Measures of Central Tendency Test

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Measures of Central Tendency Test - 35

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Question 1
1 / -0

Find the median of the following data

A
$$20$$

B
$$19$$

C
$$21$$

D
$$22$$

Solution

$$x$$ $$14$$ $$18$$
$$22$$ $$25$$
$$30$$
$$f$$ $$7$$ $$10$$ $$ 12$$ $$ 16$$ $$ 5$$
C.F. $$7$$ $$ 17$$
$$29$$
$$45$$
$$50$$

Total of frequency $$ N =50 $$

As the total of the frequencies is even, Median will be the at the mean of observations at $$ \dfrac {N}{2} $$ and $$ \dfrac {N}{2} + 1$$ position

So, Median is mean of $$ 25^{th} $$ and $$ 26^{th}$$ observation , that is Mean of $$ 22 $$ and $$ 22 = 22 $$

Question 2

1 / -0

x	5	6	10	12	13	15
f	7	8	15	13	9	8

Find the median of the following data

$$12$$

$$13$$

$$11$$

$$15$$

Solution

$$x$$	$$5$$	$$6$$	$$ 10$$	$$12$$	$$ 13$$	$$ 15$$
$$f$$	$$7$$	$$8$$	$$ 15$$	$$13$$	$$9$$	$$ 8$$
C.F.	$$7$$	$$ 7+8 = 15 $$	$$ 15+15 = 30 $$	$$ 30 + 13 = 43 $$	$$ 43 + 9 = 52 $$	$$ 52 + 8 = 60 $$

Total of frequency $$ N =60 $$

As the total of the frequencies is even, Median will be the at the mean of observations at $$ \dfrac {N}{2} $$ and $$ \dfrac {N}{2} + 1$$ position

So, Median is mean of $$ 30^{th}$$ and $$ 31^{st}$$ observation, that is Mean of $$ 10 $$ and $$ 12 = 11 $$

Question 3
1 / -0

Find the arithmetic mean for the given data:

Marks No. of students
$$0-10$$ $$5$$
$$10-20$$ $$10$$
$$20-30$$ $$25$$
$$30-40$$ $$30$$
$$40-50$$ $$20$$
$$50-60$$ $$10$$

A
$$34$$

B
$$35$$

C
$$33$$

D
$$38$$

Solution

Mid-values$$(x_i)$$ No. of students$$(f_i)$$ Product$$(f_ix_i)$$
$$5$$   $$5$$ $$25$$
$$15$$ $$10$$ $$150$$
$$25$$ $$25$$ $$625$$
$$35$$ $$30$$ $$1050$$
$$45$$ $$20$$ $$900$$
$$55$$    $$10$$ $$550$$

$$\Sigma f_i=100$$ $$\Sigma f_ix_i=3300$$

$$\therefore$$ Arithmetic Mean $$= \displaystyle \frac{\Sigma f_ix_i}{\Sigma f_i}$$

$$= \displaystyle \frac{3300}{100}$$

$$ = 33$$

Hence, the arithmetic mean is $$33.$$
Question 4
1 / -0

On Monday, a person mailed $$8$$ packages weighing an average (arithmetic mean) of $$\displaystyle 12\frac { 3 }{ 8 } $$ pounds, and on Tuesday, $$4$$ packages weighing an average of $$\displaystyle 15\frac { 1 }{ 4 } $$ pounds. What was the average weight, in pounds, of all the packages the person mailed on both days?

A
$$\displaystyle 13\frac { 1 }{ 3 } $$

B
$$\displaystyle 13\frac { 13 }{ 16 } $$

C
$$\displaystyle 15\frac { 1 }{ 2 } $$

D
$$\displaystyle 15\frac { 15 }{ 16 } $$

E
$$\displaystyle 16\frac { 1 }{ 2 } $$

Solution

Mean is the sum of the values divided by the total number of values in the data set.

