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CBSE Questions for Class 9 Maths Statistics Quiz 5 - MCQExams.com
CBSE
Class 9 Maths
Statistics
Quiz 5
In the figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table and find the total number of workers.
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0%
$$145$$
0%
$$142$$
0%
$$148$$
0%
$$146$$
Explanation
Frequency Distribution table for Daily wages & Number of workers is:
Class Interval
Frequency
$$150-200$$
$$50$$
$$200-250$$
$$30$$
$$250-300$$
$$35$$
$$300-350$$
$$20$$
$$350-400$$
$$10$$
Total Workers
$$145$$
Hence, the total number of workers is $$145$$.
Thus, option $$A$$ is correct.
The blood groups of 30 students are recorded as follows:
$$A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A,$$
$$ AB, B, A, A, O, A, AB, B, A, O, B, A, B, A$$
What are the frequency distribution for the data $$A$$ and $$AB$$ ?
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$$10, 2$$
0%
$$11, 2$$
0%
$$12, 4$$
0%
$$13, 5$$
Explanation
Frequency is the number of times the data has occured in the data set.
The frequency of the data $$A$$ is $$12$$. Since, $$A$$ has occured $$12$$ times in the given data set.
Similarly $$AB$$ has occured $$4$$ times in the data set. Therefore the frequency of $$AB$$ is $$4$$
The class marks of a continuous distribution are:
$$1.04, 1.14, 1.24, 1.34, 1.44, 1.54$$ and $$1.64$$
Is it correct to say that the last interval will be $$1.55 - 1.73$$? Justify your answer.
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0%
It is correct.
0%
It is not correct. Reason is that difference between two consecutive marks should be equal to the class size.
0%
Ambiguous
0%
Data insufficient
Explanation
Class mark means the mid point of a class interval.
Difference between two consecutive class marks is always equal to the class size.
Now $$1.14-1.04=0.1$$
$$1.24-1.14=0.1$$
$$1.34-1.24=0.1$$ and so on
Class size $$=0.1$$
But in class interval $$(1.55 - 1.73), 1.73-1.55=0.8$$ which is not equal to the class size found previously.
So class interval $$(1.55 - 1.73 )$$ cannot be the correct class interval for the given class marks.
The class mark of the class 90-120 is
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0%
90
0%
105
0%
115
0%
120
Explanation
Class mark is the mean of the upper and lower limit of the class interval = $$\frac{90+120}{2}$$ = 105
Prepare a continuous grouped frequency distribution from the following data:
Mid Point
Frequency
$$5$$
$$4$$
$$15$$
$$8$$
$$25$$
$$13$$
$$35$$
$$12$$
$$45$$
$$6$$
Also find the size of class intervals.
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Class size $$= 16$$
0%
Class size $$= 14$$
0%
Class size $$= 12$$
0%
Class size $$= 10$$
Explanation
classes
0-10
10-20
20-30
30-40
40-50
frequency
4
8
13
12
6
The mean marks (out of $$100$$) of boys and girls in an examination are $$70$$ and $$73$$, respectively. If the mean marks of all the students in that examination is $$71$$, find the ratio of the number of boys to the number of girls.
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$$1:1$$
0%
$$2:1$$
0%
$$1:2$$
0%
$$2:3$$
Explanation
Let the total number of boys be $$x$$ and the total number of girls br $$y$$.
Let the sum of the marks of boys be $$a$$ and the sum of the marks of girls be $$b$$.
