MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 6

Find the standard deviation of

$210, 240, 250, 260, 220, 230$ and

$270$ .

0%
$18$
0%
$20$
0%
$22$
0%
$25$

Explanation

Mean of the given numbers is

$\bar x =\dfrac{210+240+250+260+220+230+270}{7}=\dfrac{1680}{7}=240$

total number of values

$n=7$

Standard deviation is given by

$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$

$\implies SD=\sqrt{\dfrac{(210-240)^2+(240-240)^2+.....+(270-240)^2}{7}}$

$\implies SD=\sqrt{\dfrac{900+0+100+400+400+100+900}{7}}$

$\implies SD=\sqrt{\dfrac{2800}{7}}$

$\implies SD=\sqrt{400}=20$

Therefore the standard deviation for the given values is

$20$

The ages (in years) of a family of 6 members are 1, 5, 12, 15, 38 andThe standard deviation is found to be 15.9. After 10 years the standard deviation is

0%
increased
0%
decreased
0%
remains same
0%
none of these

Explanation

Then mean of six members=

$\frac{1+5+12+15+38+40}{6}=\frac{111}{6}=18.5$

Then Variance

$\alpha ^{2}=\frac{(18.5-1)^{2}+(18.5-5)^{2}+(18.5-12)^{2}+(18.5-15)^{2}+(18.5-38)^{2}+(18.5-40)^{2}}{5}$

$=\frac{(17.5)^{2}+(13.5)^{2}+(6.5)^{2}+(3.5)^{2}+(-19.5)^{2}+(21.5)^{2}}{6}$

$=\frac{306.25+182.25+42.25+12.25+380.25++462.25}{6}=\frac{1385.50}{6}=230.91$

So standard deviation

$\alpha =15.19$

Given after 10 year The standard deviation is 15.9

Then standard deviation remain same

Find the quartile deviation of

$4, 6, 9, 12, 18, 20, 23, 27, 34, 48$ and

$53$

0%
$12.5$
0%
$13$
0%
$12$
0%
$11.5$

Explanation

Given series of numbers is already arranged in ascending order.
Thus median is the centre value

$= 20$
First Quartile

${Q}_{1}$ will be the mid value of the observations to the left of

$20$ which is

$9$ .
Similarly, Third Quartile

${Q}_{3}$ will be the mid value of the observations to the right of

$20$ which is

$34$ .
And Quartile deviation will be

$\dfrac {{Q}_{3} - {Q}_{1}}{2} = \dfrac {34-9}{2} = \dfrac {25}{2} = 12.5$ .

The line plot below shows how students scored on last week's maths test.
How many students scored 95 or higher on the test?

0%
5 students
0%
7 students
0%
12 students
0%
16 students

Explanation

It is given that

$1$ star

$= 1$ student.

Number of students who scored

$95$ marks or more than

$95$ = number of students who scored

$95$ marks + number students who scored

$100$ marks

Students who scored

$95$ marks

$=7$

Students who scored

$100$ marks

$=5$

Total students

$=7+5=12$

The standard deviation of

$3x + 5, 3x + 7, 3x + 9, 3x + 11, 3x + 13, 3x + 15$ and

$3x + 17$ is ...........

0%
$2$
0%
$3$
0%
$4$
0%
$1$

Explanation

Mean of the given numbers is

$\bar x =\dfrac{(3x+5)+(3x+7)+(3x+9)+(3x+11)+(3x+13)+(3x+15)+(3x+17)}{7}=\dfrac{21x+77}{7}=3x+11$

total number of values

$n=7$

Standard deviation is given by

$SD=\sqrt{\dfrac{\sum_{i=1}^n (x_i-\bar x)^2}{n}}$

$\implies SD=\sqrt{\dfrac{(5-11)^2+(7-11)^2+(9-11)^2+(11-11)^2+(13-11)^2+(15-11)^2+(17-11)^2}{7}}$

$\implies SD=\sqrt{\dfrac{36+16+4+0+4+16+36}{7}}$

$\implies SD=\sqrt{\dfrac{112}{7}}$

$\implies SD=\sqrt{16}=4$

Therefore the standard deviation for the given values is

$4$

What is the total number of persons who made the journey during the week?
(Assume that one represents 10 persons)

0%
200
0%
400
0%
600
0%
900

Explanation

Given that one symbol represents 10 persons and the total number of symbols in picture =

$9+4+6+4+8+6+2= 40$

Total number of persons who made the journey during the week =

$\displaystyle 10 \times 40 = 400$

If the lower and upper quartiles of a data are 4 and
12 then the quartile deviation is_____

0%
2
0%
4
0%
6
0%
8

Explanation

Lower Quartile

${Q}_{1} = 4$

Upper Quartile

${Q}_{3} = 12$

And Quartile deviation will be

$\frac {{Q}_{3} - {Q}_{1}}{2} = \frac {12-4}{2} = \frac {8}{2} = 4$

If the lower and upper quartiles of a data are

$4$ and

$12$ , then the quartile deviation is ____

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

Quartile deviation is the half of the inter quartile deviation.

