Measures of Central Tendency Test

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Measures of Central Tendency Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The arithmetic mean of first ten natural numbers is

A
$$5.5$$

B
$$6$$

C
$$7.5$$

D
$$10$$

Solution

Formula used:
$$\text{Arithmetic mean}= \dfrac{\text{Sum of given numbers}}{\text{Total numbers}}$$

So, by using the above formula we get,
$$A.M=\dfrac{1+2+3+...+10}{10}$$
$$=\dfrac{55}{10}$$
$$=5.5$$

Hence, option $$A$$ is correct.
Question 2
1 / -0

In order to make the computation of the arithmetic mean of a set of $$50$$ numbers simpler each observation is subtracted from $$53$$ and the arithmetic mean of the set of differences is found to be $$-3.5$$. The arithmetic mean of the set of given numbers is

A
$$53.07$$

B
$$52.93$$

C
$$56.50$$

D
$$49.50$$

Solution

Mean is the sum of the values divided by the total number of values.

Given that arithmetic mean of the $$50$$ numbers when each observation is subtracted by $$53$$ is $$-3.5$$

Let the $$50$$ numbers be $$x_1,x_2,...,x_{50}$$

Therefore, $$-3.5=\dfrac{(x_1-53)+(x_2-53)+...+(x_{50}-53)}{50}$$

$$\implies {x_1+x_2+...+x_{50}-50\times53}{=-3.5\times50}$$

$$\implies x_1+x_2+...+x_{50}=2475$$ ------(1)

Mean of $$50$$ numbers is $$\dfrac{x_1+x_2+...+x_{50}}{50}$$

$$\implies mean = \dfrac{x_1+x_2+...+x_{50}}{50}$$

$$\implies mean = \dfrac{2475}{50}$$ (from (1))

$$\implies mean =49.5$$

Question 3

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Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	$$x$$	15	16	6

The missing frequency marked $$\displaystyle x$$ of the above distribution whose mean is 27 is :

$$7$$

$$8$$

$$9$$

$$10$$

Solution

Class	Frequency (f)	Class Mark (x)	fx
0-10	5	5	25
10-20	x	15	15x
20-30	15	25	375
30-40	16	35	560
40-50	6	45	270
Total	42 + x		1230 + 15x

Given, Mean $$ = \cfrac { \sum { fx } }{ \sum { f} } = 27 $$

$$ => \cfrac {1230 + 15x}{42 + x} = 27 $$

$$ => 1230 + 15x = 1134 + 27x $$

$$ 96 = 12x $$

$$ x = 8 $$

Question 4
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In a class of $$19$$ students, seven boys failed in a test. Those who passed scored $$12,15,17,15,16,15,19, 19, 17, 18, 18$$ and $$19$$ marks. The median score of the $$19$$ students in the class is

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution

The seven boys who failed the test scored the marks less than these given marks. Let the marks be $$\displaystyle { x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 },{ x }_{ 5 },{ x }_{ 6 },{ x }_{ 7 }$$
Then the marks arranged in ascending order are
$$\displaystyle { x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },{ x }_{ 4 },{ x }_{ 5 },{ x }_{ 6 },{ x }_{ 7 },12,15,15,15,16,17,17,18,18,19,19,19$$
$$\displaystyle \therefore \quad Median\quad score\quad =\left( \frac { 19+1 }{ 2 } \right) $$th item
$$\displaystyle =10$$th item=$$15$$.
Question 5
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The average income of a group of persons is $$\displaystyle \bar{x}$$ and that of another group is $$\displaystyle \bar{y}.$$ If the number of persons of both group are in the ratio 4 : 3, then average income of combined group is

A
$$\displaystyle \frac{\bar{x}+\bar{y}}{7}$$

B
$$\displaystyle \frac{3\bar{x}+4\bar{y}}{7}$$

C
$$\displaystyle \frac{4\bar{x}+3\bar{y}}{7}$$

D
None of these

Solution

Here weight ratio is, $$w_1 :w_2 =\dfrac{ 4}{3}$$ (i)
Hence required average
income of the combined group is $$=\cfrac{\sum w_ix_i}{\sum

w_i}=\cfrac{\bar{X}w_1+\bar{Y}w_2}{w_1+w_2}=\cfrac{\bar{X}\frac{w_1}{w_2}+\bar{Y}}{1+\frac{w_1}{w_2}}=\cfrac{4\bar{X}+3\bar{Y}}{7}$$

using (i)
Question 6
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If in a data, $$10$$ numbers arranged in increasing order. If the $$7$$th entry is increased by $$4,$$ then the median increases by

A
zero

B
$$4$$

C
$$6$$

D
$$5$$

Solution

The median would be average of the $$5^{th}$$ and $$6^{th}$$ observations. Since they are unaffected by increase in $$7^{th}$$ entry, the median will be unchanged.
Question 7
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Using empirical formula, calculate the mode for the following data:
$$17$$, $$16$$, $$25$$, $$23$$, $$22$$, $$23$$, $$28$$, $$25$$, $$25$$, $$23$$

A
$$14.61$$

B
$$19.6$$

C
$$23.6$$

D
$$28.3$$

Solution

Given data set is $$17,16,25,23,22,23,28,25,25,23$$

Arranging the data set in ascending order we get
$$16,17,22,23,23,23,25,25,25,28$$

Total number of values is $$10$$

Mean is $$\dfrac{16+17+22+23+23+23+25+25+25+28}{10}=\dfrac{227}{10}=22.7$$

Median is the middle value i.e., $$\dfrac{5^{th}+6^{th}}{2}$$ value is the median.

