Half of the Inter-Quartile Range

MCQ Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 4

Which of the following is true?

0%
Quartile Deviation can generally be calculated for open-ended distributions
0%
It is unduly affected by extreme values
0%
Both A and B
0%
None of these

$1,2,3,4,5,6,7$ are numbers then

${Q}_{1}$ and

${Q}_{2}$ are respectively

0%
$2,4$
0%
$2.6$
0%
$4,6$
0%
$3,5$

Explanation

The given observations are,

$1,2,3,4,5,6,7$

Number of observations

$=7$

$Q_i=\left\{i.\left(\dfrac{N+1}{4}\right)\right\}^{th}term$

$Q_1=\left\{1.\left(\dfrac{7+1}{4}\right)\right\}^{th}term$

$Q_1=2^{nd}term$

$\therefore$

$Q_1=2$

$Q_2=\left\{2.\left(\dfrac{7+1}{4}\right)\right\}^{th}term$

$Q_2=4^{th}term$

$\therefore$

$Q_2=4$

If the median of the data

$6,7,x-2,x,18,21$ written in ascending order is

$16$ , then the variance of the data is

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

Explanation

The given observations are,

$6,7,x-2,x,18,21$

Number of observations

$(n)=6$ [ even ]

$\Rightarrow$

$Median=\dfrac{3^{rd}obs+4^{th}obs}{2}$

$\Rightarrow$

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

$\Rightarrow$

$32=x-2+x$

$\Rightarrow$

$32=2x-2$

$\Rightarrow$

$34=2x$

$\therefore$

$x=17$

$\therefore$ The observations becomes,

$6,7,15,17,18,21$

$\overline{x}=\dfrac{6+7+15+17+18+21}{6}=\dfrac{84}{6}=14$

$Variance(\sigma^2)=\dfrac{(6-14)^2+(7-14)^2+(15-14)^2+(17-14)^2+(18-14)^2+(21-14)^2}{6}$

$=\dfrac{(-8)^2+(-7)^2+(1)^2+(3)^2+(4)^2+(7)^2}{6}$

$=\dfrac{64+49+1+9+16+49}{6}$

$=\dfrac{188}{6}$

$=\dfrac{94}{3}$

$=31\dfrac{1}{3}$

Statistics studies _________ information.

0%
qualitative
0%
quantitative
0%
isolated
0%
nominal

Quartile Deviation =

0%
Half of the Inter-Quartile Range
0%
[ Q(3) - Q(1) ] / 2
0%
Both A and B
0%
Q(3) - Q(1)

Inter Quartile Range =

0%
Difference between the upper and lower quartiles
0%
Q(3) - Q(1)
0%
Both A and B
0%
Q(2)-Q(1)

Days when the forecast temperature the same as the actual temperature are:

0%
Tuesday, Wednesday and Thursday
0%
Tuesday, Friday and Sunday
0%
Thursday, Friday and Saturday
0%
Tuesday, Thursday and Monday

When you are presenting a series of numbers in a business document, you should use a __________.

0%
line graph
0%
table
0%
pie chart
0%
bar chart

Fill the alternative given in the bracket:
Systematic arrangement of data in table and row is called _________.

0%
tabulation
0%
graph
0%
caption
0%
pie chart

Which of the following is true with respect to range?

0%
Higher value of Range implies higher dispersion and vice-versa
0%
Range is unduly affected by extreme values.
0%
It is not based on all the values.
0%
All of these

The arithmetic mean of the observations

$10,8,5,a,b$ is

$6$ and their variance is

$6.8$ , then

$ab=$

0%
$6$
0%
$4$
0%
$3$
0%
$12$

Explanation

$\Rightarrow$ Given observations are

$a,\,b,\,8,\,5,\,10$

According to the question,

$\Rightarrow$

$6.8=\dfrac{(6-a)^2+(6-b)^2+(6-8)^2+(6-5)^2+(6-10)^2}{5}$

$\Rightarrow$

$6.8\times 5=(6-a)^2+(6-b)^2+4+1+16$

$\Rightarrow$

$(6-a)^2+(6-b)^2=34-21$

$\Rightarrow$

$(6-a)^2+(6-b)^2=13$

$\Rightarrow$

$(6-a)^2+(6-b)^2=9+4$

$\Rightarrow$

$(6-a)^2+(6-b)^2=(3)^2+(2)^2$

$\Rightarrow$

$(6-a)^2=(3)^2$ and

$(6-b)^2=(2)^2$

$\Rightarrow$

$6-a=3$ and

$6-b=2$

$\therefore$

$a=3$ and

$b=4$

$\Rightarrow$

$ab=(3)(4)=12$

The headings of the rows given in the first column of a table are called ________.