Inter quartile deviation is the difference between upper quartile and lower quartile.

Given that, lower quartile is

$4$ and upper quartile is

$12$

Therefore, Quartile deivation is

$\dfrac{\text{upper quartile - lower quartile}}{2}$

$\implies \text{quartile deviation} =\dfrac{12-4}2=\dfrac 82 =4$

Read the graph and answer the following: The horizontal line in the graph is called:

0%
X-axis
0%
Y-axis
0%
line
0%
none

Read the graph and answer the following: The vertical line in the graph is called:

0%
Y-axis
0%
X-axis
0%
ray
0%
none of the above

How many school teachers does simile represent?

0%
$500$
0%
$600$
0%
$700$
0%
$800$

Explanation

In the table Hyderabad town has

$4200$ teachers which represented by

$7$ similes in the pictograph.

Therefore, one simile represent

$\dfrac{4200}{7} = 600$ school teachers.

How many school teachers does each smile represent?

0%
500
0%
400
0%
700
0%
none of these

Explanation

In the table Hyderabad town has

$4200$ teachers which represented by

$7$ similes in the pictograph.

Therefore, one simile represent

$\dfrac{4200}{7} = 600$ school teachers.

If the median of the data 6,7,x-2,x,18,21 written in ascending order is 16, then the variance of that data is

0%
$30\dfrac{1}{5}$
0%
$31\dfrac{1}{3}$
0%
$32\dfrac{1}{2}$
0%
$33\dfrac{1}{3}$

Explanation

$M=\dfrac{x-2+x}{2}=16$

Variance =

$\dfrac{1}{n}\sum (x_i-\overline{x})^{2}$

Required variance=

$31\dfrac{1}{3}$

Find the mean for the following data using step deviation method.

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

Answer:- By shortcut Method

Class interval width (w) = 2

$X_i$	$F_i$	$d=\cfrac{X-A}{i}$	$F_i d_i$
2	10	-1	-10
4=A	20	0	0
6	30	1	30
8	40	2	80
	$\Sigma f = 100$		$\Sigma fd = 100$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times w=4+\cfrac{100}{100} \times 2 ={4+2}=6$

C)6

From the bar graph What is total no. of students in class VIII?

0%
$45$
0%
$185$
0%
$215$
0%
$245$

Explanation

As seen in the graph, we can say the following things.

The total number of students in class VIII are

$40+48+52+45+30=215$ .

Hence, option C is correct.

Choose the formula used for arithmetic mean of grouped data by step deviation method is

0%
$\bar { x } =A-\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$
0%
$\bar { x } =A\times \dfrac { \sum { fd\prime } }{ \sum { f } } \times i$
0%
$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$
0%
$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } +i$

Explanation

The formula used for arithmetic mean of groped data by step deviation method is

$\bar { x } =A+\dfrac { \sum { fd\prime } }{ \sum { f } } \times i$

$A =$ Assumed mean of the given data

$\sum { f } =$ Summation of the frequencies given in the grouped data

$\sum { fd\prime } =$ Summation of the frequencies and deviation of a given mean data

$d\prime =\dfrac { x-A }{ i }$

$i =$ Class interval width

$\bar { x } =$ Arithmetic mean

From the given bar graph In which year the difference of the values of export and import is maximum?

0%
$1986 - 87$
0%
$1982 - 83$
0%
$1984 - 85$
0%
$1983 - 84$

Explanation

We can see the graph of years versus Export and Import.

The difference of the values of export and import is maximum in the year

$1986-87$ .

From the given bar graph. In which year the export is minimum

0%
$1984 - 85$
0%
$1986 - 87$
0%
$1983 - 84$
0%
$1982 - 83$

Explanation

The graph of years versus Export and Import in shown on

$x$ and

$y$ axis respectively.

As seen in the graph, we can say in year

$1982-83$ , the export is minimum.

From the bar graph, answer the following question What is the total no. of students in section

$A, B$ and

$C$ `.

