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CBSE Questions for Class 9 Maths Statistics Quiz 6 - MCQExams.com
CBSE
Class 9 Maths
Statistics
Quiz 6
What is frequency of the class- interval $$20-30$$ ?
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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:
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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?
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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?
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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 $$10 - 13.$$
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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:
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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.
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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)
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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
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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
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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:
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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?
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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
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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:
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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 $$x$$.
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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?
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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 $$26$$?
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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?
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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.
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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.
If $$1 - 5, 6 - 10, 11 - 15, 16 - 20, ...$$ are the classes of a frequency distribution then the lower boundary of the class $$11 - 15$$ is _____
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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
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0%
Frequency distribution
0%
Frequency
0%
Cumulative frequency
0%
Primary data
Explanation
When we arrange the given set of data in a table by writing independent data values and the number of times it appears in the data list is called frequency distribution table.
$$0 - 10, 10 - 20, 20 - 30 ,...$$ are the classes, then the lower boundary of the class $$20 - 30$$ is _____
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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 _____
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0%
$$26.5$$
0%
$$20$$
0%
$$28$$
0%
$$27$$
Explanation
Let $$M$$ be the class mark
$$ M= \dfrac { \text{Lower limit} + \text{Upper limit}}{2} \\= \dfrac {15 + 25}{2} \\= 20 $$
If $$1 - 5, 6 - 10, 11 - 15, ...$$ are the classes of a frequency distribution then the size of the class is _____
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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.
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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 _____
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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 _____
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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?
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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 $$2$$?
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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.
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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 $$ = 15 $$
$$ = \dfrac {280 + X + Y }{21} = 15 $$
$$ \Rightarrow 280 + X + Y= 15 \times 21 = 315 $$
$$ \Rightarrow X + Y = 315 - 280 = 35 $$
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