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CBSE Questions for Class 9 Maths Statistics Quiz 7 - MCQExams.com
CBSE
Class 9 Maths
Statistics
Quiz 7
In a frequency distance with classes $$0-10, 10-20$$ and so on, the size of class interval is $$10$$. lower limit of fourth class is
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0%
$$40$$
0%
$$50$$
0%
$$20$$
0%
$$30$$
Explanation
Classes - $$0-10, 10-20, 20-30, 30-40$$
Hence, lower limit of fourth class is $$30$$
The class marks of intervals $$20-30$$ and $$10-20$$ are:
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0%
$$25$$ and $$10$$
0%
$$25$$ and $$15$$
0%
$$20.5$$ and $$10.5$$
0%
$$20.5$$ and $$30.5$$
Explanation
Class mark is the average of upper limit and lower limit of the given class interval.
Therefore, class mark of $$20-30$$ interval is $$\dfrac{20+30}{2}=\dfrac{50}2=25$$
Class mark of $$10-20$$ interval is $$\dfrac{10+20}{2}=\dfrac{30}2=15$$
Total number of students in all are:
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0%
$$180$$
0%
$$150$$
0%
$$9$$
0%
$$30$$
Explanation
From the histogram:
Height (in cm)
No. of students
$$150-155 $$
$$5 $$
$$155-160$$
$$1$$
$$160-165$$
$$ 5$$
$$ 165-170$$
$$5$$
$$ 170-175$$
$$ 9$$
$$ 175-180$$
$$5$$
Total no. of students $$=n(150)+n(155)+n(160)+n(165)+n(170)+n(175)$$
$$=5+1+5+5+9+5$$
$$=30$$.
Therefore, there are $$30$$ students in total.
Hence, option $$D$$ is correct.
If the passing marks are $$41$$, the no of students that failed is
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0%
$$18$$
0%
$$33$$
0%
$$15$$
0%
$$51$$
Explanation
No. of failed student $$= 2 + 4 +12 = 18$$
Frequency of $$48$$ is
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0%
$$15$$
0%
$$18$$
0%
$$33$$
0%
$$51$$
Explanation
Observe the table ,since 48 lies in between 40-50 ,we would take this class interval and its frequency is 15.
No. of students weighing less than $$56$$kg are
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0%
$$37$$
0%
$$23$$
0%
$$46$$
0%
$$14$$
Explanation
No. of students weighing less than 56kg = 8 + 15 + 14 = 37 student
What is the minimum height possible of the tallest student?
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0%
$$170$$
0%
$$175$$
0%
$$180$$
0%
$$165$$
Explanation
From the histogram:
Height (in cm)
No. of students
$$150-155 $$
$$5 $$
$$155-160$$
$$1$$
$$160-165$$
$$ 5$$
$$ 165-170$$
$$5$$
$$ 170-175$$
$$ 9$$
$$ 175-180$$
$$5$$
The minimum height of the tallest student belongs to class interval $$175-180.$$
Thus, the minimum height possible is $$175$$ [$$\because$$ In class $$175-180, 175$$ is lower limit and can be included in class $$175-180$$].
Hence, option $$B$$ is correct.
$$1 - 5, 6 - 10, 11 - 15 ,...$$ are the classes of a distribution, the upper boundary of the class $$1 - 5$$ is _____
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0%
$$5$$
0%
$$5.5$$
0%
$$6$$
0%
$$6.5$$
Explanation
As the classes are inclusive form, upper boundary of the class $$ 1 - 5 = 5 + 0.5 = 5.5 $$
How many students are over $$170\:cm$$ tall?
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0%
$$14$$
0%
$$9$$
0%
$$5$$
0%
$$10$$
Explanation
From the histogram:
Height (in cm)
No. of students
$$150-155 $$
$$5 $$
$$155-160$$
$$1$$
$$160-165$$
$$ 5$$
$$ 165-170$$
$$5$$
$$ 170-175$$
$$ 9$$
$$ 175-180$$
$$5$$
Then, number of students whose height is over $$170 cm$$
$$=n(170)+n(175)$$
$$=9+5$$
$$=14$$.
Hence, $$14$$ students are over $$170 cm$$ tall.
Therefore, option $$A$$ is correct.
How many employees get to work in more than $$100$$ minutes?
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0%
$$20$$
0%
$$15$$
0%
$$4$$
0%
$$58$$
Explanation
From the histogram, we can observe the class width of:
$$100-120$$ minutes $$= 4$$ employees.
Therefore, number of employees get into work in more than $$100$$ minutes $$= 4$$.
