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

How many boys attended the 1995 convention?

0%
358
0%
390
0%
407
0%
540
0%
716

Explanation

Total no. of people who attended the Nation Youth Convention in 1955 =

$716$

From the graph we can see that,

Percentage of boys who attended in 1995 =

$50\%$

$\therefore$ No. of boys who attended in 1995 =

$\dfrac{50}{100} \times 716 = 358$

If the four regions shown in the graph above are the only regions in Town

$T$ , the total of which two regions accounts for exactly

$70$ percent of all cable television subscribers in Town

$T$ ?

0%
Regions $M$ and $N$
0%
Regions $M$ and $O$
0%
Regions $N$ and $O$
0%
Regions $N$ and $P$

Explanation

According to the graph,

The number of subscribers in region

$M$ is

$3000$ .

The number of subscribers in region

$N$ is

$4000$ .

The number of subscribers in region

$O$ is

$1000$ .

The number of subscribers in region

$P$ is

$2000$ .

We can see that number of subscribers in combined region of regions

$M$ and

$N$ is

$3000+4000=7000$

We can see that number of subscribers in combined region of regions

$M$ ,

$N$ ,

$O$ ,

$P$ is

$3000+4000+1000+2000=10000$ .

Therefore Region

$M+$ Region

$N$ subscribers

$=$

$70$ percent of total subscribers.

So, the correct option is

$A$ .

Variance of the first

$1$ natural numbers is.

0%
$\sqrt{5}$
0%
$\sqrt{10}$
0%
$5\sqrt{2}$
0%
$10$

Examine the graph and determine which of the following is true.

0%
(c) minus (b) equals (b) minus (d)
0%
(a) plus (d) equals (c)
0%
(c) minus (d) equals (a)
0%
(a) plus (b) equals (c) plus (d)

If the variance of

$14, 18, 22, 26, 30$ is

$32$ , then the variance of

$28, 36, 44, 52, 60$ is.

0%
$64$
0%
$128$
0%
$32\sqrt{2}$
0%
$32$

The variance of

$10, 10, 10, 10$ is.

0%
$10$
0%
$\sqrt{10}$
0%
$5$
0%
$0$

How many more students like Basketball than Cricket?

0%
$60$
0%
$120$
0%
$20$
0%
$40$

Which month maximum rainfall?

0%
April
0%
July
0%
August
0%
September

The chart below represents the average amount of rain falling each month in the town of Tegulpa. During which month of the year does the most rain fall?

0%
January
0%
April
0%
August
0%
December

Which one of the following is a false description?

0%
In a moderately asymmetrical distribution, the empirical relationship between Mean, Mode and Median suggested by Karl Pearson is Mean Mode = 3 (Mean Median)
0%
Coeffcient of variation is an absolute measure of dispersion
0%
Measure of skewness in the distribution of numerical values in the data set
0%
Kurtosis refers to the degree of flatness or peakedness in the region around the mode of a frequency curve

In _________ classification time becomes the classifying variable and data are categorized according to time.

0%
spatial
0%
quantitative
0%
qualitative
0%
temporal

Highest marks are scored in ______ .

0%
English
0%
Hindi
0%
Maths
0%
None of these

If the maximum number of marks is

$50$ in each subject, then in how many subjects student scored more than half of maximum marks?

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

Which of the following are limitations of range?

0%
It is not based on all the values.
0%
As long as the minimum and maximum values remain unaltered, any change in other values does not affect range.
0%
It can not be calculated for open-ended frequency distribution.
0%
All of these

In a frequency distribution, Range is the difference between _________ and __________

0%
upper limit of the highest class, the lower limit of the lowest class
0%
lower limit of the highest class, the upper limit of the lowest class
0%
A or B
0%
None of the above

Which of the following is another term for quartile division?

0%
Inter quartile range
0%
Semi Inter Quartile Range
0%
Inter Quartile Deviation
0%
None of these

If the minimum marks for passing in each subject is

$14$ , then in how many subject/subjects did he fail?

0%
$1$
0%
$2$
0%
$0$
0%
None of these

If the mean and standard deviation of a normal distribution are

$33$ and

$5$ respectively, what is the Inter-quartile range?

0%
$(6.67)$
0%
$(8)$
0%
$(7.74)$
0%
$(6.8)$

Karl Pearsons Coefficient of Correlation is also known as ______.

