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CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 15 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Central Tendency
Quiz 15
The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's score is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?
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0%
$$109$$
0%
$$106$$
0%
$$150$$
0%
$$160$$
Explanation
Let m be the mean and s be the standard deviation and find the z score.
$$z = (x - m) /s = (0.8 s + m - m) / s = 0.8$$
The percentage of student who scored above Jane is (from table of normal distribution).
1 - 0.7881 = 0.2119 = 21.19%
The number of student who scored above Jane is (from table of normal distribution).
21.19% 0f 500 = 106
What is the value of mean for the following data.
Class interval
Frequency
$$350-369$$
$$15$$
$$370-389$$
$$27$$
$$390-409$$
$$31$$
$$410-429$$
$$19$$
$$430-449$$
$$13$$
$$450-469$$
$$6$$
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0%
$$400$$
0%
$$400.58$$
0%
$$394$$
0%
$$394.50$$
Explanation
Class Intervals
Mid-values $$(x)$$
Frequency $$(f)$$
$$fx$$
$$350-369$$
$$359.5$$
$$15$$
$$5,392.5$$
$$370-389$$
$$379.5$$
$$27$$
$$10,246.5$$
$$390-409$$
$$399.5$$
$$31$$
$$12,384.5$$
$$410-429$$
$$419.5$$
$$19$$
$$7,970.5$$
$$430-449$$
$$439.5$$
$$13$$
$$5,713.5$$
$$450-469$$
$$459.5$$
$$6$$
$$2,757$$
$$\displaystyle\sum f = 111$$
$$\displaystyle\sum fx = 44,464.5$$
Arithmetic mean $$= \cfrac{44,464.5}{111} = 400.58$$
Find the mode of the given set;
$$3$$, $$5$$, $$9$$, $$6$$, $$5$$, $$9$$, $$2$$, $$9$$, $$3$$ and $$5$$
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0%
$$5$$
0%
$$9$$
0%
$$5$$ and $$9$$
0%
$$5$$,$$6$$ and $$9$$
Explanation
$$Mode=$$no of $$element$$ which is repeated more.
$$Mode=5$$ and $$9$$
$$\because $$ Both are repeated $$3$$ times in the series.
The median of the given data:
$$49$$, $$48$$, $$15$$, $$20$$, $$28$$, $$17$$, $$14$$ and $$10$$ is
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0%
$$18$$
0%
$$20$$
0%
$$22$$
0%
$$24$$
Explanation
$$10,14,15,17,20,28,48,49\\ n=8\\ \therefore median=\cfrac { n }{ 2 } +1\\ ={ (4+1) }^{ th }\quad obs\\ ={ (5) }^{ th }\quad obs\\ =20$$
The median of $$0.05,0.50,0.055,0.505\ and\ 0.55$$ is
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0%
$$0.05$$
0%
$$0.505$$
0%
$$0.50$$
0%
$$0.055$$
Explanation
$$\Rightarrow$$ The median is the middle number in a group of numbers.
$$\Rightarrow$$ The given numbers are $$0.05,\,0.50,\,0.055,\,0.505,\,0.55$$
$$\Rightarrow$$ Now, arranging given numbers in ascending order we get, $$0.05,\,0.055,\,0.50,\,0.505,\,0.55$$
$$\Rightarrow$$ So, here we can see middle number is $$0.50$$.
$$\therefore$$ Median is $$0.50$$
The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Time (hrs.)
$$0-2$$
$$2-4$$
$$4-6$$
$$6-8$$
$$8-10$$
No. of students
$$7$$
$$18$$
$$12$$
$$10$$
$$3$$
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0%
$$4\ hrs$$
0%
$$5\ hrs$$
0%
$$4.36\ hrs$$
0%
$$5.36\ hrs$$
Deciles divide the data into __________ equal parts.
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0%
five
0%
ten
0%
fifteen
0%
three
Explanation
Deciles basically refers to 1/10th part of a series. It divides a series into 10 equal parts.
Median of $$8,4,6,18,5,12$$ and $$100$$ is __________.
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0%
$$8$$
0%
$$6$$
0%
$$12$$
0%
$$100$$
Explanation
Arrange in ascending order $$\Rightarrow$$
$$4, 5, 6,\underset{\uparrow}{8}, 12, 18, 100$$
$$7$$ observations $$\therefore$$ Median is $$4^{th}$$ observation
Median $$=8$$ Answer.