0%
stubs
0%
captions
0%
sub titles
0%
prefatory notes

Study the following graph carefully to answer these questions.
Production (in lakh tonnes) of six units of a company in 2006 and 2007.
What is the total production of units C, D and E together for both the years? (In lakh tonnes)

0%
495
0%
595
0%
545
0%
515

The mean and standard deviation of random variable

$X$ are

$10$ and

$5$ respectively, then

$E\left (\dfrac {X - 15}{5}\right )^{2} =$ _______.

0%
$4$
0%
$3$
0%
$2$
0%
$5$

Explanation

Mean

$=10$ and standard deviation

$=5$

So, variance

$(v) = 5^{2} = 25$

$\therefore E (x)^{2} = 25 + 10^{2} = 125$

$\therefore E\left (\dfrac {x^{2} + 225 - 30x}{25}\right ) = \dfrac {1}{25} (E(x)^{2} + 225 - 30E(x))=2$

Which of the following is not two-dimensional diagram?

0%
Pie-chart
0%
Multiple bar diagram
0%
Rectangular diagram
0%
Square diagram

The following series is of the type of

Year	Population in a particular city
1995	19,00,000
1996	19,20,000
1997	25,00,000
1998	25,50,000

0%
individual series
0%
geographical
0%
discrete series
0%
time series

In tabulation, the source of the data, if any, is shown in the ________.

0%
footnotes
0%
body
0%
stub
0%
caption

What percentage of values is less than 4th decile?

0%
$50\%$
0%
$70\%$
0%
$40\%$
0%
None of the above

Choose the correct answer from the alternatives given.
Directions for questions 71 to 75: Study the graph and answer the questions.
What was the approximate percent increase in sales of Cool-sip in 1990 over its sales in 1989?

0%
200
0%
50
0%
171
0%
300

Explanation

Required percentage increase

$\displaystyle \frac{20 \, - \, 5}{5} \, \times \, 100$

$\displaystyle \frac{15}{5} \, \times \, 100$ = 300%

In an ordered series, the data are ________.

0%
in descending order
0%
in ascending order
0%
either (1) or (b)
0%
neither (a) nor (b)

The following series is of the type of

Person	Age(in years)
A	35
B	23
C	26
D	32
E	30

0%
Individual
0%
Discreate
0%
Geographical
0%
Times series

Maximum number of teachers are in between which age group?

0%
$45-50$
0%
$25-30$
0%
$40-45$
0%
$30-35$

Explanation

$\Rightarrow$ From given histogram we are going to form frequency distribution table:

$Ages$ $(Years)$	$No.\,of\,teachers$
$20-25$	$4$
$25-30$	$5$
$30-35$	$6$
$35-40$	$3$
$40-45$	$2$
$45-50$	$5$

$\Rightarrow$ From above table we can see, the maximum number of teachers are between age group

$30-35$ which is

$6$

Which of the following statements is untrue for tabulation?

0%
Statistical analysis of data requires tabulation
0%
It facilitates comparison between rows and not columns
0%
Complicated data can be presented
0%
Diagrammatic representation of data requires tabulation

Consider the following statements:
I. In a bar graph not only height but also width of each rectangle matters.
II. In a bar graph height of each rectangle matters and not its width
III. In a histogram the height as well as the width of each rectangle matters
IV. A bar graph is two dimensional
Of these statements:

0%
I alone is correct
0%
III alone is correct
0%
II and III are correct
0%
I and IV are correct