Given that mean of $$8$$ packages is $$12\dfrac 38=\dfrac{99}{8}$$

$$\implies \dfrac{\text{sum of 8 packages}}{8}=\dfrac{99}{8}$$

$$\implies \text{sum of 8 packages}={99}$$ -------(1)

Given that mean of $$4$$ packages is $$15\dfrac 14=\dfrac{61}{4}$$

$$\implies \dfrac{\text{sum of 4 packages}}{4}=\dfrac{61}{4}$$

$$\implies \text{sum of 4 packages}={61}$$ --------(2)

Total mean is $$\implies \dfrac{\text{sum of 8 packages}+\text{sum of 4 packages}}{8+4}$$

$$=\dfrac{99+61}{12}=\dfrac{160}{12}=13\dfrac 13$$

Question 5

1 / -0

Find the arithmetic mean using direct method for the following data shows distance covered by $$120$$ passengers to perform their regular work.

$$20\ km$$

$$25\ km$$

$$22\ km$$

$$15\ km$$

Solution

Answer:- Using direct method

Distance	No. of passengers $$f_i$$	$$x_i$$	$$f_i x_i$$
$$5-15$$	$$30$$	$$10$$	$$300$$
$$15-20$$	$$20$$	$$17.5$$	$$350$$
$$20-35$$	$$ 30$$	$$27.5$$	$$825$$
$$35-40$$	$$40$$	$$37.5$$	$$1500$$
	$$\Sigma f_i = 120$$		$$\Sigma f_i x_i = 2975$$

Arithmetic mean $$=\cfrac{\Sigma f_i x_i}{\Sigma f_i} =\cfrac{2975}{120} = 24.79 \simeq 25\ km$$

Question 6
1 / -0

Find the arithmetic mean of the following table using direct method.

A
$$30$$ min

B
$$50 $$ min

C
$$35 $$ min

D
$$40 $$ min

Solution

Arithmetic mean using direct method formula is $$\bar { x } = \dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } $$
$$\bar { x } = \dfrac { 35,000 }{ 1,000 } $$ min
$$\bar { x } = 35 $$ min
Question 7
1 / -0

Find the mean for the following data using shortcut method.

A
$$6.23$$

B
$$7.33$$

C
$$8.13$$

D
$$9.43$$

Solution

Answer:- By shortcut Method

Children
(x) Pocket Allowance
(f) d=x-A fd
2 10 -4 -40
4 20 -2 -40
6=A 30 0 0
8 40 2 80
10 50 4 200
$$\Sigma f = 150$$ $$\Sigma fd = 200$$
Mean = $$A+\cfrac{\Sigma fd}{\Sigma f}=6+\cfrac{200}{150} ={6+1.33}=7.33$$
B)7.33
Question 8
1 / -0

Find the mean of the following frequency distribution using direct method.

A
$$16.20$$

B
$$14.16$$

C
$$16.17$$

D
$$14.28$$

Solution

Answer:- Using direct method
Representing the data into frequency distribution table
$$x_i =\text{mid point of each class interval}$$

Class interval No. of workers
$$(f_i)$$ $$x_i$$ $$f_i x_i$$
0-5 4 2.5 10
5-10 5 7.5 37.5
10-15 6 12.5 75
15-20 7 17.5 122.5
20-25 8 22.5 180
$$\Sigma f_i=30$$ $$\Sigma f_ix_i=425$$
$$\therefore \; mean = \cfrac{\Sigma f_ix_i}{\Sigma f_i} =\cfrac{425}{30}=14.16$$

Question 9

1 / -0

A survey was conducted by a group of students in which they collected the following data regarding the number of plants in $$30$$ houses in a locality. Find the mean number of plants per house.

$$23.4$$

$$25.4$$

$$22.4$$

$$26.4$$

Solution

Given:-

Total no. of houses $$,\Sigma f_i = 30$$

Representing the data in frequency distribution table

No. of plants	No. of houses $$(f_i)$$	Classmark $$(x_i)$$	$$f_i x_i$$
$$0-6$$	$$4$$	$$3$$	$$ 12$$
$$ 6-12$$	$$6$$	$$9$$	$$54$$
$$12-18$$	$$1$$	$$15$$	$$15$$
$$18-24$$	$$3$$	$$21$$	$$63$$
$$24-30$$	$$5$$	$$27$$	$$135$$
$$30-36$$	$$6$$	$$33$$	$$198$$
$$36-42$$	$$5$$	$$39$$	$$195$$
	$$\Sigma f_i=30$$		$$\Sigma f_i x_i = 672$$

$$\therefore \; mean=\cfrac{\Sigma f_i x_i}{\Sigma f_i}=\cfrac{672}{30} = 22.4$$

Hence, the mean no. of plants per house $$= 22.4$$

Question 10
1 / -0

Calculate arithmetic mean using direct method for the following data shows distance covered by $$50$$ persons to perform their work inside the factory.