The mean and median of proper divisors of $$360$$ are connected by
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0%
median> mean
0%
median< mean
0%
median = mean
0%
median = 2(mean)
The average annual income (in Rs.) of certain agricultural workers is $$S$$ and that of other workers is $$T$$. The number of agricultural workers is $$11$$ times that of other workers. Then the average monthly income (in Rs.) of all the workers is:
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0%
$$\cfrac{S+T}{2}$$
0%
$$\cfrac{S+11T}{2}$$
0%
$$\cfrac{1}{11S}+T$$
0%
$$\cfrac{11S+T}{12}$$
Explanation
Let the number of other workers $$ =x$$
Then number of agricultural worker $$=11x$$
Total number of worker $$=12x$$
Average monthly income $$=\dfrac{S\times11x+T\times x}{12x}$$
$$=\dfrac{11S+T }{12}$$
The average age of the boys in a class is $$16$$ years and that of the girls is $$15$$ years. The average age for the whole class when there are equal number of girls and boys is:
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0%
$$15$$ years
0%
$$15.5$$ years
0%
$$16$$ years
0%
Cannot be computed with the given information
Explanation
Let No of boys = No of girls = x
Avg age of whole class = $$\dfrac{16\times x+15\times x}{x+x}$$
$$ = \dfrac{16+15}{2}$$
$$ = 15.5 $$ years
The sum of first $$n$$ natural numbers is given by the expression $$(2n^{2}+3n)$$. The mean of the given numbers is
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0%
$$\left(\dfrac{2}{n}+3\right)$$
0%
$$\left(2+\dfrac{3}{n}\right)$$
0%
$$(2+3n)$$
0%
$$(2n+3)$$
The frequency distribution of the marks obtained by $$100$$ students in a test carrying $$50$$ marks then the mean is
Marks
0-9
10-19
20-29
30-39
40-49
No. of students
8
15
20
45
12
Report Question
0%
$$28.3$$
0%
$$28$$
0%
$$27.3$$
0%
$$26.4$$
Explanation
Here the mean of given data is given by $$\mu =\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ { f }_{ i } } $$,
Where $$x_i$$ are the mid values the mark range and $$f_i$$ are the corresponding frequencies or in this case the number of students.
$$\therefore \mu =\dfrac{4.5\times 8+14.5\times 15+24.5\times 20+34.5\times 45+44.5\times 12}{8+15+20+45+12}$$
$$=\dfrac{2830}{100}$$
$$=28.30$$
Hence, the answer is $$28.30.$$
The median of $$\dfrac { x }{ 5 } ,x,\dfrac { x }{ 4 } ,\dfrac { x }{ 2 } ,\dfrac { x }{ 3 } $$ is $$8$$ then the value of $$x$$ is
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0%
24
0%
18
0%
27
0%
51
Explanation
Number of terms, n=5.
Given median = 8
$$\Rightarrow (\frac{n+1}{2})^{x} term=8$$
$$\Rightarrow (\frac{5+1}{2})^{x}term =8$$
$$\Rightarrow 3^{x}$$ term =8
$$\Rightarrow \frac{x}{4}=8$$
$$\Rightarrow x=32$$
The mean of first ten prime numbers is
Report Question
0%
$$12$$
0%
$$12.9$$
0%
$$13$$
0%
$$14$$
Explanation
First 10 prime numbers are $$2,3,5,7,11,13,17,19,23,29$$
Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$
Apply the above formula, we get
Mean=$$\frac{2+3+5+7+11+13+17+19+23+29}{10}=12.9$$
The AM of first $$5$$ whole numbers is:
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$5$$
Explanation
The first $$5$$ whole numbers are $$0, 1, 2, 3, 4$$.
We know that,
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$
Apply the above formula, we get
$$Arithmetic\ mean= \dfrac{0+1+2+3+4}{5}$$
$$= \dfrac{10}{5}$$
$$= 2$$
Hence, option $$(b)$$ is correct.
The ages of 40 girls in a class are given below:
Age (in years)
14
15
16
17
18
Number of girls
5
8
15
10
2
Find the mean age.
