MCQ Questions for Class 9 Maths Statistics Quiz 6

What is frequency of the class- interval

$20-30$ ?

0%
$25$
0%
$20$
0%
$5$
0%
$10$

Explanation

$\Rightarrow$ In the above histogram we can see that, frequency of the class interval

$20-30$ is

$20$

In the following distribution :

Monthly income range (in Rs.)	Number of families
Income more than Rs. $10000$	$100$
Income more than Rs. $13000$	$85$
Income more than Rs. $16000$	$69$
Income more than Rs. $19000$	$50$
Income more than Rs. $22000$	$33$
Income more than Rs. $25000$	$15$

The numbers of families having income range (in Rs)

$16000 - 19000$ is:

0%
$31$
0%
$26$
0%
$23$
0%
$19$

Explanation

Number of families having income range from Rs.

$16000 - 19000$

$=$ Number of families having income more than Rs.

$16000 -$ Number of families having income more than Rs.

$19000$

$= 69-50$

$= 19$

Hence, the answer is

$19$ .

The upper-class limit of 24 - 30 is?

0%
$24$
0%
$30$
0%
$24$ or $30$
0%
None of these

Explanation

Upper class limit is the maximum value of the class interval.

Therefore the upper class limit for

$24-30$ is

$30$ .

What is the mean of first 8 whole numbers?

0%
$3$
0%
$3.5$
0%
$4$
0%
$4.5$

Explanation

First

$8$ whole numbers are

$0,1,2,3,4,5,6,7$

We know that,

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$

$\therefore \ \text{Mean}=\dfrac {0+1+2+3+4+5+6+7}{8}$

$\Rightarrow \dfrac {28}{8}=3.5$

Hence, the mean of first

$8$ whole numbers is

$3.5$ .

State the number of students in the age group

$10 - 13.$

0%
$85$
0%
$113$
0%
$103$
0%
$145$

Explanation

From the table, number of students below age

$10$ is

$140$ and

the number of students below age

$13$ is

$243$ .

Therefore number of students in the age group of

$10-13$ is

$243-140=103$ .

The class mark of the third class is:

0%
$40$
0%
$42$
0%
$51$
0%
$53$

Explanation

The third class is

$40 -44$
Class Mark

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

$= \cfrac{40 + 44}{2}$
Class Mark

$= 42$

Find the class mark of the first class.

0%
$859.5$
0%
$800$
0%
$849.5$
0%
$944.5$

Explanation

The first class has the interval:

$800 - 899$
The class mark will be the mean of the lower limit and the upper limit of the class.
Class mark

$= \cfrac{\text{lower limit} + \text{upper limit}}{2}$
Class mark

$= \cfrac{800 + 899}{2}$
Class mark

$= 849.5$

The blood groups of 30 students of class VIII are recorded as follows:

$A,B,O,O,AB,O,A,O,B,A,O,B,A,O,O,$

$A,AB,O,A,A,O,O,AB,B,A,O,B,A,B,O.$
Which is the most common and which is the rarest blood group among these students?

Answer is in form of (Rarest, Common)

0%
$(AB,O)$
0%
$(O,AB)$
0%
$(A,O)$
0%
$(O,A)$

Explanation

Most common is the one which has occurreded very frequently and the rarest the one which has occurred very few times.

From the given data set

$O$ has occurred red very frequently and

$AB$ has occurred very few times.

Therefore, the rarest is

$AB$ and common is

$O$ .

Hence,

$(\text{Rarest, Common})=(AB,O)$

The A.M. of the first ten odd numbers is

0%
$10$
0%
$100$
0%
$1000$
0%
$1$

Explanation

First ten odd numbers are

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$ respectively.

$A.M. (\overline {x}) = \dfrac{\text {sum of all the terms}}{\text {total number of terms}}$

$A.M. \left(\overline { x } \right) = \displaystyle\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$

$= \displaystyle\frac{100}{10}$

$=10$

$\therefore$ the

$A.M$ of first ten odd numbers is

$10$

The mean of first five prime numbers is

0%
$5.6$
0%
$3.6$
0%
$6.83$
0%
$5.2$

Explanation

First five prime numbers are:

$2,3,5,7,11$
Mean

$= \cfrac{\text{Sum}}{\text{Count of Numbers}}$
Mean

$= \cfrac{2 + 3 + 5 + 7 + 11}{5}$
Mean

$= \cfrac{28}{5}$
Mean

$= 5.6$

The horizontal line in a bar graph is called:

0%
$x$ -axis
0%
$y$ -axis
0%
Imaginary axis
0%
None of these

Explanation

In the Cartesian coordinate system, the horizontal reference line is called

$x$ -axis.