Mean marks of boys are $$=70$$
Mean marks of girls are $$=73$$
Therefore,
$$\Rightarrow \dfrac{a}{x}=70$$
$$\Rightarrow a=70x$$
and
$$\Rightarrow \dfrac{b}{y}=73$$
$$\Rightarrow b=73y$$
Now, mean marks of all the students is $$71$$
Therfore,
$$\Rightarrow \dfrac{sum\ of\ the\ marks\ of\ all\ the\ students}{x+y}=71$$
$$\Rightarrow sum\ of\ the\ marks\ of\ all\ the\ students=a+b$$
$$\Rightarrow a+b=70x+73y$$
Therefore,
$$\Rightarrow \dfrac{70x+73y}{x+y}=71$$
$$\Rightarrow 70x+73y=71x+71y$$
$$\Rightarrow2y=x$$
$$\Rightarrow \dfrac{x}{y}=\dfrac{2}{1}$$
$$Hence\ the\ ratio\ is\ 2:1$$
In the class intervals $$10-20, 20-30$$, the number $$20$$ is included in:
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0%
$$10-20$$
0%
$$20-30$$
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Both the intervals
0%
None of these intervals
Explanation
The number is always included in the lower limit of the class interval. Hence, $$20$$ is included in $$20-30$$
For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are the:
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0%
upper limits of the classes
0%
lower limits of the classes
0%
class marks of the classes
0%
upper limits of perceeding classes
Explanation
For drawing a frequency polygon of a continuous frequency distribution,
we draw the class marks of the classes on the abscissae and the frequency of the classes on the ordinates.
We know, class marks are the mean of lower and upper limit of the class intervals.
Hence, option $$C$$ is correct.
In a frequency distribution, the mid value of a class is $$10$$ and the width of the class is $$6$$. The lower limit of the class is:
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$$6$$
0%
$$7$$
0%
$$8$$
0%
$$12$$
Explanation
$$\textbf{Step1: Find the values}$$
$$\text{Let $$x$$ and $$y$$ be the upper and lower class limit in a frequency distribution.}$$
$$\text{Given, mid value of a class is 10}$$
$$\Rightarrow \dfrac{x+y}{2}=10$$
$$\Rightarrow x+y=20 \ldots (1)$$
$$\text{Also given that, width of the class is 6}$$
$$x-y=6 \ldots(2)$$
$$\text{On Adding equation $$(1)$$ and $$(2)$$, we get}$$
$$\Rightarrow 2 x=20+6 $$
$$\Rightarrow 2 x=26 $$
$$\Rightarrow x=13 $$
$$\text{On putting $$x=13$$ in (1) we get}$$
$$\Rightarrow 13+y=20 $$
$$\Rightarrow y=7 $$
$$\textbf{Hence the lower limit of the class is $$7$$.}$$
To draw a histogram to represent the following frequency distribution:
Class Interval
$$5-10$$
$$10-15$$
$$15-25$$
$$25-45$$
$$45-75$$
Frequency
$$6$$
$$12$$
$$10$$
$$8$$
$$15$$
The adjusted frequency for the class $$25-45$$ is:
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$$6$$
0%
$$5$$
0%
$$3$$
0%
$$2$$
Explanation
The frequency distribution will consist of classes with an interval of $$5$$. The class $$25-45$$ includes four such classes with an interval of $$5$$. Thus the adjusted frequency will be $$\dfrac{8}{4}=2$$
A grouped frequency table with class intervals of equal sizes using $$250-270$$ ($$270$$ not included in this interval) as one of the class interval is constructed for the following data :
$$268, 220, 368, 258, 242, 310, 272, 342,$$
$$310, 290, 300, 320, 319, 304, 402, 318,$$
$$406, 292, 354, 278, 210, 240, 330, 316,$$
$$406, 215, 258, 236.$$
The frequency of the class $$310-330$$ is:
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0%
$$4$$
0%
$$5$$
0%
$$6$$
0%
$$7$$
Explanation
Numbers between $$310$$ and $$330$$ are $$310, 310, 320, 319, 318, 316.
\space 330$$ will not be included in this interval. Thus, frequency is $$6$$.