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0%
13.9
0%
14.9
0%
15.9
0%
16.9
Which of the following pairs of numbers has an average (arithmetic mean) of $$2$$?
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0%
$$1 - \sqrt {2}, 3 + \sqrt {2}$$
0%
$$2\sqrt {3}, 2 - 2\sqrt {3}$$
0%
$$\dfrac {1}{0.5}, \dfrac {2.4}{1.6}$$
0%
$$\sqrt {5}, \sqrt {3}$$
0%
$$\dfrac {1}{\dfrac {2}{3}}, \dfrac {1}{\dfrac {2}{5}}$$
Which measures of central tendency get affected if the extreme observations on both the ends of a data arranged in descending order are removed ?
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0%
Mean and mode
0%
Mean and Median
0%
Mode and Median
0%
Mean,Median and Mode
Explanation
Mean and Mode are affected if the observation of the ends are removed.
Median Is not affected as the observation a the middle remains at the middle.
Hence, option (A) is the correct answer.
In the formula $$x=a+h(f_i u_i/ f_i)$$, for finding the mean of grouped frequency distribution, $$u_i=$$
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0%
$$(x_i +a)/h$$
0%
$$h(x_i -a)$$
0%
$$(x_i -a)/h$$
0%
$$(a-x_i )/h$$
Explanation
According to the question,
$$x= a+h(f_i u_i/f_i)$$,
Above formula is a step deviation formula.
In the above formula,
$$x_i$$ is data values,
$$a$$ is assumed mean,
$$h$$ is class size,
When class size is same we simplify the calculations of the mean by computing the coded mean of $$u_1, u_2, u_3.....$$,
Where $$u_i= (x_i -a)/h$$
Hence, option (C) is correct.
Read the following news report and answer on the basis of the same: For an examination, 60 students of a school appeared. Marks obtained by the 12 students were 25 each. The number of students who scored 20 and 5 each was three more than the students who scoredAlso, 8 students who scored 10 marks each. The students who scored 15 marks each were half the students who scored 10 each and remaining students had scored 30 each. Answer the following question:
The value of mean will be:
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0%
18.583
0%
17.583
0%
16.583
0%
15.583
Explanation
The average marks of 100 students in a class areBut while calculating it, the marks of a student were written 73 instead ofFind out the corrected arithmetic mean.
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0%
76
0%
48
0%
48.80
0%
47.80
Explanation
Hence, correct answer is option D.
The marks obtained by $$19$$ students of a class are
$$27, 36, 22, 31, 25, 26, 33, 24, 37, 32, 29, 28, 36, 35, 27, 26, 32, 35$$ and $$28$$. Find the upper quartile.
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0%
$$35$$
0%
$$36$$
0%
$$37$$
0%
None of these
Explanation
Arrange the marks in increasing order. we get,
$$22,24,25,26,26,27,27,28,28,29,31,32,32,33,35,35,36,36,37$$
Since, total no. of terms $$(n)=19$$
Upper quartile (Q3)$$=3( \cfrac {n+1}{4})^{th}$$term
$$=3(\cfrac {19+1}{4})^{th}$$term
$$=3\times 5=15^{th}$$term$$=35$$
Find
the value of $$n$$, if the
median of the numbers $$n + 3, n + 7, .... n + 31, n + 35$$ is $$30$$.