0%
$40$
0%
$140$
0%
$52$
0%
$215$

Explanation

As seen from the graph, the total number of students in sections

$A,B$ and

$C$ are

$A+B+C$

$=40+48+52=140$

Hence, option B is correct.

Study the bar graph and answer the question given below:

What was the percentage decline in the production of oil from

$2007$ to

$2008$ ?

0%
$30\%$
0%
$33\%$
0%
$33.33\%$
0%
$36.3\%$

Explanation

Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage

$= \dfrac{20 - 30}{30}\times 100\%$

$= -33.33%$
So there was decline of

$33.33\%$ from the year

$2007$ and

$2008$

Study the bar graph and answer the question given below:

In how many of the given years was the production of oil more than average production of the given years?

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Average production (in

$1000$ tonnes) over the given years will be

$\cfrac{1}{7}[15 + 30 + 20 + 40 + 50 + 30 + 60]$
=

$\cfrac{245}{7} = 35$
So, the production in the years

$2009, 2010, 2012$ are more than the average production.

Study the above graph and calculate the difference between the number of cars manufactured by company

$Y$ in

$2001$ and

$2000$ ?

0%
$20000$
0%
$30000$
0%
$40000$
0%
$50000$

Explanation

Number of cars manufactured by company

$Y$ in

$2001$

$=80000$

Number of cars manufactured by company

$Y$ in

$2000$

$= 100000$

The difference is

$100000-80000 = 20000$

Find the mean for the following data using step deviation method.

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

Answer:- By shortcut Method

Class interval width (w) = 3

X	F	$d=\cfrac{X-A}{i}$	Fd
3	5	-1	-5
6=A	10	-0	-0
9	15	1	15
12	20	2	40
	$\Sigma f = 50$		$\Sigma fd = 50$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times i=6+\left(\cfrac{50}{50} \times 3\right) = 6+3 = 9$

A) 9

Find the mean for the following data using step deviation method.

0%
$43$
0%
$44$
0%
$45$
0%
$46$

Explanation

Answer:- By shortcut Method

Class interval width (i)= 24-12 = 12

X	F	$d=\cfrac{x-A}{i}$	Fd
12=A	1	0	0
24	2	1	2
36	3	2	6
48	4	3	12
60	5	4	20
	$\Sigma f = 15$		$\Sigma fd = 40$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times i=12+\left(\cfrac{40}{15} \times 12\right)={12+32}=44$

B)44

Study the above graph and calculate the difference between the number of cars manufactured by Company

$X$ in

$2001$ and

$2002$ ?

0%
$20000$
0%
$30000$
0%
$40000$
0%
$50000$

Explanation

Number of cars manufacturd by

$X$ in

$2001 = 120000$

Number of cars manufactured by

$X$ in

$2002=80000$

The difference is

$120000-80000=40000$

If the mean of the data :

$7,8,9,7,8,7 , \lambda , 8$ is

$8 ,$ then the variance of this data is :

0%
$\dfrac { 7 } { 8 }$
0%
$1$
0%
$\dfrac { 9 } { 8 }$
0%
$2$

Study the bar graph and answer the question given below:

What was the percent increase in production of oil in

$2006$ compared to that in

$2012$ ?

0%
$100\%$
0%
$200\%$
0%
$300\%$
0%
$400\%$

Explanation

Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage

$= \dfrac{60 - 15}{15}\times 100\%$

$= 300%$
So there was an increase of

$300\%$ from the year

$2006$ and

$2012$

Study the bar graph and answer the question given below:

The average production of

$2007$ and

$2009$ was exactly to the average production of which of the following pair of the years?

0%
$2007$ and $2006$
0%
$2007$ and $2008$
0%
$2009$ and $2011$
0%
$2011$ and $2012$

Explanation

Average production of

$2007$ and

$2009$ (in

$1000$ tonnes) was

$\dfrac{30 + 40}{2}= 35$
Now for other given years it was
Option

$A$ for

$2007$ and

$2006 = \cfrac{30 + 15}{2}= 22.5$
Option

$B$ for

$2007$ and

$2008 = \cfrac{30 + 20}{2}= 25$
Option

$C$ for

$2009$ and

$2011 = \cfrac{40 + 30}{2}= 35$
Option

$A$ for

$2011$ and

$2012 = \cfrac{30 + 60}{2}= 45$
So it is equal to the year

$2009$ and

$2011$

Study the bar graph and answer the question given below:

What was the percentage increase in the production of oil from

$2009$ to

$2010$ ?

