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CBSE Questions for Class 11 Commerce Economics Measures Of Central Tendency Quiz 12 - MCQExams.com
CBSE
Class 11 Commerce Economics
Measures Of Central Tendency
Quiz 12
The ages of the $$115$$ people who live on an apartment are grouped as follows$$:$$
The mean length of the plants is $$33.43$$years using direct method. Find y in the table.
Age(years)
Number of people
$$0-10$$
$$10$$
$$10-20$$
$$15$$
$$20-30$$
$$26$$
$$30-40$$
Y
$$40-50$$
$$23$$
$$50-60$$
$$16$$
$$60-70$$
$$3$$
$$70-80$$
$$1$$
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$$23$$
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$$28$$
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$$15$$
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$$21$$
Explanation
Length(cm)
No. of plants$$(f_i)$$
Mid point $$x_i$$
$$f_ix_i$$
$$0-10$$
$$10$$
$$5$$
$$10\times 5=50$$
$$10-20$$
$$15$$
$$15$$
$$15\times 15=225$$
$$20-30$$
$$26$$
$$25$$
$$26\times 25=650$$
$$30-40$$
$$y$$
$$35$$
$$y\times 35=35y$$
$$40-50$$
$$23$$
$$45$$
$$23\times 45=1035$$
$$50-60$$
$$16$$
$$55$$
$$16\times 55=880$$
$$60-70$$
$$3$$
$$65$$
$$3\times 65=195$$
$$70-80$$
$$1$$
$$75$$
$$1\times 75=75$$
$$\Sigma f_i=115$$ $$\Sigma f_ix_=3110+35y$$
Arithmetic mean using direct method formula is
$$\overline{x}=\displaystyle\frac{\Sigma f_i x_i}{\Sigma f_i}$$
$$33.43 = (3110 + 35y)/115$$
$$433.43 \times 115 = 3110 + 35y$$
$$3844.45 - 3110 = 35y$$
$$y = 734.45/35$$
$$y = 20.9$$
Therefore, approximately the value of $$y$$ is $$21.$$
Using step deviation method find the mean.
X
20-40
40-60
60-80
80-100
frequency
4
8
12
16
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$$40$$
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$$50$$
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$$60$$
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$$70$$
Explanation
X
Mid point (X)
Frequency (F)
i=class interval width
A=50 Assumed Means
u=$$\dfrac{x-A}{i}$$
fu
20-40
30
4
20
50
-1
-4
40-60
50=A
8
20
50
0
0
60-80
70
12
20
50
1
12
80-100
90
16
20
50
2
32
$$\Sigma f=40$$
$$\Sigma f u=40$$
The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u = \frac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}$$ = arithmetic mean
$$\overline {X} =50 +\dfrac{40}{40}\times 20$$
$$= 50 + 20$$
$$= 70$$
Find the measures of central tendency fro the data set $$2, 4, 5, 1, 7, 2, 3$$.
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mean $$= 3.42$$, median $$= 3$$ and mode $$= 2$$
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mean $$= 3$$, median $$= 3$$ and mode $$= 2$$
0%
mean $$= 3.42$$, median $$= 3$$ and mode $$= 3$$
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mean $$= 3$$, median $$= 3$$ and mode $$= 3$$
Explanation
Mean, median and mode of a data set are the measures of central tendency.
Step 1: Mean $$=$$ $$\dfrac{\sum X}{n}$$
$$\sum X$$ $$=$$ sum of the data values.
$$n =$$ number of the data values
Mean $$=$$ $$\dfrac{2 + 4+5+1+7+2+3}{7}=3.42$$
Step 2: The data set in the ascending order is $$1, 2, 2, 3, 4, 5, 7 $$
So, Median of the set is $$3$$. [Median is the middle data value of the ordered set.]
Step 3: Mode is the data value(s) that appear most often in the data set. So, the modes of the data set are $$2$$.
So, the measures of central tendency of the given set of data are mean $$= 3.42$$, median $$= 3$$ and mode $$= 2$$.
The measure of central tendency of mean is affected by
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variables
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distributions
0%
outliers
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range
Explanation
The measure of Central tendency of mean is affected by $$outlers$$ ,
because outliers are observation that lies on abnormal distance from
other values in a random sample data.