If the mean of the observations:
$$x, x + 3, x + 5, x + 7, x + 10$$ is $$9$$, the mean of the last three observations is
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0%
$$10\dfrac{1}{3}$$
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$$10\dfrac{2}{3}$$
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$$11\dfrac{1}{3}$$
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$$11\dfrac{2}{3}$$
Explanation
Mean of
$$x, x + 3, x + 5, x + 7, x + 10 = 9$$
Sum of the terms $$= 5x + 25$$
Mean = $$\dfrac{5x+25}{5} =9$$
$$x +5 =9$$
$$x =4$$
Last three observations are $$9,11,14$$
Mean $$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$$
The width of each of five continuous classes in a frequency distribution is $$5$$ and the lower class-limit of the lowest class is $$10$$. The upper class-limit of the highest class is:
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0%
$$15$$
0%
$$25$$
0%
$$35$$
0%
$$40$$
Explanation
Lower class limit $$= 10$$
Width of each class $$= 5$$
Width till the upper class limit for a frequency distribution having $$5$$ classes $$= 5\times 5=25$$
Hence, the upper class limit of highest class $$= 10 +25 =35$$
A grouped frequency distribution table with classes of equal sizes using $$63-72$$ $$(72$$ included$$)$$ as one of the class is constructed for the following data:
$$30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,\\
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,\\
102, 110, 88, 74, 112, 14, 34, 44$$
The number of classes in the distribution will be:
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$$9$$
0%
$$10$$
0%
$$11$$
0%
$$12$$
Explanation
If one the classes is $$63-72$$ then the range of the class is $$10$$ $$(72$$ included$$)$$
Range of the given data points $$= 112-30 = 82$$
Number of classes $$= \dfrac{82}{10} = 8.2$$
Classes required will be $$9$$ after round off.
The class marks of a frequency distribution are given as follows: $$15, 20, 25, ...$$
The class corresponding to the class mark $$20$$ is:
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$$12.5 , 17.5$$
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$$17.5 , 22.5$$
0%
$$18.5 , 21.5$$
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$$19.5 , 20.5$$
Explanation
Class marks are: $$15, 20, 25, ...$$
Lower limit of class $$20$$ is mean of $$15$$ and $$20 = \cfrac{15+20}{2} = 17.5$$
Upper limit of class $$20$$ is the mean of $$20$$ and $$25= \cfrac{20+25}{2} = 22.5 $$
$$20$$ is the class mark for the class interval $$17.5-22.5$$
According to above histogram, which group has the maximum number of workers?
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$$810$$
0%
$$820$$
0%
$$830$$
0%
$$840$$
Explanation
From the given histogram,
$$800-810$$ rupees was given to $$3$$ workers,
$$810-820$$ rupees was given to $$2$$ workers,
$$820-830$$ rupees was given to $$1$$ workers,
$$830-840$$ rupees was given to $$9$$ workers,
$$840-850$$ rupees was given to $$5$$ workers,
$$850-860$$ rupees was given to $$1$$ workers,
$$860-870$$ rupees was given to $$3$$ workers,
$$870-880$$ rupees was given to $$1$$ workers,
$$880-890$$ rupees was given to $$1$$ workers,
$$890-900$$ rupees was given to $$4$$ workers.
Hence, $$830-840$$ rupees was given to maximum number of workers, i.e. $$9$$.
Therefore, $$830$$ group has the maximum number of workers.
Thus, option $$C$$ is correct.
What is the sum of the frequencies of the intervals 40-50 and 50-60 ?
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0%
$$15$$
0%
$$25$$
0%
$$45$$
0%
$$20$$
Explanation
$$\Rightarrow$$ In above graph we can see, frequencies of the intervals $$40-50$$ and $$50-60$$ are $$10$$ and $$5$$.
$$\therefore$$ The sum of the frequencies of the intervals $$40-50$$ and $$50-60$$ $$=10+5=15$$
According to above histogram,
how many workers earn less than $$Rs. 850$$?
