MCQ Questions for Class 9 Maths Statistics Quiz 3

How many students have heights between

$180 - 200$ ?

0%
$10$
0%
$24$
0%
$15$
0%
$7$

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,

$10$ is the number of student's height between the interval

$180-200$ .

Hence, option

$A$ is correct.

Find the class interval with lowest frequency.

0%
$318-335$
0%
$354-371$
0%
$372-389$
0%
$426-443$

Explanation

In the given data, we observe the lowest frequency

$=1$

So, the class interval for the lowest frequency

$1$ is

$426-443.$

Fill in the blank

$:$

$___________$ frequency is defined as a running total of frequencies.

0%
Variable
0%
Accelerate
0%
Data
0%
Cumulative

The following frequency distribution gives the monthly consumption of electricity of

$68$ consumers of a locality. In which class interval, the number of consumers are the highest?

0%
$105-125$
0%
$125-145$
0%
$145-165$
0%
$185-205$

Explanation

The highest number of consumers are at the class interval

$125-145.$
So, the highest frequency is

$20$ at the interval

$125-145.$

How many ponds had

$20-39$ ducks?

0%
$25$
0%
$30$
0%
$20$
0%
$15$

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,

$20$ ponds had

$20-39$ number of ducks.

Hence, option

$C$ is correct.

12	24	36	48	60	72
24	48	60	72	50	30
50	45	50	75	50	34

The number of students in a quiz competition scores are in the table. How many students scored

$50$ marks?

0%
$4$
0%
$5$
0%
$8$
0%
$3$

Explanation

Scores	Frequency
12	1
24	2
30	1
34	1
36	1
45	1
48	2
50	4
60	2
72	2
75	1

So, there are

$4$ students scored

$50$ marks.

The annual maintenance cost of a machine in a factory over a seven years period is represented in the histogram.
In which year the maintenance is very low?

0%
$1995-1996$
0%
$1996-1997$
0%
$1997-1998$
0%
$1998-1999$

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 clearly, the maintenance is very low in the year

$1997-1998$ .

Hence, option

$C$ is correct.

In which hour interval, the children are

$55$ ?

0%
$0-10$
0%
$10-20$
0%
$20-30$
0%
$30-40$

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,

$55$ children entered into the library between the interval

$10-20$ .

Hence, option

$B$ is correct.

Which is the lowest ticket collected by Judah?

0%
$5$
0%
$11-13$
0%
$14-16$
0%
$10$

Explanation

The lowest height of the bar represents the lowest ticket collected by Judah i.e.

$5$ .

Hence, the answer is

$5$ .

Find the total number of tickets.

0%
$60$
0%
$80$
0%
$90$
0%
$100$

Explanation

The total number of tickets

$= 25 + 15 + 10 + 5 + 20 + 25 = 100$

Rahul made a frequency table of height for his students. In which class interval, the number of students is less?

0%
$5-10$
0%
$10-15$
0%
$15-20$
0%
$35-40$

How many tickets were collected in the interval

$11-13$ ?

0%
$25$
0%
$15$
0%
$10$
0%
$5$

Explanation

Since the height of the bar in the interval

$11-13$ is

$5$ .

So,

$5$ tickets were collected in the interval

$11-13$ .

Hence, option D is correct.

Find the number of children in the interval

$20-30$ hours?

0%
$10$
0%
$15$
0%
$25$
0%
$45$

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

$= 10$ .

The number of children entered into the library in

$20-30$ hours

$= 15$ .

Hence, option

$B$ is correct.

The annual maintenance cost of a machine in a factory over a seven years period is represented in the histogram.
In which year the maintenance cost is

$2500$ ?

0%
$1998-1999$
0%
$2000-2001$
0%
$2001-2002$
0%
$1995-1996$

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 is Rs.

$2500$ in the year

$1998-1999$ .

Hence, option

$A$ is correct.

In which frequency the number of tickets were same?

0%
$2-4$ and $17-19$
0%
$5-7$ and $8-10$
0%
$8-10$ and $11-13$
0%
$14-16$ and $17-19$

Explanation

The number of tickets are same in the frequency

$2-4$ and

$17-19$ .

Judah recorded the number of football tickets collected by his friends, on the histogram.

What is the total number of tickets collected by Judah between the interval

$8-10$ ?

0%
$5$
0%
$15$
0%
$10$
0%
$40$

Explanation

The total number of tickets collected by Judah between the interval

$8-10 = 10$

From the graph, find the number of tickets collected between the interval

$14-16$ .

0%
$25$
0%
$20$
0%
$15$
0%
$10$

Explanation

From the graph, the number of tickets collected between the interval

$14-16$ is

$20$ .

How many music directors required to create

$(9-10)$ song?