So, $$4$$ employees get into work more than $$100$$ minutes.
Hence, option $$C$$ is correct.
Convert the following data into grouped frequency table. Find the total frequency:
$$2, 3, 4, 12, 11, 2, 4, 3, 15, 15, 16, 2, 3, 20, 20$$
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0%
$$10$$
0%
$$15$$
0%
$$20$$
0%
$$25$$
Explanation
Arrange the data in ascending order: $$2, 2, 2, 3, 3, 3, 4, 4, 11, 12, 15, 15, 16, 20, 20$$
Range $$=$$ largest number $$-$$ smallest number
Range $$= 20 - 2 = 18$$
Let us form $$4$$ groups from the data, class width $$= \dfrac{\text{range}}{4} = \dfrac{18}{4} = 4.5$$ round off the number, we get $$5$$.
Form a table,
Class interval
Frequency
$$0-5$$
$$8$$
$$6-11$$
$$1$$
$$12-17$$
$$4$$
$$18-23$$
$$2$$
So, the total frequency $$=8+1+4+2=15$$.
In which class interval number of ducks is equal to $$20$$ ponds?
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0%
$$0-19$$ and $$80-99$$
0%
$$19-39$$ and $$80-99$$
0%
$$20-39$$ and $$80-99$$
0%
$$40-59$$ and $$100-119$$
Explanation
From the histogram, the class width of number of ducks:
$$0 - 19 = 30$$ number of ponds.
$$20 - 39 = 20$$ number of ponds.
$$40 - 59 = 25$$ number of ponds.
$$60 - 79 = 10$$ number of ponds.
$$80 - 99 = 20$$ number of ponds.
$$100 - 119 = 15$$ number of ponds.
Therefore, $$20-39$$ and $$80-99$$ is the two class interval equal to $$20$$ ponds.
Hence, option $$C$$ is correct.
How many employees get to work in less than $$20$$ minutes?
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0%
$$4$$
0%
$$6$$
0%
$$10$$
0%
$$15$$
Explanation
From the histogram, we can observe the class width of:
$$0-20$$ minutes $$= 4$$ employees.
Therefore, the number of employees get into work in less than $$20$$ minutes $$= 4$$.
So, $$4$$ employees get into work less than $$20$$ minutes.
Hence, option $$A$$ is correct.
How many employees get to work in less than $$60$$ minutes?
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0%
$$10$$
0%
$$5$$
0%
$$15$$
0%
$$20$$
Explanation
From the histogram, we can observe the class width of:
$$0-20$$ minutes $$= 4$$ employees,
$$20-40$$ minutes $$= 6$$ employees,
$$40-60$$ minutes $$ = 10$$ employees.
Therefore, number of employees get into work in less than $$60$$ minutes $$= 4 + 6 + 10 = 20$$.
So, $$20$$ employees get into work less than $$60$$ minutes.
Hence, option $$D$$ is correct.
How many number of ducks are there in less than $$60$$ ponds?
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0%
$$59$$
0%
$$39$$
0%
$$29$$
0%
$$19$$
Explanation
From the histogram, the class width gives the number of ducks:
$$0 - 19 = 30$$ number of ponds.
$$20 - 39 = 20$$ number of ponds.
$$40 - 59 = 25$$ number of ponds.
$$60 - 79 = 10$$ number of ponds.
$$80 - 99 = 20$$ number of ponds.
$$100 - 119 = 15$$ number of ponds.
Therefore, $$30+20+25=59$$ number of ducks are there in less than $$60$$ ponds.
Hence, option $$A$$ is correct.
Convert the following data into grouped frequency table with class interval length of $$5$$. Find the third interval frequency:
$$12, 12, 11, 12, 2, 3, 3, 3, 4, 4, 2, 2, 5, 6, 7, 4, 7, 8, 15, 15, 20, 20$$
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0%
$$2$$
0%
$$5$$
0%
$$10$$
0%
$$4$$
Explanation
Arrange the data in increasing order: $$2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 7, 7, 8, 11, 12, 12, 12, 15, 15, 20, 20$$
Form the required frequency distribution table with interval of $$5$$
Class interval
Frequency
$$0-5$$
$$10$$
$$6-11$$
$$5$$
$$12-17$$
$$5$$
$$18-23$$
$$2$$
Arrange the data in order: $$2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 7, 7, 8, 11, 12, 12, 12, 15, 15, 20, 20$$
So, the third interval frequency is $$5$$.
How many students have heights less than $$220$$ cm?