If the data set is perfectly normal, the mean, median and mode
are
_____to each other.
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Equal
0%
Unequal
0%
Similar
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Not comparable
Explanation
Since , when data is Normal it will be in $$bell shape$$ and also
$$symmetrical$$. In that case $$Median$$ , $$Mean$$ & $$Mode$$ of data are $$Equal$$.
When the mean is the best measure of central tendency?
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continuous and symmetrical data
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discontinuous and symmetrical data
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skewed data only
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continuous data only
Explanation
$$\Rightarrow$$ The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed.
$$\Rightarrow$$ However, it all depends on what you are trying to show from your data.
$$\Rightarrow$$ The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$$\overline{x}=\dfrac{x_1+x_2+x_3+...x_n}{n}$$
If the average (arithmetic mean) of $$t$$ and $$(t+2)$$ is $$x$$ and if the average of $$t$$ and $$(t-2)$$ is $$y$$, what is the average of $$x$$ and $$y$$?
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$$1$$
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$$\cfrac{t}{2}$$
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$$t$$
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$$t+\cfrac{1}{2}$$
Explanation
Given, the average (arithmetic mean) of $$t$$ and $$t+2$$ is $$x$$ and the average of $$t$$ and $$ t−2$$ is $$y$$
Then $$x=\dfrac{t+(t+1)}{2}=\dfrac{2t+1}{2}$$
And $$y=$$ $$y=\dfrac{t+(t-1)}{2}=\dfrac{2t-1}{2}$$
Then average $$x$$ and $$y =$$ $$\dfrac{x+y}{2}=\dfrac{\frac{2t+1}{2}+\frac{2t-1}{2}}{2}=\dfrac{2t+1+2t-1}{2\times 2}=\dfrac{4t}{4}=t$$
What is the weighted arithmetic mean of the first $$n$$ natural numbers whose weights are equal to the corresponding numbers?
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$$\dfrac { 1 }{ 2 } \left( n+1 \right) $$
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$$\dfrac { 1 }{ 2 } \left( n+2 \right) $$
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$$\dfrac { 1 }{ 3 } \left( n+1 \right) $$
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$$\dfrac { 1 }{ 3 } \left( 2n+1 \right) $$
Explanation
Weight Arithmetic mean $$= \cfrac { 1*1+2*2+3*3+....{ n }^{ 2 } }{ 1+2+3+....n } $$
$$=\cfrac { n(2n+1)(n+1) }{ 6*\cfrac { n(n+1) }{ 2 } } \\ =\cfrac { 2n+1 }{ 3 } \\ =\cfrac { 1 }{ 3 } (2n+1)$$.
X, Y, Z are three sets of values.
Which of the following statements is true?
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Mean of X is equal to Mode of Z.
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Mean of Z is equal to Mode of Y.
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Median of Y is equal to Mode of X.
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Mean, Median and Mode of X are same.
Explanation
Arranging the data ascending order we get
$$X:1,2,2,3,3,3,7$$; $$Y:3,5,5,7,8,9,12$$ and $$Z:2,3,4,4,4,7,11$$
Mean of $$X=\cfrac{1+2+2+3+3+3+7}{7}=3$$
Mean of $$Y=\cfrac{3+5+5+7+8+9+12}{7}=7$$
Mean of $$Z=\cfrac{2+3+4+4+47+11}{7}=5$$
Mode of $$X=3$$
Mode of $$Y=5$$
Mode of $$Z=4$$
Median of $$X=3$$
Median of $$Y=7$$
Median of $$Z=4$$
The table below shows the number of cars Jing sold each month last year. What is the median of the data in the table?
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$$13$$
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$$16$$
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$$19$$
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$$20.5$$
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$$23.5$$
Explanation
Number of cars sold are $$25,15,22,19,16,13,19,25,26,27,28,29$$
In ascending order, $$13,15,16,19,19,22,25,25,26,27,28,29$$
Middle most values are $$22$$ and $$25$$
Median is a middle most value or average of two middle most values.
$$\Rightarrow M=\dfrac {(22+25)}{2}=23.5$$
The average (arithmetic mean) of $$t$$ and $$y$$ is $$15$$, and the average of $$w$$ and $$x$$ is $$15$$. What is the average of $$t, w, x$$ and $$y$$?