0%
$8$
0%
$5$
0%
$4$
0%
$6$

Explanation

By looking the graph, we get

$8$ music directors required to create

$(9-10)$ song

In the graph you can find the takings of a chocolate bar for

$8$ months. Answer the following question:

In which two months,

$40000$ chocolate bars are taken out?

0%
Jan and Feb
0%
Feb and May
0%
May and Apr
0%
Jun and Jul

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

$=60000$ .

Clearly, we can say Feb and May months shows the same number of chocolates, i.e.

$40000$ chocolates.
So, in the months of Feb and May,

$40000$ chocolate bars are taken out.

Hence, option

$B$ is correct.

Calculate the number of songs made during the interval

$9-10$ .

0%
$14$
0%
$6$
0%
$8$
0%
$10$

Explanation

By looking the graph, we get the number of songs made during the interval

$9-10$ is

$8$ .

In which interval the number of music directors made more songs?

0%
$1-2$
0%
$3-4$
0%
$5-6$
0%
$7-8$

How many songs were made between the interval

$3-4$ ?

0%
$20$
0%
$18$
0%
$6$
0%
$14$

Explanation

By looking the graph, we get the number of songs made between the interval

$3-4$ is

$18$ .

Upto

$6$ songs are totally made by how many music directors?

0%
$20$
0%
$18$
0%
$10$
0%
$50$

Explanation

$6$ songs made by total number of music directors

$= 20 + 18 + 12 = 50$

How many students got more than

$70$ marks?

0%
$16$
0%
$8$
0%
$47$
0%
$18$

Explanation

From the given frequency polygon, we get the students who got more than

$70$ marks

$= 16 + 18 + 13 = 47$ .

Hence, option

$C$ is correct.

Mean of

$40$ observations was given as

$160$ .It was detected that

$125$ was misread as

$160$ . Find the correct mean.

0%
$158$
0%
$169$
0%
$159$
0%
$161$

Explanation

Formula used:

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

Sine, number of observations

$,n=40$
Mean

$=160$ (initial wrong mean)
Incorrected sum

$=n \times u=40 \times 160=6400$
Now

$125$ was read as

$165$ , so
Corrected sum

$=6400-165+125=6360$

$\therefore$ Corrected mean

$=\dfrac{6360}{40} =159$

Length of the class

$11-20$ is ______.

0%
$9$
0%
$10$
0%
$11$
0%
$20$

Explanation

Length of a class is given by

$b-a$ , where

$b$ and

$a$ are the upper and lower limits of the class respectively.

$\therefore$ length

$=20-11=9$
Option

$A$ is correct.

A factory kept track of the number of broken plates per shipment last month. Which range of broken plates per shipment occurred more frequently?

0%
$10-19$
0%
$40-49$
0%
$30-39$
0%
$20-29$

Explanation

From the histogram, the number of shipment per range of broken plates:

$0-9$ is

$8$

$10-19$ is

$7$

$20-29$ is

$1$

$30-39$ is

$6$

$40-49$ is

$10$

$50-59$ is

$6$

$60-69$ is

$7$

$70-79$ is

$7$ .

The range of broken plates per shipment occurred more frequently

$= 40-49$ , i.e.

$10$ (The longest bar has the most frequency).

Hence, option

$B$ is correct.

Find the number of songs made during the interval

$5-6$ .

0%
$20$
0%
$15$
0%
$6$
0%
$8$

Explanation

The height of the bar represents the number of songs.

So,

$20$ songs were made during the interval

$5-6$ .

Find the average of the expressions

$2x+4, 5x-1$ and

$-x+3$ .

0%
$x+2$
0%
$x-2$
0%
$2x+2$
0%
$2x-2$

Explanation

Average of the expression

$=$

$\dfrac{sum \quad of \quad expression }{total \quad no. \quad of \quad expressions}$

$\Rightarrow \dfrac{2x+4+5x-1-x+3}{3}$

$\Rightarrow \dfrac{6x+6}{3}$

$\Rightarrow \dfrac{3(2x+2)}{3}=2x+2$

$12-n, 12, 12+n$
What is the average (arithmetic mean) of the

$3$ quantities in the list above?

0%
$4$
0%
$12$
0%
$18$
0%
$4+\dfrac{n}{3}$
0%
$12+\dfrac{n}{3}$

Explanation

Given

$3$ Quantities:

$12$

$-$

$n$ ,

$12$ ,

$12$

$+$

$n$

Average of the above

$3$ Quantities

$=$

$\dfrac {\text{Sum of the above}\: 3\: \text{Quantities}} {\text{Number of Quantities}}$

$=$

$\dfrac { 12 \space - \space n \space + \space 12 \space + \space 12 \space + \space n }{3}$

$=$

$\dfrac {12 \times 3}{3}$

$=$

$12$

Therefore, average of the Quantities listed above is

$12$ .

CBSE Questions for Class 9 Maths Statistics Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Statistics Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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