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0%
$$30$$
0%
$$20$$
0%
$$10$$
0%
$$40$$
Explanation
From the given histogram,
$$800-810$$ rupees was given to $$3$$ workers,
$$810-820$$ rupees was given to $$2$$ workers,
$$820-830$$ rupees was given to $$1$$ workers.
$$830-840$$ rupees was given to $$9$$ workers,
$$840-850$$ rupees was given to $$5$$ workers,
$$850-860$$ rupees was given to $$1$$ workers,
$$860-870$$ rupees was given to $$3$$ workers.
$$870-880$$ rupees was given to $$1$$ workers,
$$880-890$$ rupees was given to $$1$$ workers,
$$890-900$$ rupees was given to $$4$$ workers.
Then, the number of workers who earn less than rupees $$850$$
$$=n(800)+n(810)+n(820)+n(830)+n(840)$$
$$=3+2+1+9+5=20$$.
Hence, option $$B$$ is correct.
The number of hours for which students of a particular class watched television during holidays are shown through the given graph.
For how many hours did the maximum number of students watch TV?
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0%
$$5-6$$
0%
$$3-4$$
0%
$$4-5$$
0%
$$2-3$$
Explanation
Class Interval
Frequency
No. of hours
No. of Students
$$1 -2$$
$$4$$
$$2 - 3$$
$$8$$
$$3 - 4$$
$$22$$
$$4 - 5$$
$$32$$
$$5 - 6$$
$$8$$
$$6 - 7$$
$$6$$
Total-
$$80$$
The highest frequency is $$32$$, and the corresponding class interval is $$4-5$$
Hence, maximum number of students watched TV for $$4-5$$ hours,
Which class-interval has the maximum frequency?
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$$0-10$$
0%
$$30-40$$
0%
$$10-20$$
0%
$$40-50$$
Explanation
We will form table from above histogram:
$$CI$$
$$Frequency$$
$$0-10$$
$$15$$
$$10-20$$
$$10$$
$$20-30$$
$$20$$
$$30-40$$
$$25$$
$$40-50$$
$$10$$
$$50-60$$
$$5$$
$$\Rightarrow$$ From above table table we can see, $$30-40$$ has frequency $$25$$ which is maximum.
$$\therefore$$ The class-interval has the maximum frequency is $$30-40$$
How many students watched TV for less than 4 hours?
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0%
$$34$$
0%
$$32$$
0%
$$24$$
0%
$$30$$
Explanation
From the given diagram, we have
$$4$$ students watch television for $$1-2$$ hours
$$8$$ students watch television for $$2 - 3$$ hours
$$22$$ students watch television for $$3-4$$ hours
Thus, in total $$4+8+22 = 34$$ students watch television for less than $$4$$ hours
If the class intervals in a frequency distribution are $$(72 - 73.9), (74 - 75.9), (76 - 77.9), (78 - 79.9)$$ etc., then the mid-point of the class $$(74 - 75.9)$$ is
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$$74.50$$
0%
$$74.90$$
0%
$$74.95$$
0%
$$75.00$$
Explanation
Midpoint $$= \cfrac{\text{lower limit} + \text{upper limit}}{2}$$
$$= \cfrac{74 +75.9}{2} = 74.95$$
How many class-intervals have equal frequency ?
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$$2$$
0%
$$1$$
0%
$$3$$
0%
None
Explanation
$$\Rightarrow$$ In above histogram we can see, class intervals $$10-20$$ and $$40-50$$ have same frequency which is $$10$$.
$$\therefore$$ The class intervals have equal frequency are $$2$$.
If the upper and the lower limits of a class interval are $$16$$ and $$10$$, the class-interval is?
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0%
$$10-16$$
0%
$$16-10$$
0%
$$10-16$$ or $$16-10$$
0%
None of these
Explanation
The upper limit of a class interval is the maximum value of that interval and the lower limit of a class interval is the minimum value of that interval.
Given that upper limit is $$16$$ and lower limit is $$10$$.
Therefore the class interval is $$10-16$$.
The upper limit of $$5 -12.5$$ is?
Report Question
0%
$$5$$
0%
$$12.5$$
0%
$$5$$ or $$12.5$$
0%
None of these
Explanation
The upper limit of the class interval is the maximum value of that interval.