$\sum\limits_{i = 1}^{18} {({x_i} - 8) = 9}$ and

$\sum\limits_{i = 1}^{18} {{{({x_i} - 8)}^2} = 45}$ , then standard deviation of

${x_1},{x_2},...,{x_{18}}$ is

0%
$\dfrac{4}{9}$
0%
$\dfrac{9}{4}$
0%
$\dfrac{3}{2}$
0%
$\dfrac{3}{4}$

Explanation

Given: $\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right) = 9}$

$\displaystyle\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2} = 45}$

Let $X$ be a random variable taking values ${x_1},........{x_{18.}}$

Then $X-8$ has the values ${x_1} - 8,....,{x_{18}} - 8.$

Now, $E\left( {X - 8} \right) =\displaystyle {{\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right)} } \over {18}}$

$= \displaystyle{9 \over {18}} = {1 \over 2}$

And $E\left[ {{{\left( {X - 8} \right)}^2}} \right] = \displaystyle{{\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2}} } \over {18}}$

$=\displaystyle {{45} \over {18}} = {5 \over 2}$

Thus, $Var\left( {X - 8} \right) = E\left[ {{{\left( {X - 8} \right)}^2}} \right] - \left[ {E{{\left( {X - 8} \right)}}} \right]^2$

$=\displaystyle {5 \over 2} - {\left( {{1 \over 2}} \right)^2}$

$= \displaystyle{5 \over 2} - {1 \over 4}$

$= \displaystyle{9 \over 4}$

We know $Var\left( {1.X - 8} \right) = {1^2}Var\left( X \right)$ , thus,

$Var\left( X \right) = \displaystyle{9 \over 4}$

Standard deviation of $X = \sqrt {Var\left( X \right)}$

$= \displaystyle\sqrt {{9 \over 4}} = {3 \over 2}$

The horizontal line in a bar graph is called:

0%
$x$ -axis
0%
$y$ -axis
0%
Imaginary axis
0%
None of these

Explanation

In the Cartesian coordinate system, the horizontal reference line is called

$x$ -axis.

Therefore, the horizontal line in a bar graph is called

$x$ -axis.

Hence, option

$A$ is correct.

$\displaystyle \sum _{ r=1 }^{ 9 }{ \left( { x }_{ 1 }-5 \right) } =9$ and

$\displaystyle \sum _{ r=1 }^{ 9 }{ \left( { x }_{ 1 }-5 \right) ^{ 2 } } =45$ then the standard deviation of the

$9$ items

$x_{1},x_{2}....x_{9}$ is:

0%
$3$
0%
$9$
0%
$4$
0%
$2$

Coefficient of skewnwss for the value

$Median = 18.8,{Q_1} = 14.6,{Q_3} = 25.2$ is

0%
$0.2$
0%
$0.5$
0%
$0.7$
0%
None

Explanation

$\Rightarrow$ Given data :

$M_d=18.8,Q_1=14.6,Q_3=25.2$

The coefficient of skewness based on quartile is known as Bowley's coefficient of skewness is given by

$S_k(B)=\dfrac{Q_3+Q_1-2M_d}{Q_3-Q_1}$

$=\dfrac{25.2+14.6-(2\times 18.8)}{25.2-14.6}$

$=\dfrac{39.8-37.6}{10.6}$

$=\dfrac{2.2}{10.6}$

$=0.2$

The mean and standard deviation of a random variable

$x$ is given by

$5$ and

$3$ respectively. The standard deviation of

$2-3x$ is _____

0%
$-7$
0%
$81$
0%
$34$
0%
$9$

Explanation

$\sigma(2-3x)=\sigma(-3x)=|-3|\sigma(x)=3\times 3=9$

Solve:

$\log_5 \dfrac{(25)^4}{\sqrt{625}}$

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

Explanation

We have,

$\log_5 \dfrac{(25)^4}{\sqrt{625}}$

$\Rightarrow \log_5 \dfrac{(25)^4}{(25)}$

$\Rightarrow \log_5 (25)^3$

$\Rightarrow \log_5 5^6$

$\Rightarrow 6\log_5 5$

$\therefore \log m^n=n\log m$

$\Rightarrow 6\times 1$

$\therefore \log_a a=1$

$\Rightarrow 6$

Hence, this is the answer.

Use the graph to answer the question.How many students have more than 100 rupees?