0%
simple correlation coefficient
0%
product moment correlation
0%
both A and B
0%
none of the above

The Multiplicative Model of a time series is expressed is :

0%
$Y_{t} = T_{t} \times C_{t} \times S_{t} \times R$
0%
$Y_{t} = T_{t} - C_{t} - S_{t} - R_{t}$
0%
$Y_{t} = T_{t} + C_{t} + S_{t} + R_{t}$
0%
$Y_{t} = T_{t}\cap C_{t}\cup S_{t} \cup R_{t}$

Which of the following relations among the location parameters does not hold?

0%
$Q_2 = Median$
0%
$P_50 = Median$
0%
$D_5 = Median$
0%
$D_4 = Median$

__________ gives a precise numerical value of the degree of linear relationship between two variables X and Y.

0%
Scatter diagram
0%
Spearmans coefficient of correlation
0%
Karl Pearsons Coefficient of Correlation
0%
None

The

$P_82$ for the following distribution is

X	2	3	4	5	6	7	8	9	10	11
Y	3	6	9	18	20	14	10	10	7	2

0%
$6$
0%
$5$
0%
$8$
0%
$9$

Inter-Quartile Range is based upon ________.

0%
First 50% of the values
0%
middle 50% of the values
0%
Last 50 % of the values
0%
25% of the values

Time series is an important statistical technique which is widely used by _________.

0%
Scientists
0%
Sociologists
0%
Research workers
0%
All of above

Study the following graph carefully and answer the question given.
In which of the following pairs of companies the difference between average productions for the three years is maximum?

0%
E and F
0%
D and C
0%
E and C
0%
A and E

Explanation

Total production of A

$=145$

Total production of B

$=160$

Total production of C

$=105$

Total production of D

$=185$

Total production of E

$=220$

Total production of F

$=115$

Therefore, difference between E and C production is the least.

Study the following graph carefully and answer the question given.
What is the percentage decline in production by company C from

$1995$ to

$1996$ ?

0%
$5$
0%
$20$
0%
$15$
0%
$12.50$

Explanation

Production of company C in

$1995=40$ lakhs tonne

Production of company C in

$1996=35$ lakhs tonne

% decline in production

$=\cfrac { 40-35 }{ 40 } \times 100$

$=12.5$ %

Choose the correct answer from the alternatives given.
Directions for questions 71 to 75: Study the following bar-graph and answer the questions.
The average electric consumption by the family during these 5 months in 2013 is

0%
470 units
0%
440 units
0%
450 units
0%
460 units

Explanation

Average units consumption in the year 2013

$\displaystyle = \, \frac{550 \, + \, 500 \, + \, 400 \, + \, 350 \, + \, 500}{5} \, = \frac{2300}{5}$ =

$460$ units

Choose the correct answer from the alternatives given.
Directions for questions 71 to 75: Study the following bar-graph and answer the questions.
For how many months in 2012, the consumption of electric units was more than the average units consumption?

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

Explanation

Average units consumption in the year 2012

$\displaystyle \frac{600 \, + \, 700 \, + \, 400 \, + \, 300 \, + \, 200}{5} \, = \, \frac{2200}{5}$ = 440 units
Hence, required months are July and August.

In a group,

$2$ students spend Rs

$8$ daily,

$3$ students spend Rs

$10$ daily and

$5$ students spend Rs

$6$ daily. The average spending of all

$10$ students is _______.

0%
$7.6$
0%
$5.8$
0%
$8.5$
0%
$6.7$

If the sum of mean and variance of

$B.D$ for

$5$ trials is

$1.8$ , the binomial distribution is

0%
$(0.8+0.2)^5$
0%
$(0.2+1.8)^5$
0%
${\left( {0.8 + 0.2} \right)^{10}}$
0%
${\left( {0.2 + 1.8} \right)^{10}}$

Explanation

a binomial distribution is characterized by

$2$ parameters:

$n$ - number of trials

$p$ - probability of success in one trail.

We there by know n=5, we need to find

$p$ .

$Mean$ of the binomial distribution is given by,

$mean = np$

variance is given by

$variance\quad =\quad np(1-p)\\ \therefore \quad mean\quad +\quad variance\quad =\quad 1.8\\ =>\quad 5p\quad +\quad 5p(1-p)\quad =\quad 1.8\\ =>\quad 5{ p }^{ 2 }-10p+1.8=0\\ =>\quad p=\quad 0.2\quad or\quad 1.8$

$=>\quad p=\quad 0.2\ [\because 1\ge p\ge 0]$

$q=0.8$

Binomial distribution is

$(0.2+0.8)^n$ .