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$$7.5$$
0%
$$15$$
0%
$$22.5$$
0%
$$30$$
Explanation
Given,
Average of $$t$$ and $$y$$ $$=$$ $$15$$
$$\Rightarrow \dfrac {t \space + \space y}{2}$$ $$=$$ $$15$$
$$\Rightarrow t$$ $$+$$ $$y$$ $$=$$ $$15$$ $$\times$$ $$2$$
$$\Rightarrow t$$ $$+$$ $$y$$ $$=$$ $$30$$
Average of$$w$$ and $$x$$ $$=$$ $$15$$
$$\Rightarrow \dfrac {w \space + \space x}{2}$$ $$=$$ $$15$$
$$\rightarrow w$$ $$+$$ $$x$$ $$=$$ $$15$$ $$\times$$ $$2$$
$$\Rightarrow w$$ $$+$$ $$x$$ $$=$$ $$30$$
Now, a
verage of $$t$$, $$y$$, $$w$$ and $$x$$ $$=$$ $$\dfrac {(t \space + \space y \space + \space w \space + \space x)}{4}$$
$$=$$ $$\dfrac {(t \space + \space y) \space + \space (w \space + \space x)}{4}$$
$$=$$ $$\dfrac {30 \space + \space 30}{4}$$
$$=$$ $$\dfrac {60}{4}$$
$$=$$ $$15$$
Therefore, average of $$t$$, $$y$$, $$w$$ and $$x$$ is $$'15'$$.
The mode of the observations $$5, 4, 4, 3, 5, x, 3, 4, 3, 5, 4, 3,$$ and $$5$$ is $$3$$. What is the median?
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$$3$$
0%
$$4$$
0%
$$5$$
0%
$$4.5$$
Explanation
Since frequency of mode $$(3)$$ should be maximum $$x=3$$
Now, arranging data in ascending order we get
3,3,3,3,3,4,4,4,4,,5,5,5,5
Median = Middle term $$=4$$
The mean of $$48, 27, 36, 24, x$$, and $$2x$$ is $$37.x =$$
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$$13\dfrac {1}{3}$$
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$$16\dfrac {2}{3}$$
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$$29$$
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$$33\dfrac {3}{4}$$
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$$40$$
Explanation
Mean $$=$$ $$\dfrac{sum \quad of \quad no.}{total \quad no}$$
$$\Rightarrow 37=\dfrac{48+27+36+24+x+2x}{6}$$
$$\Rightarrow 135+3x=222$$
$$\Rightarrow 3x=222-135$$
$$\Rightarrow 3x=87$$
$$\Rightarrow x=\dfrac{87}{3}=29$$
Given, $$10, 18, 4, 15, 3, 21, x $$
If $$x$$ is the median of the $$7$$ numbers listed above, which of the following could be the value of $$x$$?
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$$5$$
0%
$$8$$
0%
$$9$$
0%
$$14$$
0%
$$16$$
Explanation
Given numbers
$$10,$$ $$18,$$ $$4,$$ $$15,$$ $$3,$$ $$21,$$ $$x$$
Median is the middle value in the list of numbers. To find the median, the numbers have to be in the numerical order.
As $$x$$ is the given median, $$x$$ should be placed in the middle of the list in numerical order. The list contains $$7$$ numbers. So, $$x$$ should be placed in the $${4}^{th}$$ position.
Rewriting the list in increasing order,
$$3,$$ $$4,$$ $$10,$$ $$x,$$ $$15,$$ $$18,$$ $$21$$
So, $$x$$ should lie between $$10$$ and $$15$$.
Possible values of $$x$$, $$10$$ $$\leq$$ $$x$$ $$\leq$$ $$15$$
From the given options,
Only $$14$$ satisfies the above condition.
Therefore, the value of $$x$$ is $$14$$.
The average of six numbers isIf the average of two of those numbers is 2, what is the average of the other four numbers?
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5
0%
6
0%
7
0%
8
0%
9
The median of $$2, 7, 4, 8, 9, 1$$ is
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$$4$$
0%
$$6$$
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$$5.5$$
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$$7$$
Explanation
$$1,2,4,7,8,9$$
No. of terms $$=6$$
Median $$=\dfrac{6+1}{2}=3.5$$ term
(4)th Mean = Median $$\dfrac{4+7}{2}=5.5$$
For a collection of $$11$$ items, sum of all items is $$132$$, then the arithmetic mean is.