A
$$20.5 km$$

B
$$32.8 km$$

C
$$25.2 km$$

D
$$27.6 km$$

Solution

Answer:- Using direct method
Representing the data into frequency distribution table
$$x_i =\text{mid point of each class interval}$$

Class interval No. of workers
$$(f_i)$$ $$x_i$$ $$f_i x_i$$
1-11 4 6 24
11-21 6 16 96
21-31 10 26 260
31-41 12 36 432
41-51 18 46 828
$$\Sigma f_i=50$$ $$\Sigma f_ix_i=1640$$
$$\therefore \; mean = \cfrac{\Sigma f_ix_i}{\Sigma f_i} =\cfrac{1640}{50}=32.8$$
Average distance covered by 50 persons = 32.8
B)32.8

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$$x$$	$$14$$	$$18$$	$$22$$	$$25$$	$$30$$
$$f$$	$$7$$	$$10$$	$$ 12$$	$$ 16$$	$$ 5$$
C.F.	$$7$$	$$ 17$$	$$29$$	$$45$$	$$50$$

Marks	No. of students
$$0-10$$	$$5$$
$$10-20$$	$$10$$
$$20-30$$	$$25$$
$$30-40$$	$$30$$
$$40-50$$	$$20$$
$$50-60$$	$$10$$

Children (x)	Pocket Allowance (f)	d=x-A	fd
2	10	-4	-40
4	20	-2	-40
6=A	30	0	0
8	40	2	80
10	50	4	200
	$$\Sigma f = 150$$		$$\Sigma fd = 200$$

Class interval	No. of workers $$(f_i)$$	$$x_i$$	$$f_i x_i$$
0-5	4	2.5	10
5-10	5	7.5	37.5
10-15	6	12.5	75
15-20	7	17.5	122.5
20-25	8	22.5	180
	$$\Sigma f_i=30$$		$$\Sigma f_ix_i=425$$

Class interval	No. of workers $$(f_i)$$	$$x_i$$	$$f_i x_i$$
1-11	4	6	24
11-21	6	16	96
21-31	10	26	260
31-41	12	36	432
41-51	18	46	828
	$$\Sigma f_i=50$$		$$\Sigma f_ix_i=1640$$

Measures of Central Tendency Test - 35

Result

Measures of Central Tendency Test - 35

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SHARING IS CARING

Question 1

$$20$$

$$19$$

$$21$$

$$22$$

Solution

Question 2

$$12$$

$$13$$

$$11$$

$$15$$

Solution

Question 3

$$34$$

$$35$$

$$33$$

$$38$$

Solution

Question 4

$$\displaystyle 13\frac { 1 }{ 3 } $$

$$\displaystyle 13\frac { 13 }{ 16 } $$

$$\displaystyle 15\frac { 1 }{ 2 } $$

$$\displaystyle 15\frac { 15 }{ 16 } $$

$$\displaystyle 16\frac { 1 }{ 2 } $$

Solution

Question 5

$$20\ km$$

$$25\ km$$

$$22\ km$$

$$15\ km$$

Solution

Question 6

$$30$$ min

$$50 $$ min

$$35 $$ min

$$40 $$ min

Solution

Question 7

$$6.23$$

$$7.33$$

$$8.13$$

$$9.43$$

Solution

Question 8

$$16.20$$

$$14.16$$

$$16.17$$

$$14.28$$

Solution

Question 9

$$23.4$$

$$25.4$$

$$22.4$$

$$26.4$$

Solution

Question 10

$$20.5 km$$

$$32.8 km$$

$$25.2 km$$

$$27.6 km$$

Solution

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