Therefore, the upper limit of $$5-12.5$$ is $$12.5$$.
Which class-interval has the minimum frequency ?
Report Question
0%
$$20-30$$
0%
$$40-50$$
0%
$$50-60$$
0%
$$10-20$$
Explanation
We will form a table from the above histogram:
Class Interval
Frequency
$$0-10$$
$$15$$
$$10-20$$
$$10$$
$$20-30$$
$$20$$
$$30-40$$
$$25$$
$$40-50$$
$$10$$
$$50-60$$
$$5$$
$$\Rightarrow$$ From above table table we can see, $$50-60$$ has frequency $$5$$ which is minimum.
$$\therefore$$ The class-interval that has the minimum frequency is $$50-60$$
A histrogram consists of
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0%
sectors
0%
rectangles
0%
triangle
0%
squares
Explanation
A histogram is an area diagram. It can be defined as a set of
rectangles
with bases along with the intervals between class boundaries and with areas proportional to frequencies in the corresponding classes.
If $$25$$ is the arithmetic mean of $$ X$$ and $$46,$$ then find $$X.$$
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0%
$$2$$
0%
$$4$$
0%
$$8$$
0%
$$16$$
Explanation
Since $$25$$ is the arithmetic mean of $$X$$ and $$46$$,
$$\therefore 25=\displaystyle \frac{(X+46)}{2}$$
$$\therefore X+46=50$$
$$\therefore X=4$$
If the range of $$14, 12, 17, 18, 16, x$$ is $$20$$ and $$x>0$$, the value of $$x$$ is:
Report Question
0%
$$2$$
0%
$$28$$
0%
$$32$$
0%
cannot be determined
Explanation
The range is the difference between the smallest value and the largest value of the data set.
Given data set is $$14, 12, 17,18,16,x$$ and t
he range is given as $$20$$.
In the given set the smallest value is $$12$$ and the largest value is $$18$$.
Therefore the range is $$18-12=6\neq20$$
Hence, $$18$$ is not the largest value. Variable $$x$$ is the largest value.
Therefore, range is
$$x-12=20$$
$$\implies x=20+12=32$$
Following are the marks obtained by 30 students in an examination :
15
16
44
27
7
37
20
17
47
25
11
47
8
20
38
28
21
23
9
24
36
30
31
20
10
30
40
19
41
17
Taking class intervals 0-10, 10-20, ..................., 40-50, construct a frequency table.
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0%
C.I.
Tally Marks.
Frequency
0-10
10-20
20-30
30-40
40-50
|||
$${||||}\! \! \! \! \! \! {\diagdown}\:\: ||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: ||||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: |$$
$${||||}\! \! \! \! \! \! {\diagdown}$$
3
7
9
6
5
0%
C.I.
Tally Marks.
Frequency
0-10
10-20
20-30
30-40
40-50
|||
$${||||}\! \! \! \! \! \! {\diagdown}\:\: ||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: ||||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: |$$
$${||||}\! \! \! \! \! \! {\diagdown}$$
9
7
3
6
7
0%
C.I.
Tally Marks.
Frequency
0-10
10-20
20-30
30-40
40-50
|||
$${||||}\! \! \! \! \! \! {\diagdown}\:\: ||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: ||||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: |$$
$${||||}\! \! \! \! \! \! {\diagdown}$$
5
7
3
6
9
0%
None of these
Explanation
Create class intervals $$0 - 10$$, $$11 - 20$$, $$21- 30$$ an so on.and count the frequency in each interval:
C.I.
Tally Marks.
Frequency
0-10
10-20
20-30
30-40
40-50
|||
$${||||}\! \! \! \! \! \! {\diagdown}\:\: ||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: ||||$$
$${||||}\! \! \! \! \! \! {\diagdown}\: |$$
$${||||}\! \! \! \! \! \! {\diagdown}$$
3
7
9
6
5
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