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$$11$$
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$$12$$
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$$14$$
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$$13$$
The mode of the data $$5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4$$ is
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$$2$$
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$$3$$
0%
$$4$$
0%
$$5$$
Explanation
$$\textbf{Step 1: Find the mode.}$$
$$\text{The given observations are : } 5, 5, 5, 5, 5, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4$$
$$\text{In any series of observations, mode is the observation which appears the maximum}$$
$$\text{number of times.} $$
$$\text{Given data is 5,5,5,5,5,1,2,2,3,3,3,4,4,4,4 }$$
$$\text{In the given observations, 5 appears the maximum number of times. } $$
$$\text{So, the mode is 5.} $$
$$\textbf{Hence, the correct option is D.}$$
A random sample of 20 people is classified in the following table according to their ages
Age
Frequency
15-25
2
25-35
4
35-45
6
45-55
5
55-65
3
What is the mean age of this group of people?
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$$41.0$$
0%
$$41.5$$
0%
$$42.0$$
0%
$$42.5$$
Explanation
$$Mean=\ \dfrac{\sum { (frequency\times Mid-Point)_i }}{\sum{(frequency)_i}} $$
$$Mean=\ \dfrac{830}{20}=41.5$$
The following observations have been arranged in the ascending order. If the median of the data $$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$$ is $$63$$, then the value of $$x$$ is ___________.
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$$63$$
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$$62$$
0%
$$61$$
0%
$$60$$
Explanation
Given data is $$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$$
Also given median of data $$=63$$.
Since, total number of observation are even.
So,
$$\text{median}=\dfrac{x+(x+2)}{2}$$
$$=\dfrac{2x+2}{2}=x+1$$
Thus $$63=x+1$$
$$\Rightarrow x=62$$
What is the mode for the data 20, 20, 20,21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 25 ?
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7
0%
21
0%
22
0%
25
Explanation
Mode is the value that appears in a given data more often than the others. If two numbers appear maximum no of times we take average of both.
So no of 20's=3
no of 21's=5
no of 22's=7
no of 23's=5
no of 24's=2
no of 25's=1
Since 22 appear more no of times, mode of given data is 22.
Find the mean of the first five prime numbers.
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3.3
0%
7.8
0%
5.6
0%
12.1
Explanation
First five prime numbers are $$2, 3, 5, 7$$ and $$11.$$
Number of observations $$=5$$
$$Mean=\dfrac{Sum\,of\,the\,numbers}{No.\,of\,observations}$$
$$=\dfrac{2+3+5+7+11}{5}$$
$$=\dfrac{28}{5}$$
$$=5.6$$
The mean of the values $$0, 1, 2, ..... , n$$ having corresponding weight $$^nC_0, ^nC_1, ^nC_2, ..... , ^nC_n$$ respectively, is?
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$$\displaystyle\frac{2^n}{(n+1)}$$
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$$\displaystyle\frac{2^{n+1}}{n(n+1)}$$
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$$\displaystyle\frac{n+1}{2}$$
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$$\displaystyle\frac{n}{2}$$
Explanation
The solution mean is $$\bar{x}=\displaystyle\frac{0\cdot 1+1\cdot \, ^nC_1+2\cdot \,^nC_2+3\cdot \,^nC_3+....+n\cdot \,^nC_n}{1+\,^nC_1+\,^nC_2+....+\,^nC_n}$$
$$=\frac{\displaystyle \sum_{r=0}^n\,r\cdot\, ^nC_r}{\displaystyle \sum_{r=0}^n\,^nC_r} = \frac{\displaystyle \sum_{r=1}^n \,r\cdot \frac{n}{r}\,\cdot \,^{n-1}C_{r-1}}{\displaystyle \sum_{r=0}^n\,^nC_r}$$
$$=\frac{n\displaystyle \sum_{r=1}^n\,^{n-1}C_{r-1}}{\displaystyle \sum_{r=0}^n \, ^nC_r}$$
$$=\displaystyle\frac{n\cdot 2^{n-1}}{2^n}=\frac{n}{2}$$
The 'less than' ogive curve and the 'more than' ogive curve intersect at.
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0%
Median
0%
Mode
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Arithmetic mean
0%
None of the above
Explanation
The less than ogive curve gives cumulative frequency (probability) for $$x <= a.$$
The more than ogive curve gives cumulative frequency (probability) for $$x >= a.$$
They intersect exactly at the median, as cumulative frequency will be % of the total at median.
