MCQ Questions for 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

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:

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:

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

0%
$18$
0%
$33$
0%
$15$
0%
$51$

Explanation

No. of failed student

$= 2 + 4 +12 = 18$

Frequency of

$48$ is

0%
$15$
0%
$18$
0%
$33$
0%
$51$

No. of students weighing less than

$56$ kg are

0%
$37$
0%
$23$
0%
$46$
0%
$14$

What is the minimum height possible of the tallest student?

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 _____

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?

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?

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:

$2, 3, 4, 12, 11, 2, 4, 3, 15, 15, 16, 2, 3, 20, 20$

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?

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?

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?

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?

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:

$12, 12, 11, 12, 2, 3, 3, 3, 4, 4, 2, 2, 5, 6, 7, 4, 7, 8, 15, 15, 20, 20$

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?

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

$60-80$ ?

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

$100-120$ ?

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?

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, number 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?

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

$2000$ .

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?

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?

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?

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

$1997-1998$ ?

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?

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

$6-20$ ?

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.

1023028_463517_ans_dc43e936045a433fb7295c1fe4b41c28.PNG

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.

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.

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.

CBSE Questions for Class 9 Maths Statistics Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Statistics Quiz 7 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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