Measures of Central Tendency Test

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Measures of Central Tendency Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Find the first quartiles for the following data.
$$8, 11, 20, 10, 2, 17, 15, 5, 16, 15, 25, 6, 5$$

A
$$2$$

B
$$6$$

C
$$8$$

D
$$5$$

Solution

First arrange the data set in order.
$$2, 5, 5, 6, 8, 10, 11, 15, 15, 16, 17, 20, 25$$
Here the median is the middle number $$= 11$$
So, the first quartile is the median of the lower half of the data.
Therefore, the first quartile is $$5$$.
Question 2
1 / -0

Find the weighted mean of marks obtained by a student in certain examination if he has secured $$75,80,67,90$$ marks in English, Tamil, Mathematics, social Science respectively, having corresponding weights $$3, 2, 2, 1.$$

A
$$75.16$$

B
$$76.12$$

C
$$62.17$$

D
$$78.12$$

Solution

Subject
Marks obtained
Weights $$w_i$$
$$x_iw_i$$
English
$$75$$
$$3$$
$$225$$
Tamil
$$80$$
$$2$$
$$160$$
Mathematics
$$67$$
$$2$$
$$134$$
Social science
$$90$$
$$1$$
$$90$$

$$\Sigma w_i=8$$
$$\Sigma x_iw_i=609$$
Weighted Arithmetic mean $$=$$ $$\displaystyle\dfrac {\Sigma x_iw_i}{\Sigma w_i}$$ $$=609/8$$
Weighted Arithmetic mean $$= 76.125$$

Question 3

1 / -0

Marks	10-20	20-30	30-40	40-50	50-60
Frequency	4	10	20	40	50

The following frequency distribution shows the marks obtained by $$124$$ students in Economy at a certain school. Find the arithmetic mean using the direct method.

$$50$$ marks

$$80$$ marks

$$30$$ marks

$$45$$ marks

Solution

Answer:- Direct Method:

Marks	Frequency	$$x_i$$	$$f_i x_i$$
$$10-20$$	$$4$$	$$15 $$	$$60$$
$$20-30$$	$$10 $$	$$25$$	$$250$$
$$30-40$$	$$20$$	$$35$$	$$700$$
$$ 40-50$$	$$40$$	$$45$$	$$1800 $$
$$50-60$$	$$50 $$	$$55$$	$$2750$$
	$$\Sigma f_i = 124$$		$$\Sigma f_i x_i = 5560$$

$$Mean = \cfrac{\Sigma f_i x_i}{\Sigma f_i} $$

$$= \cfrac{5560}{124} =44.83 \simeq 45marks.$$

Question 4
1 / -0

During the first period in the class, Elsa's math quiz scores were $$90, 92, 93, 88, 95, 88, 97, 87$$ and $$98$$. What was the median quiz score?

A
$$90$$

B
$$91$$

C
$$88$$

D
$$92$$

Solution

The median of a set of data is the middlemost number in the set.
So, first arrange the data in order.
$$87, 88, 88, 90, 92, 93, 95, 97, 98$$
The median quiz score is $$92$$.
Question 5
1 / -0

Find the mode of the following data
$$ 58, 35, 54, 32, 54, 58, 34, 35, 35, 32, 58, 35, 35, 89, 58, 32, 32, 89.$$

A
$$35$$

B
$$58$$

C
$$32$$

D
$$89$$

Solution

Given, $$ 58, 35, 54, 32, 54, 58, 34, 35, 35, 32, 58, 35, 35, 89, 58, 32, 32, 89.$$
The mode is the value which appears most often. In the above data the value which appears most often is $$35$$, since it appears $$5$$ times. \
So, mode is $$35$$.
Question 6
1 / -0

Choose the formula used for arithmetic mean of grouped data by Assumed mean method is

A
$$\bar { x } = A - \dfrac { \sum { fd } }{ \sum { f } } $$

B
$$\bar { x } = A + \dfrac { \sum { fd } }{ \sum { f } } $$

C
$$\bar { x } = A \times \dfrac { \sum { fd } }{ \sum { f } } $$

D
$$\bar { x } = { A } / { \dfrac { \sum { fd } }{ \sum { f } } }$$

Solution

The formula used for arithmetic mean of grouped data by shortcut method is $$\bar { x } = A + \dfrac { \sum { fd } }{ \sum { f } } $$
$$A =$$ Assumed mean of the given data
$$\sum { f } =$$ Summation of the frequencies given in the grouped data
$$\sum { fd } =$$ Summation of the frequencies and deviation of a given mean data
$$d =$$ deviation of a mean data
$$\bar { x } =$$ arithmetic mean
Question 7
1 / -0

Identify in which ideal measure of central tendency used to find the middle value, if the data are ordinal?

