0%
$40-50$

Explanation

From the given table, the highest frequency is

$17$

The highest frequency belongs to the class interval

$30-40$

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

0%
$\displaystyle 2 m + l$
0%
$\displaystyle 2 m - l$
0%
$\displaystyle m - l$
0%
$\displaystyle m - 2l$

Explanation

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.

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

0%
$10.5$
0%
$20$
0%
$20.5$
0%
$17.5$

Explanation

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.

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

0%
6
0%
7
0%
5.5
0%
11

Explanation

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

$=x_l-x_s=7$

The statistical data are of two types. These types are

0%
technical data and presentation data
0%
primary data and secondary data
0%
primary data and personal data
0%
none of the above

The size of the each class intervals is:

0%
$5$
0%
$0$
0%
$10$
0%
None of these

Explanation

The size of the each class intervals is the difference between the upper and lower limits of any class

$= 10-0 = 10.$

The classmark of a class interval is

0%
Lower limit + Upper limit
0%
Upper limit - Lower limit
0%
$\displaystyle \frac{1}{2}$ (Lower limit + Upper limit)
0%
$\displaystyle \frac{1}{4}$ (Lower limit + Upper limit)

Explanation

Class mark is the average of lower limit and upper limit.
Thus, Class mark

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

The arithmetic mean of first five natural number is

0%
$2$
0%
$3$
0%
$4$
0%
$8$

Explanation

Formula used:

$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$

Apply the above formula, we get

$\text{Arithmetic mean}=\cfrac{1+2+3+4+5}{5}$

$=3$

The class mark of the class intervaI

$40-50$ is _____.

0%
$40$
0%
$45$
0%
$50$
0%
Data insufficient

Explanation

The number in the middle of the class is called class mark.
i.e,

$45$

In a bar chart:

0%
rectangles are of equal lengths
0%
width of rectangles is fixed
0%
rectangles are drawn with unequal shapes between them
0%
width of rectangles carry important meaning

Histogram are a great way to show results of _________.

0%
categories
0%
continuous data
0%
both $A$ and $B$
0%
None of these

Explanation

We know, histogram is a graphical representation of data using bars with no gap in between, i.e. it shows continuous data.

Also, the height of the bar shows the frequency of the class-interval.

Hence, option

$B$ is correct.

Which type of data can be represented by histogram?

0%
Grouped data
0%
Ungrouped data
0%
Continuous data
0%
None of these

What is the upper and lower limit of the class

$55- 60$ ?

0%
$55, 60$
0%
$55, 55$
0%
$60, 60$
0%
$60, 55$

Explanation

$60$ is the upper limit and

$55$ is the lower limit of the class

$55-60$

CBSE Questions for Class 9 Maths Statistics Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Statistics Quiz 2 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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