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0%
$$11$$
0%
$$14$$
0%
$$19$$
0%
$$27$$
0%
$$32$$
Explanation
Given series is $$n+35 = n+3 + (t-1)4$$
$$\therefore 35=3+4t-4$$
$$\therefore 36=4t$$
$$\therefore t = 9$$ (where $$t$$ is number of terms)
Therefore the middle term is $$5^{th}$$ term
$$5^{th}$$ term is $$ n+3 + 16 = 30$$
$$\therefore n+19=30$$
$$\therefore n=11$$
The variate of a distribution takes the values $$1, 2, 3, ...n$$ with frequencies $$n, n -1, n -2,... 3, 2, 1.$$ then mean value of the distribution is
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0%
$$ \displaystyle \frac{n\left ( n+2 \right )}{3} $$
0%
$$ \displaystyle \frac{n\left ( n+1 \right )\left ( n+2 \right )}{6} $$
0%
$$ \displaystyle\frac{n+2}{3} $$
0%
$$ \displaystyle\frac{\left ( n+1 \right )\left ( n+2 \right )}{6} $$
Explanation
$$\displaystyle x_{i}=1, 2,3, 4 ...,n-1, n $$
$$\displaystyle f_{i}=n,n-1,n-2,n-3,...2,1$$
Now $$\displaystyle \sum_{i=1}^n x_if_i=
\sum_{i=1}^{n} [ n+2\left ( n-1 \right )+3\left ( n-2 \right
)...$$ $$\displaystyle +\left ( n-2 \right )2+n.1 ] $$ $$\displaystyle
=\sum_{r=1}^{n}[ ( n+1 )r-r^{2}]$$$$\displaystyle = \left ( n+1 \right
)\sum_{r=1}^{n}r-\sum_{r=1}^{n}r^{2}$$ $$\displaystyle =\frac{\left (
n+1 \right )\left ( n \right )\left ( n+1 \right )}{2}-\frac{n\left (
n+1 \right )\left ( 2n+1 \right )}{6}$$$$\displaystyle = \frac{n\left (
n+1 \right )\left ( n+2 \right )}{6}$$
Also $$\displaystyle \sum f_{i}=
1+2+...+n=\frac{n\left ( n+1 \right )}{2}$$
$$\displaystyle \therefore
\mbox{Mean}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}=\frac{n\left ( n+1 \right )\left
( n+2 \right )}{6}\times \frac{2}{n\left ( n+1 \right
)}$$ $$\displaystyle = \frac{n+2}{3}$$
$$X$$
$$0$$
$$1$$
$$2$$
$$3$$
$$4$$
Frequency
$$4$$
$$f$$
$$9$$
$$g$$
$$4$$
The table above gives the frequency distribution of a discrete variable $$X$$ with two missing frequencies. If the total frequency is $$25$$ and the arithmetic mean of $$X$$ is $$2$$, then what is the value of the missing frequency $$f$$?
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0%
$$4$$
0%
$$5$$
0%
$$6$$
0%
$$7$$
Consider the following data:
$$X$$
$$1$$
$$2$$
$$3$$
$$4$$
$$5$$
$$f$$
$$3$$
$$5$$
$$9$$
-
$$2$$
The arithmetic mean of the above distribution is $$2.96$$. What is the missing frequency?
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0%
$$4$$
0%
$$6$$
0%
$$7$$
0%
$$8$$
If Mean deviation about median is $$7.60$$ coefficient of mean deviation about median is $$12.45$$. Find the median____.
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0%
$$63.75$$
0%
$$61$$
0%
$$62.21$$
0%
$$59.75$$
Explanation
Median is the middle most value of a given series that represents the whole class of the series. When mean deviation about the median and its coefficient is given, then
Median = mean deviation about the median/ coefficient of mean deviation about the median
=> Median = 7.60 / 0.1245
= 61
Calculate Median salary from the following frequency distribution:
X (Rs. in 000)
$$10-20$$
$$20-30$$
$$30-40$$
$$40-50$$
$$50-60$$
$$60-70$$
Y (Frequency)
$$2$$
$$3$$
$$6$$
$$5$$
$$2$$
$$2$$
Report Question
0%
$$38.5$$
0%
$$36.98$$
0%
$$38.33$$
0%
$$39.00$$
Explanation
Median is the middle most value of a given series that represents the whole class of the series. For a group data,
Median =
L +
[{
(n/2) – B}/
G
]
× w where
L
is the lower class boundary of the group containing the median
n
is the total number of values
B
is the cumulative frequency of the groups before the median group
G
is the frequency of the median group
w
is the group width
Since the median is the middle value, which in this case is the 10
th
one, which is in the 30-40 group. Therefore , 30-40 is the median group so
L
= 30
n
= 20
B
= 2 + 3 = 5
G
= 6
w
= 10
Median
= 30 + [{
(20/2) – 5}/
6]
× 10
= 30 + 8.33
=
38.33
The median of $$14, 12, 10, 9, 11$$ is
Report Question
0%
$$11$$
0%
$$10$$
0%
$$9.5$$
0%
$$10.5$$
Explanation
Median = Arranging no. in increasing order
$$9,10,11,12,14$$
Median = $$\dfrac{n+1}{2}=\frac{5+1}{2}$$
$$=3rd$$ term $$=11$$
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