Hence the answer is A

Study the following graph carefully and answer the question given.
What was the difference between the number of B type tyres manufactured in

$1994$ and

$1995$ ?

0%
$1,00,000$
0%
$20,00,000$
0%
$10,00,000$
0%
$12,50,000$

Explanation

Number of B type tyres in

$1995=35$ lakhs

Number of B type tyres in

$1994=25$ lakhs

difference

$=35-25=10$ lakhs

$=1000000$

If each observation of a dist. whose

$S.D$ is

$\sigma$ , is increased by

$\lambda$ then the variance of the new observations is

0%
$\sigma$
0%
$\sigma+\lambda$
0%
$\sigma^ {2}$
0%
$\sigma^ {2}+\lambda$

Study the following graph carefully and answer the question given.
Which of the following companies recorded the minimum percentage growth from

$1994$ to

$1995$ ?

0%
A
0%
B
0%
C
0%
F

Explanation

order of company growth from

$1994$ to

$1995$ is

$E<A<B<C<F<D$

Study the following graph carefully and answer the question given.
The total number of all the three types of tyres manufactured was the least in which of the following years?

0%
$1995$
0%
$1996$
0%
$1992$
0%
$1994$

Explanation

Number of

$3$ type tyres in

$1992=75$ lakhs

Number of

$3$ type tyres in

$1993=75$ lakhs

Number of

$3$ type tyres in

$1994=77.5$ lakhs

Number of

$3$ type tyres in

$1995=85$ lakhs

Number of

$3$ type tyres in

$1996=65$ lakhs

Therefore, total number of tyres is least in

$1996$

Study the following graph carefully and answer the question given.
In which of the following years was the percentage production of B type of C type tyres the maximum?

0%
$1994$
0%
$1992$
0%
$1995$
0%
$1993$

Explanation

$1995$ , number of type B tyres are far greater than type C tyres

% production of B on % production on C is maximum in

$1995$

Standard deviation of

$3, 5, 7, 9, 11, 13$ is?

0%
$12$
0%
$11$
0%
$11.66$
0%
None

The standard deviation of the data:

$x:$

$1$

$a$

${a}^{2}$ ........

${a}^{n}$

$f:$

${ _{ }^{ n }{ C } }_{ 0 }$

${ _{ }^{ n }{ C } }_{ 1 }$ .....

${ _{ }^{ n }{ C } }_{ n }$ is

0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }-{ \left( \cfrac { 1+{ a }^{ } }{ 2 } \right) }^{ n }$
0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ a }^{ } }{ 2 } \right) }^{ n }$
0%
${ \left( \cfrac { 1+{ a }^{ 2 } }{2 } \right) }^{ 2n }-{ \left( \cfrac { 1+{ a }^{ 2 } }{ 2 } \right) }^{ n }$
0%
None of these

Standard deviation of the first

$2$ n +

$1$ natural numbers is

0%
$\sqrt{\dfrac {n(n+1)}{2}}$
0%
$\sqrt{\dfrac {n(n+1)(2n + 1)}{3}}$
0%
$\sqrt{\dfrac {n(n+1)}{3}}$
0%
$\sqrt{\dfrac {n^2(n+1)}{2}}$

For a series the information available is

$n=10$ ,

$\displaystyle\sum x=60, \displaystyle\sum x^2=1000$ . The standard deviation is?

0%
$8$
0%
$64$
0%
$24$
0%
$128$

The standard deviation of

$15$ terms is

$6$ and each item is decreased by

$1$ . Then the standard deviation of new data is?

0%
$5$
0%
$7$
0%
$\dfrac{91}{15}$
0%
$6$

Explanation

If each term of the data set is increased or decreased by the same amount, the standard deviation remains unchanged.

$\therefore$ Standard deviation of the new data set is also 6.
.