A
mean

B
median

C
mode

D
range

Solution

Median is the ideal measure of central tendency used to find the middle value, if the data are ordinal.
Example: Consider the data: $$1, 2, 3, 4, 5, 5, 6, 3, 3, 2, 2, 1, 2 $$
First arrange the data in ordinal.
So, $$1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 6$$
So, $$3$$ is the median.
Median is the middle number.
Question 8
1 / -0

Which ideal measure of central tendency used to find the middle value, if the data are categorical?

A
mean

B
median

C
mode

D
range

Solution

Mode is the ideal measure of central tendency used to find the middle value, if the data are categorical.
Example: Discrete data: $$1, 2, 3, 4, 5, 5, 6, 3, 3, 2, 2, 1, 2$$
Mode is the most occurring value.
Here $$2$$ is the mode.
Question 9
1 / -0

The lower quartile is the value of the middle of the first set, where $$25\%$$ of the values are _____ than $$Q_1$$ and $$75\%$$ are larger.

A
longer

B
larger

C
smaller

D
higher

Solution

The lower quartile is the value of the middle of the first set, where $$25\% $$ of the values are smaller than $$Q_1$$ and $$ 75\%$$ are larger.
Example: $$2, 5, 5, 6, 8, 10, 11$$
Here the median is the middle number $$= 6$$
So, the first quartile is the median of the lower half of the data.
Then, the lower quartile of the first half will be $$2, 5, 5$$ which is $$25\%$$ of the given data
So, the remaining $$75\%$$ numbers can be considered as larger.
Question 10
1 / -0

The ______ is the median of of the lower half of the data set.

A
quartile

B
upper quartile

C
range

D
lower quartile

Solution

The lower quartile is the median of of the lower half of the data set.
Example: consider the data set: $$10, 20, 30, 40, 50, 60, 70, 80, 90, 100$$
The median of the lower half of the data set is $$10, 20, 30, 40, 50$$
So, the lower quartile $$= 30$$.

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Subject	Marks obtained	Weights $$w_i$$	$$x_iw_i$$
English	$$75$$	$$3$$	$$225$$
Tamil	$$80$$	$$2$$	$$160$$
Mathematics	$$67$$	$$2$$	$$134$$
Social science	$$90$$	$$1$$	$$90$$
		$$\Sigma w_i=8$$	$$\Sigma x_iw_i=609$$

Measures of Central Tendency Test - 17

Result

Measures of Central Tendency Test - 17

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Question 1

$$2$$

$$6$$

$$8$$

$$5$$

Solution

Question 2

$$75.16$$

$$76.12$$

$$62.17$$

$$78.12$$

Solution

Question 3

$$50$$ marks

$$80$$ marks

$$30$$ marks

$$45$$ marks

Solution

Question 4

$$90$$

$$91$$

$$88$$

$$92$$

Solution

Question 5

$$35$$

$$58$$

$$32$$

$$89$$

Solution

Question 6

$$\bar { x } = A - \dfrac { \sum { fd } }{ \sum { f } } $$

$$\bar { x } = A + \dfrac { \sum { fd } }{ \sum { f } } $$

$$\bar { x } = A \times \dfrac { \sum { fd } }{ \sum { f } } $$

$$\bar { x } = { A } / { \dfrac { \sum { fd } }{ \sum { f } } }$$

Solution

Question 7

mean

median

mode

range

Solution

Question 8

mean

median

mode

range

Solution

Question 9

longer

larger

smaller

higher

Solution

Question 10

quartile

upper quartile

range

lower quartile

Solution

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