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0%
$$40$$
0%
$$63$$
0%
$$48$$
0%
$$53$$
Explanation
From the histogram, the class width of number of students:
$$120 - 140$$ is $$4$$
$$140 - 160$$ is $$10$$
$$160 - 180$$ is $$24$$
$$180 - 200$$ is $$10$$
$$200 - 220$$ is $$15$$
$$220 - 240$$ is $$7$$.
Therefore, the number of students having heights less than $$220$$ cm $$= 15 + 10 + 24 + 10 + 4 = 63$$.
Hence, option $$B$$ is correct.
The histogram shows that the level of diabetic (in mg per dl) of $$105$$ children. How many children have a level of diabetic between $$60-80$$?
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0%
$$20$$
0%
$$15$$
0%
$$25$$
0%
$$35$$
Explanation
By referring the histogram, number of students with diabetic level:
$$20-40$$ is $$5$$ students
$$40-60$$ is $$20$$ students
$$60-80$$ is $$35$$ students
$$80-100$$ is $$15$$ students
$$100-120$$ is $$10$$ students
$$120-140$$ is $$20$$ students.
Therefore, $$35$$ children have a level of diabetic between $$60-80$$.
Hence, option $$D$$ is correct.
The histogram shows that the level of diabetic (in mg per dl) of $$105$$ children. How many children have a level of diabetic between $$100-120$$?
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0%
$$15$$
0%
$$10$$
0%
$$20$$
0%
$$35$$
Explanation
By referring the histogram, number of students with diabetic level:
$$20-40$$ is $$5$$ students
$$40-60$$ is $$20$$ students
$$60-80$$ is $$35$$ students
$$80-100$$ is $$15$$ students
$$100-120$$ is $$10$$ students
$$120-140$$ is $$20$$ students.
Therefore, $$10$$ children have a level of diabetic between $$100-120$$.
Hence, option $$B$$ is correct.
$$70$$ number of student's height are measured in cm as shown in the histogram. How many students have heights more than $$180$$ cm?
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0%
$$40$$
0%
$$63$$
0%
$$38$$
0%
$$32$$
Explanation
The given histogram represents the height of $$70$$ students in $$cm$$.
For finding the number of students having a height of more than $$180\ cm$$, we need to add the heights of the bars representing the bars $$180-200,\ 200-220,\ 220-240$$
The height of the bar, representing the number of students is:
$$180-200\rightarrow 10$$
$$200-220\rightarrow 15$$
$$220-240\rightarrow 7$$
Therefore, n
umber of students having height more than $$180\ cm= 10 + 15 + 7$$
$$ = 32$$
Hence, $$32$$ students have height more than $$180\ cm$$.
How many students have heights less than $$180$$ cm?
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0%
$$40$$
0%
$$63$$
0%
$$38$$
0%
$$53$$
Explanation
From the histogram, the frequency associated with the class width is as follows:
$$120 - 140$$ is $$4$$
$$140 - 160$$ is $$10$$
$$160 - 180$$ is $$24$$
$$180 - 200$$ is $$10$$
$$200 - 220$$ is $$15$$
$$220 - 240$$ is $$7$$.
Therefore, number of students who have heights less than $$180$$ cm $$= 24 + 10 + 4 = 38$$.
Hence, option $$C$$ is correct.
Find the decreased maintenance cost in the year $$2000-2001$$ when compared to $$1999$$ to $$2000$$.
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0%
$$100$$
0%
$$120$$
0%
$$1500$$
0%
$$150$$
Explanation
From the histogram, the maintenance cost in the year:
$$1995-1996$$ is Rs. $$2000$$
$$1996-1997$$ is Rs. $$3000$$
$$1997-1998$$ is Rs. $$1500$$
$$1998-1999$$ is Rs. $$2500$$
$$1999-2000$$ is Rs. $$4850$$
$$2000-2001$$ is Rs. $$5000$$
$$2001-2002$$ is Rs. $$7200$$.
Then, the decreased maintenance cost in the year $$2000-2001$$ when compared to $$1999 - 2000 = 5000 - 4850 = 150$$ rupees.
Hence, option $$D$$ is correct.
How many total number of children entered into the library between $$0$$ to $$20$$ hours?
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0%
$$45$$
0%
$$55$$
0%
$$100$$
0%
$$85$$
Explanation
From the histogram,
The number of children entered during $$0-10$$ hours $$= 45$$
The number of children entered during $$10-20$$ hours $$= 55$$
The number of children entered during $$20-30$$ hours $$= 15$$
The number of children entered during $$30-40$$ hours $$= 25$$
The number of children entered during $$40-50$$ hours $$= 85$$
The number of children entered during $$50-60$$ hours $$= 10$$.
