Measures of Central Tendency Test

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Measures of Central Tendency Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Mean of $$9$$ observations was founded to be $$35$$. Later on, it was detected that an observation $$81$$ was misread as $$18$$, then the correct mean of the observations is ____________.

A
$$40$$

B
$$41$$

C
$$42$$

D
$$43$$

Solution

Mean of $$9$$ observations $$= 35$$
Sum of $$9$$ observations $$= 35\times9 = 315$$

Since $$81$$ was misread as $$18$$
New sum $$= 315-18+81= 378$$

The correct mean $$= \dfrac{378}{9} = 42$$
Question 2
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The median of the observations $$30, 91, 0, 64, 42, 80, 30, 5, 117, 71$$ is __________.

A
$$52$$

B
$$51$$

C
$$53$$

D
$$52.5$$

Solution

Arranging the given numbers in ascending order
$$0, 5,30,30,42,64,71,80,91,117$$
Median of the above numbers is:
$$=\dfrac{42+64}{2}$$
$$=\dfrac{106}{2}=53$$
Question 3
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The mode of the following numbers is-
$$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$$.

A
$$14$$

B
$$16$$

C
$$19$$

D
$$15$$

Solution

Number Frequency
14 4
15 5
16 1
17 1
18 1
19 2
20 1
Mode is the number in a set that occurred mostly or present in maximum number in that set of data
Frequency of the number $$15$$ is maximum
so mode is $$15$$
Question 4
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The ______ is the middle element when the data set is arranged in order of the magnitude.

A
median

B
mean

C
mode

D
quartile

Solution

The median is the middle value of the data set (that is arranged in ascending order). It is the value that divides a data set into two equal parts.

In an individual data set it is calculated by using the formula:
(i) If the data has an odd number of figures
$$\text{Median} = \dfrac{N+1}{2}$$ th value

(ii) If the data has an even number of figures
$$\text{Median} = \dfrac{N}{2}$$ th value
Question 5
1 / -0

The mean salary paid per week to $$1000$$ employees of an establishment was found to be Rs. $$900$$. Later on, it was discovered that the salaries of two employees were wrongly recorded as Rs. $$750$$ and Rs. $$365$$ instead of Rs. $$570$$ and Rs. $$635$$. Find the corrected mean salary.

A
$$900.90$$

B
$$1,115$$

C
$$1,225$$

D
$$900.09$$

Solution

We first find the corrected sum of observation.
Corrected sum of observations$$=$$(Sum of total incorrect observation)$$-$$(Sum of incorrect data)$$+$$(Sum of correct data).
$$=900\times 1,000-(750+365)+(570+635)=9,00,090$$
$$\therefore$$ Corrected Mean $$=Rs.\dfrac{9,00,090}{1,000}=Rs.900.09$$
Question 6
1 / -0

Mode of $$2,4,6,3,4,3,3,4,4,2$$ will be:

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

The mode is the value which appears most often in the data.
In this case, $$4$$ is appearing most often, therefore $$4$$ is the mode.
Question 7
1 / -0

Find the arithmetic mean of numbers from $$1$$ to $$9$$.

A
$$9$$

B
$$5$$

C
$$8$$

D
$$3$$

Solution

The numbers from $$1$$ to $$9$$ are $$1,2,3,4,5,6,7,8,9$$
$$ A.M. =\dfrac{\text{Sum of observations}}{\text{Number of observations}}= \dfrac {1+2+3+4+5+6+7+8+9}{9} = \dfrac {45}{9} = 5 $$

Question 8

1 / -0

Find the median of the following data.

Age greater than (Years)	No. of Persons
$$0$$	$$230$$
$$20$$	$$218$$
$$30$$	$$200$$
$$40$$	$$165$$
$$50$$	$$123$$
$$60$$	$$73$$
$$70$$	$$288$$

$$31.5$$

$$41.5$$

$$41.6$$

$$31.6$$

Solution

Note that it is a greater than type cumulative frequency distribution. First we convert it into a less than type form.

Age greater than(Years)	Greater than Cumulative frequency	Frequency	Less than Cumulative frequency
$$0-10$$	$$230$$	$$12$$	$$12$$
$$10-20$$	$$218$$	$$18$$	$$30$$
$$20-30$$	$$200$$	$$35$$	$$65$$
$$30-40$$	$$165$$	$$42$$	$$107$$
$$40-50$$	$$123$$	$$50$$	$$157$$
$$50-60$$	$$73$$	$$45$$	$$202$$
$$60-70$$	$$28$$	$$20$$	$$222$$
$$70-80$$	$$8$$	$$8$$	$$230$$

$$N/2=230/2=115$$, therefore median class is $$40-50$$
Also $$L_m=40, f_m=50, h=10, C=107$$
$$\therefore M_d=40+\displaystyle\frac{115-107}{50}\times 10=41.6$$.

Question 9
1 / -0

In a moderately a symmetrical distribution- The mean and mode are $$50$$ and $$37.5$$ respectively, find median.

A
$$45.50$$

B
$$45.86$$

C
$$52.5$$

D
$$52.86$$

Solution

Using the empirical relation between mean, median and mode, we can write X$$-M_0=3(X-M_d)$$ or $$2X=3M_d-M_0$$; $$45.83$$.
Question 10
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It is not uncommon in a median:

A
To locate it graphically

B
To use it then data is quantitive

C
To use it when data is rigidly defined

D
All of the above

Solution

It is not uncommon in a median
To locate it graphically
to use it then data is quantitive
to use it when data is rigidly defined

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Re-Attempt Weekly Quiz Competition

Number	Frequency
14	4
15	5
16	1
17	1
18	1
19	2
20	1

Measures of Central Tendency Test - 21

Result

Measures of Central Tendency Test - 21

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Question 1

$$40$$

$$41$$

$$42$$

$$43$$

Solution

Question 2

$$52$$

$$51$$

$$53$$

$$52.5$$

Solution

Question 3

$$14$$

$$16$$

$$19$$

$$15$$

Solution

Question 4

median

mean

mode

quartile

Solution

Question 5

$$900.90$$

$$1,115$$

$$1,225$$

$$900.09$$

Solution

Question 6

$$2$$

$$3$$

$$4$$

$$6$$

Solution

Question 7

$$9$$

$$5$$

$$8$$

$$3$$

Solution

Question 8

$$31.5$$

$$41.5$$

$$41.6$$

$$31.6$$

Solution

Question 9

$$45.50$$

$$45.86$$

$$52.5$$

$$52.86$$

Solution

Question 10

To locate it graphically

To use it then data is quantitive

To use it when data is rigidly defined

All of the above

Solution

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