Therefore median is $$\dfrac{23+33}{2}=23$$

The empirical formula is $$mode =3median-2mean$$

$$\implies mode=3(23)-2(22.7)=69-45.4=23.6$$

Question 8

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If the mean of following frequency distribution is $$188$$ find the missing frequencies $$\displaystyle f_{1}\: and\: f_{2}$$

C.I.	0-80	80-160	160-240	240-320	320-400	Total
Freq.	20	25	$$\displaystyle f_{1}$$	$$\displaystyle f_{2}$$	10	100

$$\displaystyle f_{1}$$ = 51, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 20, $$\displaystyle f_{2}$$ = 40

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 20

Solution

Given : Mean = 188

C.L.	$$\displaystyle f_{i}$$	$$\displaystyle x_{i}$$	$$\displaystyle f_{i}x_{i}$$
0 - 80	20	40	800
80 - 160	25	120	3000
160 - 240	$$\displaystyle f_{1}$$	200	200 $$\displaystyle f_{1}$$
240 - 320	$$\displaystyle f_{2}$$	280	280 $$\displaystyle f_{2}$$
320 - 400	10	360	3600
Total	$$\displaystyle \sum f_{i}=100=55+f_{1}+f_{2}$$		$$\displaystyle \sum x_{i}f_{i}=7400+200f_{1}+2800f_{2}$$

$$\displaystyle \overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\: \: \Rightarrow 188=\frac{7400+200f_{1}+280f_{2}}{100}$$
$$\displaystyle 18800=7400+200f_{1}+280f_{2}\Rightarrow 18800-7400=200f_{1}+280f_{2}$$
$$\displaystyle 11400=200f_{1}+280f_{2}\Rightarrow 1140=20f_{1}+28f_{2}$$
$$\displaystyle 285=5f_{1}+7f_{2}............(i)$$
Also $$\displaystyle f_{1}+f_{2}+55=100\: \: \Rightarrow f_{1}+f_{2}=100-55\: \: \Rightarrow f_{1}+f_{2}=45....................(ii)$$
Solving equation (i) and (ii) $$\displaystyle f_{1}=15,f_{2}=30$$

Question 9

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Consider the table given below

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of Students	12	18	27	20	17	6

The arithmetic mean of the marks given above is

$$18$$

$$28$$

$$27$$

$$6$$

Solution

$$ x_i = \cfrac{\text{upper limit + lower limit}}{2} $$

Marks	No. of students $$ \left( f_i \right) $$	$$ x_i $$	$$ x_i f_i $$
0-10	12	5	60
10-20	18	15	270
20-30	27	25	675
30-40	20	35	700
40-50	17	45	765
50-60	6	55	330
	$$ \Sigma{f_i} = 100 $$		$$ \Sigma{x_i f_i} = 2800 $$

$$ \therefore \; Mean=\cfrac { \Sigma { x_{ i }f_{ i } } }{ \Sigma { f_{ i } } } = \cfrac{2800}{100} = 28$$

Hence, the option $$(B)$$ is correct

Question 10
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A distribution consists of three components with frequencies $$45, 40$$ and $$15$$ having their means $$2, 2.5, 2$$ respectively. The mean of the combined distribution is

A
$$2.1$$

B
$$2.2$$

C
$$2.3$$

D
$$2.4$$

Solution

Given, $$f_1 = 45$$, $$f_2 = 40$$, $$f_3 = 15$$, $$x_1 = 2$$, $$x_2 = 2.5$$ and $$x_3 = 2$$
Now, mean $$=$$ $$\dfrac{x_1 f_1 + x_2 f_2 + x_3 f_3}{f_1 + f_2 + f_3}$$
$$=$$ $$\dfrac{45 \times 2 + 40 \times 2.5 + 15 \times 2}{45 + 40 + 15}$$
$$=$$ $$\dfrac{90 + 100 + 30}{100}$$
$$=$$ $$\dfrac{220}{90} = 2.2$$

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Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 32

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Measures of Central Tendency Test - 32

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SHARING IS CARING

Question 1

$$5.5$$

$$6$$

$$7.5$$

$$10$$

Solution

Question 2

$$53.07$$

$$52.93$$

$$56.50$$

$$49.50$$

Solution

Question 3

$$7$$

$$8$$

$$9$$

$$10$$

Solution

Question 4

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 5

$$\displaystyle \frac{\bar{x}+\bar{y}}{7}$$

$$\displaystyle \frac{3\bar{x}+4\bar{y}}{7}$$

$$\displaystyle \frac{4\bar{x}+3\bar{y}}{7}$$

None of these

Solution

Question 6

zero

$$4$$

$$6$$

$$5$$

Solution

Question 7

$$14.61$$

$$19.6$$

$$23.6$$

$$28.3$$

Solution

Question 8

$$\displaystyle f_{1}$$ = 51, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 20, $$\displaystyle f_{2}$$ = 40

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 30

$$\displaystyle f_{1}$$ = 15, $$\displaystyle f_{2}$$ = 20

Solution

Question 9

$$18$$

$$28$$

$$27$$

$$6$$

Solution

Question 10

$$2.1$$

$$2.2$$

$$2.3$$

$$2.4$$

Solution

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