Therefore, the total number of children during $$0-20$$ hours $$= 45 + 55 = 100$$.
Hence, option $$C$$ is correct.
What is the total number of children entered in to the library between $$0-30$$ hours?
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0%
$$45$$
0%
$$55$$
0%
$$100$$
0%
$$115$$
Explanation
From the histogram,
The number of children entered during $$0-10$$ hours $$= 45$$
The number of children entered during $$10-20$$ hours $$= 55$$
The number of children entered during $$20-30$$ hours $$= 15$$
The number of children entered during $$30-40$$ hours $$= 25$$
The number of children entered during $$40-50$$ hours $$= 85$$
The number of children entered during $$50-60$$ hours $$= 10$$.
Therefore,
the total number of children during $$0-30$$ hours $$= 45 + 55 + 15 = 115$$
.
Hence, option $$D$$ is correct.
Which one of the following graph is an example of frequency polygon?
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0%
0%
0%
0%
Explanation
We know, the frequency polygon is formed by joining the mid-points of each upper sides of adjacent rectangles of the histogram by line segments.
Then clearly, the option $$B$$ is an example of frequency polygon.
How much did the maintenance cost decreased in $$1996-1997$$ when compared to $$1997-1998$$?
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0%
$$1200$$
0%
$$1500$$
0%
$$2500$$
0%
$$4500$$
Explanation
From the histogram, the maintenance cost in the year:
$$1995-1996$$ is Rs. $$2000$$
$$1996-1997$$ is Rs. $$3000$$
$$1997-1998$$ is Rs. $$1500$$
$$1998-1999$$ is Rs. $$2500$$
$$1999-2000$$ is Rs. $$4850$$
$$2000-2001$$ is Rs. $$5000$$
$$2001-2002$$ is Rs. $$7200$$.
Then, the maintenance cost decreased in $$1996-1997$$ when compared to $$1997-1998 = 3000 - 1500 = 1500$$
rupees
.
Hence, option $$B$$ is correct.
In the graph you can find the takings of a chocolate bar for $$8$$ months.
Answer the following question:
What is the average chocolate bars taken out from the month January to May?
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0%
$$23000$$
0%
$$35000$$
0%
$$50000$$
0%
$$42000$$
Explanation
From the given frequency polygon,
Number of chocolates taken in Jan $$=50000$$
Number of chocolates taken in Feb $$=40000$$
Number of chocolates taken in Mar $$=35000$$
Number of chocolates taken in Apr $$=45000$$
Number of chocolates taken in May $$=40000$$
Number of chocolates taken in June $$=50000$$
Number of chocolates taken in Jul $$=60000$$
Number of chocolates taken in Aug $$=60000$$
.
Then, Average of chocolate bars taken out $$=$$ $$\dfrac{50000 + 40000+35000+45000+40000}{5}$$
$$=$$ $$\dfrac{210000}{5}$$
$$= 42000$$.
Therefore, $$42000$$ chocolate bars are taken out from the month January to May.
Hence, option $$D$$ is correct.
The architecture student counted the number of bricks in each building in his neighbourhood. How many buildings are there in the interval $$6-20$$?
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0%
$$2$$
0%
$$6$$
0%
$$10$$
0%
$$14$$
Explanation
Number of buildings in interval $$6-10=2$$
Number of buildings in interval $$11-15=2$$
Number of buildings in interval $$16-20=6$$.
Therefore, the total number of buildings in $$6-20=2+2+6$$
$$=10$$.
Hence, option $$C$$ is correct.
The number of hours each student of a class spends studying is $$4, 6, 5, 8, 9, 5, 7, 4$$, and $$8$$ hours. Identify the frequency polygon for the given data.
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0%
figure $$1$$
0%
figure $$3$$
0%
figure $$4$$
0%
figure $$2$$
Use the histogram and find the sum of the number of students studying for $$4 - 5$$ and $$8 - 9$$ hours.
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0%
5
0%
8
0%
4
0%
2
Explanation
From the histogram,
The number of students studying for $$4 - 5$$ $$= 5 $$.
The number of students studying for $$6 - 7$$ $$= 4 $$.
The number of students studying for $$8 - 9$$ $$= 3 $$.
Therefore, the sum of the number of students studying for $$4 - 5$$ and $$8 - 9$$ hours $$= 5 + 3 = 8$$.
Hence, option $$B$$ is correct.
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