The 'less than' ogive curve and the 'more than' ogive curve intersect at Median.
Option A is correct Answer.
A variate X takes values 2, 9, 3, 7, 5, 4, 3, 2,What is the median?
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2
0%
4
0%
7
0%
9
Explanation
The data point which is in middle of ascending or descending series is called median.
So, for data $$2,9,3,7,5,4,3,2,10$$
Arranging in ascending order $$2,2,3,3,4,5,7,9,10$$
So, middle value is $$4.$$
Median of the given data set is $$4.$$
Which of the following are positional measure of dispersion?
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Standard Deviation, Variance
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Percentile, Variance
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Quartile, Variance
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Percentile,Quartile
Explanation
Percentile,Quartile are positional measure of dispersion because it tells about the position of a particular data value has within a data set.
Standard deviation, Variance are computational measure of dispersion.
Which of the following is true?
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The arithmetic mean is affected by the presence of extreme values in the data.
0%
Median is that positional value of the variable which divides the distribution into two equal parts
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The median is not affected by the presence of extreme values in the data.
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All of these
Explanation
Arithmetic mean refers to the average amount in a given group of data. So arithmetic mean can be calculated by adding the first term and the last term of the series and then dividing it by 2. So arithmetic mean is affected by the presence of extreme values in the data.
Median is the middle most value of a given series that represents the whole class of the series.So since it is a positional average, it is calculated by observation of a series and not through the extreme values of the series which. Therefore, median is not affected by the extreme values of a series.
The mean, median and mode respectively of the numbers $$7, 4, 3, 5, 6, 3, 3, 2, 4, 3, 4, 3, 3, 4, 4, 3, 2, 2, 4, 3, 5, 4, 3, 4, 3, 4, 3, 1, 2, 3$$ are __________.
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$$3.47, 3, 3$$
0%
$$3, 3, 3$$
0%
$$4, 3, 3$$
0%
$$5, 4, 3$$
Explanation
Sum of numbers $$=7+4+3+5+6+3+3+2+4+3+4+3+3+4+4+3+2+2+4+3$$
$$+5+4+3+4+3+4+3+1+2+3$$
$$=104$$
Total numbers are $$30$$
$$\text{mean}= \dfrac{104}{30}=3.46667=3.47$$
The mode is the number which appeared most.
Therefore, $$\text{mode}=3$$
For median arrange the numbers in ascending order:
$$1,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,5,5,6,7$$
Median $$=\dfrac{(15^{th}+16^{th}) \text{term}}{2}$$
$$=\dfrac{3+3}{2}=3$$
In how many ways can 5 persons be seated at a round table so that two particular persons will be together?
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0%
12
0%
120
0%
720
0%
180
Explanation
The number of ways in which n number of people can arrange themselves at a round table is (n-1)! ways.
The number of ways in which 5 people can arrange themselves at a round table is 4! ways. But considering 2 persons a single unit, t
he number of ways in which 5 people can arrange themselves at a round table is 3! ways. Now those two people can arrange themselves in 2! ways. So the answer will be 3! * 2! = 12 ways.
The sum of the deviations of a set of values $$x_1, x_2$$, ...... $$x_n$$ measured from $$50$$ is $$-10$$ and the sum of deviations of the values from $$46$$ is $$70$$. The mean is ___________.
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$$49$$
0%
$$49.5$$
0%
$$49.75$$
0%
$$50$$
Explanation
Sum of deviations from $$50$$ is $$-10$$
$$\Rightarrow \sum(xi - 50) = -10$$
$$\sum x_i - 50\sum1 = -10$$
$$\sum x_i - 50n = -10$$
$$\therefore y-50n=-10.....(1)$$
Sum of deviations from $$46$$ is $$70$$
$$\Rightarrow \sum(x_i - 46) = 70$$
$$\sum x_i - 46\sum1 = 70$$
$$\sum x_i - 46n = 70$$
$$\therefore y-46n=70.....(2)$$
Solving $$(1)$$ and $$(2)$$, we get
$$4n = 80$$ i.e. $$n=20$$
Putting value of $$n$$ in $$(1)$$, we get
$$y=990$$
Mean $$= \dfrac{\sum x_i}{n} = \dfrac{y}{n} = \dfrac{990}{20} = 49.5$$
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