0%
7
0%
6
0%
100
0%
8

Explanation

From the graph

Number of students have more than 100 rupees

$=4+3+1=8$

The variance of the data

$6,\ 8,\ 10,\ 12,\ 14,\ 16,\ 18,\ 20,\ 22,\ 24$ is

0%
$15$
0%
$20$
0%
$30$
0%
$33$

Explanation

$\overline{x}=\dfrac{6+8+10+12+14+16+18+20+22+24}{10}=\dfrac{150}{10}=15$

$Variance=\dfrac{\sum(x-\overline{x})^2}{10}$

$=\dfrac{(6-15)^2+(8-15)^2+(10-15)^2+(12-15)^2+(14-15)^2+(16-15)^2+(18-15)^2+(20-15)^2+(22-15)^2+(24-15)^2}{10}$

$=\dfrac{(-9)^2+(-7)^2+(-5)^2+(-3)^2+(-1)^2+(1)^2+(3)^2+(5)^2+(7)^2+(9)^2}{10}$

$=\dfrac{81+49+25+9+1+1+9+25+49+81}{10}$

$=\dfrac{330}{10}$

$=33$

$\therefore$ Variance

$=33$

If the sum of squares of deviations of $15$ observations from their mean $20$ is $240$ , then what is the value of coefficient of variation (CV)?

0%
$20$
0%
$21$
0%
$22.36$
0%
$24.70$

Explanation

$CV =\dfrac{sd}{mean} \times 100$

Standard deviation

$sd = \sqrt {\dfrac{240}{15}}=\sqrt {16} = 4$

Mean

$= 20$

$CV = \dfrac{4}{20} \times 100 = 0.2 \times 100 = 20$

Coefficient of variation

$= 20$

$V$ is the variance and

$M$ is the mean of first

$15$ natural numbers, then what is

$V+M^2$ equal to ?

0%
$\dfrac{124}{3}$
0%
$\dfrac{148}{3}$
0%
$\dfrac{248}{3}$
0%
$\dfrac{124}{9}$

Explanation

first 15 natural numbers

$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$N=5$

mean,

$\mu =\dfrac{1}{N}\sum_{i=1}^{N}x_i$

$\mu =\dfrac{1}{15}(1+2+3+4+5+6+7+8+9+10+11+12+13+14+15)$

$=\dfrac{120}{5}=8$

variance,

$\sigma ^2=\dfrac{1}{N}\sum_{i=1}^{N}(x_i-\mu )^2$

$=\dfrac{1}{15}\sum_{i=1}^{15}(x_i-8 )^2$

$=\dfrac{1}{15}(280)$

Given,

$V+M^2$

$=\dfrac{280}{15}+8^2$

$=\dfrac{1240}{15}$

$=\dfrac{248}{3}$

A graph that displays data that changes continuously over periods of time is

0%
Bar graph
0%
pie chart
0%
Histogram
0%
Line graph

State whether the following statements are true (T) or false (F):
In a bar graph, width of a bar has no significance. It is only for eye attraction.

0%
True
0%
False

Explanation

We know,

A bar graph is a graphical representation of data using bars of equal width drawn vertically or horizontally with equal spacing between them and different heights.

But, width of a bar do not have any significance and is only for eye attraction.

Hence, the statement is true.

Therefore, option

$A$ is correct.

State whether the following statements are true (T) or false (F):
In a bar graph, bars can be drawn either vertically or horizontally.

0%
True
0%
False

Explanation

We know,

A bar graph is a graphical representation of data using bars of equal width drawn vertically or horizontally with equal spacing between them and different heights.

Then clearly, in a bar graph, bars can be drawn either vertically or horizontally.

That is, the statement is true and option

$A$ is correct.

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
The maximum number of ice cream cones were sold on:

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

Study the graph carefully and answer the questions given below it. What is the total wheat import during all the given years (in thousand tonnes)?

0%
$27420$
0%
$28240$
0%
$29240$
0%
$28420$

Explanation

Total wheat import during all the given years

$=3465+1811+2413+4203+7016+5832+2000+2500=29240$ Thousand tonnes

Observe the following pictograph which shows the number of ice cream cones sold by school canteen during a week. Chose the correct answer from the given options:
The minimum number of ice cream cones were sold on:

0%
Monday
0%
Saturday
0%
Tuesday
0%
Thursday

Explanation

After observing the pictograph, we can say that

Monday

$= 2\times 5 = 10$ ice creams

Tuesday

$= 2\times 8 = 16$ ice creams

Wednesday

$= 2\times 6 = 12$ ice creams

Thursday

$= 2\times 3 + 1 = 7$ ice creams

Friday

$= 2\times 7 = 14$ ice creams

Saturday

$= 2\times 4 = 8$ ice creams

Hence, the minimum number of ice cream were sold on Thursday i.e.