Therefore, the horizontal line in a bar graph is called

$x$ -axis.

Hence, option

$A$ is correct.

What temperature does the given thermometer shows?

0%
$35^{\circ}C$
0%
$77^{\circ}F$
0%
$95^{\circ}C$
0%
$122^{\circ}F$

Explanation

Thermometer shows

$35^o C$ or

$95^o F$

Hence the correct answer is option A.

The arithmetic mean of first ten natural numbers is

0%
$5.5$
0%
$6$
0%
$7.5$
0%
$10$

Explanation

Formula used:

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

So, by using the above formula we get,

$A.M=\dfrac{1+2+3+...+10}{10}$

$=\dfrac{55}{10}$

$=5.5$

Hence, option

$A$ is correct.

The class-mark of class

$50-60$ is:

0%
$50$
0%
$60$
0%
$55$
0%
None of these

Explanation

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

$55$

The range of

$15, 14, x, 25, 30, 35$ is

$23$ . Find the least possible value of

$x$ .

0%
$14$
0%
$12$
0%
$13$
0%
$11$

Explanation

Let

$x$ be the least value and

$35$ be the Highest value.
Since, Range

$=$ Highest value

$-$ Lowest value

$\Rightarrow 23 = 35 - x$

$\Rightarrow x = 12$

$\therefore$ Option

$B$ is correct.

The following marks were obtained by the students in a test

$81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62$ .
What is the range of the marks obtained?

0%
$9$
0%
$17$
0%
$27$
0%
$33$

Explanation

The given observation are:

$81, 72, 90, 90, 86, 85, 92, 70, 71, 83, 89, 95, 85, 79, 62$ .
The minimum marks obtained are:

$62$
Maximum marks obtained are:

$95$
Thus, range

$=$ maximum marks

$-$ minimum marks

$= 95 - 62 = 33$

The distance (in km) of

$40$ engineers from their residence to their place of work were found as follows:

$5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8,$

$3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12$
How many engineers travelled distance more than

$14$ and less than

$26$ ?

0%
$9$
0%
$10$
0%
$11$
0%
$8$

Explanation

Distance of engineers from their residence to their work place is given by

$5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8,$

$3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12$

Now, the distances less than more than

$15$ and less than

$25$ are

$15,16,17,18,19,20,21,22,23,24,25$

The distance

$(>14\ and\ <26)$ travelled by engineers are

$20,25,19,17,18,17,16,15,18,15,15$

$\therefore$ The number of engineers who travelled distance more than

$14$ and less than

$26$ is

$11$ .

Hence, option C is correct.

The bar graph shows the number of cakes sold at a shop in four days. What is the difference in number of cakes between the highest and the lowest daily sale?

0%
$20$
0%
$35$
0%
$30$
0%
$40$

Explanation

From the above bar graph, the highest number of cakes sold is

$45$ (i.e., on Sunday).

and the lowest number of cakes sold is

$25$ (i.e., on Monday).

Therefore, the difference between the highest and lowest number of cakes

$=45-25=20$ .

Hence, option

$A$ is correct.

Given above is a bar graph showing the heights of six mountain peaks. Read the above diagram and answer the following:

Write the ratio of the heights of highest peak and the lowest peak.

0%
$22 : 15$
0%
$15 : 22$
0%
$20 : 13$
0%
$13 : 22$

Explanation

From the above bar graph,

Height of highest peak

$= 8800\ m$
Height of lowest peak

$= 6000\ m$

Hence the required ratio will be,
Ratio

$=8800 : 6000$

$= 22 :15$

Hence, option

$A$ is correct.