Variance of

$^{10}C_{0}, ^{10}C_{1}, ^{10}C_{2},^{10}C_{10}$ is:

0%
$\dfrac{10.^{20}C_{10}-2^{10}}{100}$
0%
$\dfrac{11.^{20}C_{10}-2^{10}}{11}$
0%
$\dfrac{10.^{20}C_{10}-2^{20}}{100}$
0%
$\dfrac{10.^{20}C_{10}-2^{10}}{121}$

The standard deviation of

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

0%
$4$
0%
$2$
0%
$\sqrt{7}$
0%
None of the above

Explanation

Given observation is

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

Mean=

$\cfrac { 1+2+3+4+5+6+7 }{ 7 } =\cfrac { 28 }{ 7 } =4$

Variance=

$\sum _{ +r=1 }^{ n }{ \cfrac { \left( x-{ x }_{ i } \right) ^{ 2 } }{ n } } =\cfrac { \left( 1-4 \right) ^{ 2 }+\left( 2-4 \right) ^{ 2 }+\left( 3-4 \right) ^{ 2 }+\left( 4-4 \right) ^{ 2 }+\left( 5-4 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 }+\left( 7-4 \right) ^{ 2 } }{ 7 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 }+0+{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 7 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 9+4+1+1+4+9 }{ 7 } =\cfrac { 28 }{ 7 } =4$

standard deviation=

$\sqrt { variance } \\ \sqrt { 4 } =2$

What is standard deviation of the set of observations

$32,28,29,30,31$ ?

0%
$1.6$
0%
$\sqrt { 2 }$
0%
$2$
0%
$none\ of\ these$

Explanation

$\sigma =\displaystyle \sum _{ i=1 }^{ n }{ \sqrt { \dfrac { \left( { x }_{ i }-{ x }_{ av } \right) }{ m-1 } } }$

${ x }_{ au }=\dfrac { 32+28+29+30+31 }{ 5 } =30$

$\therefore =\sqrt { \dfrac { { \left( 32-30 \right) }^{ 2 }+{ \left( 28-30 \right) }^{ 2 }+{ \left( 29-30 \right) }^{ 2 }+{ \left( 30-30 \right) }^{ 2 }+{ \left( 31-30 \right) }^{ 2 } }{ 5-1 } }$

$=\sqrt { \dfrac { 4+4+1+1 }{ 4 } } =\sqrt { \dfrac { 10 }{ 4 } } =\sqrt { \dfrac { 10 }{ 2 } } =1.58$

In an experiment with

$10$ observations on

$x,\displaystyle \sum { x=60, } \displaystyle \sum { { x }^{ 2 } } =1000$ one observation that was

$20$ was found to be wrong and was replaced by the correct value

$30$ then the correct variance is

0%
$64$
0%
$111$
0%
$52$
0%
$101$

Variance of the data given below

Size of item	3.5	4.5	5.5	6.5	7.5	8.5	9.5
Frequency	3	7	22	60	85	32	8

0%
1
0%
1.231
0%
1.321
0%
1.312

The variance of first

$20$ - natural numbers is?

0%
$\dfrac{399}{4}$
0%
$\dfrac{379}{12}$
0%
$\dfrac{133}{2}$
0%
$\dfrac{133}{4}$

Explanation

$Var[X]=E[X^{2}]-(E[X])^{2}=\cfrac{1}{20}[(1^{2}+2^{2}+.....+20^{2})]-(E[X])^{2}$

Now,

$E[X]=\sum_{i=1}^{20}P_{i}X_{i}=\cfrac{1}{20}(1+2+3+...+20)=10.5$

$(P_{1}=P_{2}=P_{3}=.......P_{20}=\cfrac{1}{20})$

$\therefore Var[X]=\cfrac{1}{20}[(1^{2}+2^{2}+3^{2}+....+20^{2})]-10.5^{2}=\cfrac{1}{20}\times 2870-110.25$

$=143.5-110.25=33.25=\cfrac{133}{4}$

The mean and standard deviation of

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

0%
$\dfrac{7}{2},\sqrt{\dfrac{35}{12}}$
0%
$\dfrac{7}{3} ,\sqrt{3}$
0%
$3,3$
0%
$3,35/12$

For r.v. X whose p.m.f. is

X	0	1	2
p(X=)x	${ b }^{ 2 }$	2ab	${ a }^{ 2 }$

Std. deviation of

$X i.e.\sigma =$

0%
$\sqrt { ab }$
0%
$\sqrt { 2ab }$
0%
$\sqrt { 3ab }$
0%
$2\sqrt { ab }$

The average marks of 10 students in a class is 60 with standard deviation 4 while the average marks of other 10 students is 40 with standard deviationIf all the 20 students are taken together, their standard deviation will be

0%
$\sqrt {109}$
0%
12
0%
$\sqrt {116}$
0%
$\sqrt {126}$

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

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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