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

Explanation

Percentage we can get easily by getting the difference between two and then by finding that difference is what percent of the initial value.
So Percentage

$= \dfrac{50- 40}{40}\times 100\%$

$= 25\%$
So there was increase of

$25\%$ from the year

$2009$ and

$2010$

Find the mean for the following data using step deviation method.

0%
$15.833$
0%
$16.833$
0%
$17.833$
0%
$18.833$

Explanation

Answer:- By shortcut Method

Class interval width (i) =

$5-0 = 5$

X	C. I. mid point	F	$d=\cfrac{X-A}{i}$	Fd
0-5	2.5	4	-1	-4
5-10	7.5=A	8	0	0
10-15	12.5	12	1	12
15-20	17.5	16	2	32
20-25	22.5	20	3	60
		$\Sigma F = 60$		$\Sigma Fd=100$

Mean =

$A+\cfrac{\Sigma fd}{\Sigma f} \times i=7.5+\left(\cfrac{100}{60} \times 5\right)={7.5+8.33}=15.83$

$15.83$

If the interest on loans amounted to Rs.

$2.5$ lakhs then the total amount of expenditure on advertisement, transport and infrastructure is

$5.625$ lakhs.

0%
True
0%
False

Explanation

Let the total amount of expenditure on advertisement, transport and infrastructure is

$x$ .

Given, the expenditure on loans= 2.5 lakhs

So the ratio of total expenditure on advertisement, transport and infrastructure to the expenditure on loans

$=$

$\dfrac{x}{2.5}$

Also the ratio of total expenditure on advertisement, transport and infrastructure to the expenditure on loans according to the bar graph

$=$

$\dfrac{25+10+10}{20}$

$=$

$\dfrac{45}{20}$

$\therefore$

$\dfrac{x}{2.5}$ =

$\dfrac{45}{20}$

$\Rightarrow x=\dfrac{2.5\times 45}{20}=5.625$ lakhs

So, the total amount of expenditure on advertisement, transport and infrastructure is

$= 5.625$ lakhs

The ratio of the total expenditure on R & D and transport to the total expenditure on infrastructure and salary is

$17:30$

0%
True
0%
False

Explanation

Let the total amount of expenditures be

$x$ .
Then, the total expenditure on R & D and transport:

$=$

$\dfrac{24+10}{10+14+24+10+25+20+40}\times x$

$= \dfrac{34 x}{143}$

And total expenditure on infrastructure and salary:

$=$

$\dfrac{10+40}{10+14+24+10+25+20+40}\times x$

$= \dfrac{50 x}{143}$

Required ratio

$=$

$\dfrac{\dfrac{34x}{1143}}{\dfrac{50x}{143}}$

$=$

$\dfrac{17}{25}$

The total amount of expenditures of the company is

$10.21$ times of expenditure on Taxes

0%
True
0%
False

Explanation

Let the total expenditures be

$x$ .
Then, the expenditure on Taxes

$=$

$\dfrac{14}{10+14+24+10+25+20+40}\times x$

$=$

$\dfrac{14}{143}\times x$
Ratio of the total expenditure to the expenditure on Taxes:

$=$

$\dfrac{x}{x\times 0.0979}$

$= 10.214$
Then, the total expenditure is

$10.21$ times the expenditure of Taxes.

Study the above graph and calculate the average numbers of cars manufactured by Company

$X$ over the given period.

0%
$200{,}000$
0%
$100{,}000$
0%
$300{,}000$
0%
$400{,}000$

Explanation

Average number of cars manufactured by Company

$X$

$=\cfrac{1}{6}[100000 + 80000 + 60000 + 120000 + 80000 + 160000]$

$=\cfrac{1}{6}[600000]$

$= 100,000$

Study the above graph and calculate the average numbers of cars manufactured by Company

$Y$ over the given period.

0%
$96{,}666.66$
0%
$86{,}666.66$
0%
$76{,}666.66$
0%
$66{,}666.66$

Explanation

Average number of cars manufactured by Company

$Y$

$=\cfrac{1}{6}[80000 + 60000 + 100000 + 80000 + 120000 + 140000]$

$= \cfrac{1}{6}[580000]$

$= 96{,}666.66$

for two data sets,each of size 5, the variances are given to be 4 and 5 andthe correspondings means are given to be 2 and 4 respectively. the variances of the combined data set is

0%
11/2
0%
6
0%
13/2
0%
5/2

How many bottles of pickles were exported on Friday and Tuesday altogether?