$1 - 5, 6 - 10, 11 - 15, 16 - 20, ...$ are the classes of a frequency distribution then the lower boundary of the class

$11 - 15$ is _____

0%
$11$
0%
$11.5$
0%
$10.5$
0%
$10$

Explanation

In order to make class intervals even we need to add

$0.5$ and reduce

$0.5$ from class boundaries

$1-5$ =

$0.5 - 5.5$

$6-10$ =

$5.5 - 10.5$

$11 - 15$ =

$10.5 - 15.5$
so , lower boundary of class

$11-15$ =

$10.5$

An orderly distribution of the raw data into certain specified categories is known as

0%
Frequency distribution
0%
Frequency
0%
Cumulative frequency
0%
Primary data

$0 - 10, 10 - 20, 20 - 30 ,...$ are the classes, then the lower boundary of the class

$20 - 30$ is _____

0%
$10$
0%
$10.5$
0%
$20$
0%
$20.5$

The lower limit and upper limit of an exclusive class interval are

$15$ and

$25$ respectively. Then the class mark is _____

0%
$26.5$
0%
$20$
0%
$28$
0%
$27$

Explanation

Let

$M$ be the class mark

$M= \dfrac { \text{Lower limit} + \text{Upper limit}}{2} \\= \dfrac {15 + 25}{2} \\= 20$

$1 - 5, 6 - 10, 11 - 15, ...$ are the classes of a frequency distribution then the size of the class is _____

0%
$4$
0%
$5$
0%
$3$
0%
$2.5$

Explanation

As the classes are of inclusive form,

Size of the class is

$=(Upper\ limit - Lower\ limit + 1 )$
So, size of the class

$=( 5 - 1 + 1) = 5$

The arithmetic mean of 5 numbers islf one of the number is excluded the mean of the remaining number isFind the excluded number.

0%
27
0%
25
0%
30
0%
35

Explanation

Sum of 5 numbers =27

$\times$ 5 =135
When one of the numbers is excluded.
Sum of remaining 4 numbers = 4

$\times$ 25 = 100
Excluded number = 135 -100 = 35.

The class mark of a class is

$25$ and if the upper limit of that class is

$40$ then its lower limit is _____

0%
$30$
0%
$20$
0%
$15$
0%
$10$

Explanation

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

$\Rightarrow 25 = \cfrac {\text{Lower limit} + 40}{2}$

$\Rightarrow \text{Lower limit} = 50 - 40 = 10$

The range of

$8, 17, 28, 16, 30, 28, 15, 5, 19$ and

$35$ is _____

0%
$20$
0%
$18$
0%
$30$
0%
$16$

Explanation

The difference between the maximum and minimum data entries is called the range
So, Range = maximum number - minimum number
Range =

$35 -5$
Range =

$30$

Refer the figure given above and answer the following questions.

How many class-intervals have equal frequency?

0%
$2$
0%
$3$
0%
$1$
0%
None

Explanation

We can see that two class intervals have bars drawn to the same height of

$10 .$ Hence, two is the correct answer.

Which of the following pairs of numbers has an average (arithmetic mean) of

$2$ ?

0%
$1 - \sqrt {2}, 3 + \sqrt {2}$
0%
$2\sqrt {3}, 2 - 2\sqrt {3}$
0%
$\dfrac {1}{0.5}, \dfrac {2.4}{1.6}$
0%
$\sqrt {5}, \sqrt {3}$
0%
$\dfrac {1}{\dfrac {2}{3}}, \dfrac {1}{\dfrac {2}{5}}$

The mean of

$20$ observations is

$15$ . One observation

$20$ is deleted and two more observations are included to the data. If the mean of new set of observations is

$15$ , then find the sum of the two new observations included.

0%
$30$
0%
$35$
0%
$33$
0%
$32$

Explanation

Mean

$= \dfrac { \text{Sum of observations}}{\text{Total number of observations}}$
Given, mean of

$20$ observations

$= 15$ .
Thus, sum of

$20$ observations

$= 15 \times 20 = 300$
When

$20$ is deleted and two more observations are included in the data, sum of observation

$= 300 - 20 + X + Y = 280 + X + Y$ .

And number of observations is

$20 - 1 + 2 = 21$
Given, Mean of wages of new set of observations

$= 15$

$= \dfrac {280 + X + Y }{21} = 15$

$\Rightarrow 280 + X + Y= 15 \times 21 = 315$

$\Rightarrow X + Y = 315 - 280 = 35$

CBSE Questions for Class 9 Maths Statistics Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Statistics Quiz 6 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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