$7$ ice cream.

So, Option ( D ) is the correct answer.

To get an interest of Rs

$280$ per year, how much money should be deposited?

0%
$2500$
0%
$3000$
0%
$3500$
0%
$400$

Explanation

From the above graph, we can notice that, to get an interest of

$280$ rupees per year, the money to be deposited is

$3500$ rupees (approximately).

Use the graph to find the interest on Rs

$2500$ for a year.

0%
$Rs. 100$
0%
$Rs.200$
0%
$Rs.300$
0%
$Rs.400$

Explanation

From the above graph, we can observe that simple interest on

$Rs. 2500$ rupees is

$Rs. 200$ rupees which is marked as black point in the graph.

The frequency distribution has been represented graphically as follows:

marks	$0-20$	$20-40$	$40-60$	$60-100$
No of students	$10$	$15$	$20$	$25$

Do you think this representation is correct or incorrect?

0%
Correct
0%
Incorrect
0%
Ambiguous
0%
Data insufficient

Explanation

From the graph,

$10$ students got

$0-20$ marks .

$15$ students got

$20-40$ marks.

$20$ students got

$40-60$ marks and

$25$ students got

$60-100$ marks.

The same is shown in the frequency distribution table.

During which periods did the patient's temperature showed an upward trend?

0%
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 2 p.m. to 3 p.m.
0%
9 a.m. to 10 a.m., 10.30 a.m. to 11.30 a.m., 2 p.m. to 3 p.m.
0%
9 a.m. to 10 a.m., 10 a.m. to 11.30 a.m., 1 p.m. to 2 p.m.
0%
9 a.m. to 10 a.m., 10 a.m. to 11 a.m., 1 p.m. to 2 p.m.

Explanation

Whenever there is an increase in the value of

$y - axis$ as the time(on

$x - axis$ ) passes by, there is an upward trend.
Thus, the patient shows an upward trend between

$9 am - 10 am$ ,

$10 am - 11 am$ and

$2 pm - 3 pm$

Hence, option A is correct.

When was the patients temperature

$38.5$ C?

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

Explanation

From the given graph, we can observe that the patient temperature

$38.5^0C$ was at

$12$ noon

The patients temperature at 1 p.m. is:

0%
$37.0$ C
0%
$39.5$ C
0%
$36.5$ C
0%
$36.0$ C

Explanation

From the given graph, we can observe that at

$1 p.m.$ the patient temperature is

$36.5^0C$

How much distance did the car cover during the period 7.30 a.m. to 8 a.m?

0%
$20$ km
0%
$40$ km
0%
$80$ km
0%
$100$ km

Explanation

Using the information provided in the table, we can plot the graph shown above, taking the time on the

$x-axis$ and the distance (in km) on the

$y-axis$ .

From the above graph, we can observe that at

$7:30\ am$ , the distance traveled is

$100\space km$ and at

$8:00\ am$ , the distance traveled is

$120 \space km$ .

Therefore, the distance covered during the period

$7:30$ and

$8:00$ is

$120-100=20\space km$

Hence, option A is correct.

Draw a graph for the given information with suitable scales on the axes and answer the following question.

The time when the car had covered a distance of

$100$ km since its start is:

0%
$7:00$ am
0%
$7:30$ am
0%
$8:00$ am
0%
$8:30$ am

Draw a graph for the following:

Side of square (in cm)	$2$	$3$	$3.5$	$5$	$6$
Perimeter (in cm)	$8$	$12$	$14$	$20$	$24$

State the typre of graph.

0%
Linear graph
0%
Bar graph
0%
Pie graph
0%
None of these

During which week did Plant B grow least?

0%
1st week
0%
2nd week
0%
3rd week
0%
None of these

Explanation

Growth of plant B during 1st week

$= 1 - 0 = 1.$

Growth of plant B during 2nd week

$= 7 - 1 = 6.$

Growth of plant B during 3rd week

$= 10 - 7 = 3.$

Plant B grew least in the first week.

CBSE Questions for Class 11 Commerce Applied Mathematics Descriptive Statistics Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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