0%
$4500$
0%
$1500$
0%
$5000$
0%
$5500$

Explanation

$1$ pickle jar

$= 500$ bottles
Friday

$= 6 \times 500 = 3000$
Tuesday

$= 5 \times 500 = 2500$
Altogether

$= 3000 + 2500 = 5500$ .

Subtracting two of the officers earnings, the difference is

$1800$ . The name of officers are

0%
Amit and Bala
0%
Frank and Thomas
0%
Thomas and Amit
0%
Amit and Madesh

Explanation

Given: Subtracting two of the officers earnings, the difference is

$1800$ .
Amit

$= 6$
Thomas

$= 2$
So, the difference

$= 6 - 2 = 4$
1 money bag

$= 450$
Then,

$4 \times 450 = 1800$
Therefore, the two officers are Thomas and Amit.

How much money did Frank earn in a month?

0%
$2450$
0%
$900$
0%
$4050$
0%
$3400$

Explanation

From the figure, we cay say

Frank earnings in a month is

$= 9 \times 450 = 4050$

If the expenditure on advertisement is

$5.5$ lakhs then the difference between the transport and R & D is

$3.8$ lakhs

0%
True
0%
False

Explanation

Let the total expenditures be

$x$ .

Then, the expenditure on advertisement

$=$

$\dfrac{25}{10+14+24+10+25+20+40}\times x$

$=$

$\dfrac{25}{143}\times x$

$= 5.5$ lakhs

$\Rightarrow x=\dfrac{5.5\times 143}{25}=31.46$ lakhs

So, the difference between the expenditures on transport and R&D is,

$= \dfrac{24 - 10}{143}\times 31.46$ lakhs

$= 3.08$ lakhs

His earning is one-half of what Bala earns. Who is he?

0%
Amit
0%
Madesh
0%
Thomas
0%
Frank

Explanation

Bala earnings

$= 4 \times 450 = 1800$
Likewise Thomas earnings

$= 2 \times 450 = 900$
So, Thomas is the officers who earns one-half of what Bala earns.
Therefore, Thomas is the correct officer.

Using the line graph, find the average income of sportsman during the period

$2000-2003$ .

0%
$187.5$ crores
0%
$287.5$ crores
0%
$387.5$ crores
0%
$487.5$ crores

Explanation

Average income os sportsman from the year

$2000-2003 =$

$\dfrac{210 + 440 + 300 + 200}{4}$

$=$

$\dfrac{1150}{4}$

$= 287.5$ crores

The number of washing machines purchased in the past

$35$ weeks is given in a line graph. Find the total number of washing machines purchased in

$20$ weeks.

0%
$140$
0%
$150$
0%
$160$
0%
$170$

Explanation

The total number of washing machines purchased in

$20$ weeks

$= 10 + 20 + 40 + 90 = 160$

How many books were lent in first

$3$ days?

0%
$15$
0%
$5$
0%
$200$
0%
$225$

Explanation

First

$3$ days books were lent

$= 6 + 5 + 4 = 15$

$1$ book

$= 15$ books
So,

$15 \times 15 = 225$

How much more money did Frank earn than Amit?

0%
$1350$
0%
$1200$
0%
$1690$
0%
$1540$

Explanation

Frank earnings

$= 9 \times 450 = 4050$
Amit earnings

$= 6 \times 450 = 2700$
So, the difference

$= 4050 - 2700$

$= 1350$

How many books were checked out on Friday?

0%
$75$ `
0%
$90$
0%
$60$
0%
$6$

Explanation

Books were checked out on Friday

$= 6 \times 15 = 90$ books (

$1$ book

$= 15$ books)

Which day fewest books were checked out?

0%
Friday
0%
Tuesday
0%
Saturday
0%
Wednesday

How many fewer books were checked out on Saturday than Tuesday?

0%
$15$
0%
$30$
0%
$45$
0%
$10$

Explanation

Tuesday

$= 5$

Saturday

$= 3$
So, the difference is

$5 - 3 = 2 \times 15 = 30$ books were checked out on Tuesday than Saturday.

How much less money did Thomas earn than Bala?

0%
$1800$
0%
$900$
0%
$1000$
0%
$1200$

Explanation

Thomas earnings

$= 2 \times 450 = 900$
Bala earnings

$= 4 \times 450 = 1800$
So, the difference

$= 1800 - 900$

$= 900$

In which day, where the number of checkouts are

$75$ books?

0%
Thursday
0%
Friday
0%
Tuesday
0%
Wednesday

Explanation

In figure, there are total

$5$ book on Tuesday.

$\therefore$ on Tuesday the number of checkouts

$= 5 \times 15 = 75$ books.

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 6 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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