Measures of Central Tendency Test

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Measures of Central Tendency Test - 37

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Weekly Quiz Competition

Question 1

1 / -0

Estimate the mean of the given data using Assumed meant method.

$$50$$

$$49.14$$

$$51.83$$

$$61.37$$

Solution

Answer:- Using Shortcut method

Interval	$$x = \cfrac{U.L. + L.L.}{2}$$	Frequency f	d=x-A	fd
22-32	27	12	-30	-360
32-42	37	7	-20	-140
42-52	47	11	-10	-110
52-62	57=A	3	0	0
62-72	67	19	10	190
72-82	77	6	20	120
		$$\Sigma f=58$$		$$\Sigma fd = -300$$

Mean = $$A + \cfrac{\Sigma fd}{\Sigma f} = 57+\cfrac{-300}{58} = 57 - 5.17 = 51.827 \simeq 51.83$$

C)51.83

Question 2
1 / -0

The marks obtained by $$40$$ students of Class XII of a certain school in a Science paper consisting of $$200$$ marks are presented in table below. Find the mean of the marks obtained by the students using step deviation method.

A
$$153.5$$

B
$$154.5$$

C
$$150.2$$

D
$$156.5$$

Solution

Answer:- Using Shortcut Method
Class interval width (i) = $$120-100 = 20$$

Marks X Students
(F) $$d=\cfrac{X-A}{i}$$ Fd
100-120 110 4 -2 -8
120-140 130 5 -1 -5
140-160 150 = A 11 0 0
160-180 170 20 1 20
$$\Sigma F = 40$$ $$\Sigma Fd = 7$$
Mean ( By step deviation formula) = $$A+\cfrac{\Sigma Fd}{\Sigma F} \times i = 150+\left(\cfrac{7}{40} \times 20 \right) = 150+3.5+ = 153.5$$
A) Mean = $$153.5$$

Question 3

1 / -0

Class interval	$$4-8$$	$$8-12$$	$$12-16$$	$$16-20$$
No. of employees	$$5$$	$$10$$	$$15$$	$$20$$

Find the mean of the following frequency distribution using step deviation method.

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Answer:- By shortcut Method(Step deviation mean method)

Class interval width $$(w) =8-4 = 4$$

Class interval $$(i)$$	$$X $$	No. of employees $$F $$	$$d=\cfrac{X-A}{i}$$	$$Fd $$
$$4-8$$	$$6$$	$$5$$	-$$1 $$	$$-5$$
$$8-12$$	$$10=A$$	$$ 10$$	$$0$$	$$0$$
$$12-16$$	$$14$$	$$15$$	$$1 $$	$$15$$
$$16-20$$	$$18$$	$$20$$	$$ 2$$	$$40$$
		$$\Sigma F = 50$$		$$\Sigma Fd = 50$$

Mean = $$A+\cfrac{\Sigma Fd}{\Sigma f} \times i=10+\left(\cfrac{50}{50} \times 4\right) = 10+4 = 14$$

Hence option (D) is correct

Question 4

1 / -0

The marks scored by $$25$$ students in the exam are as follows. Find the average using Assumed mean method.

$$73.1$$

$$75.8$$

$$77.4$$

$$76.9$$

Solution

Marks	$$X =$$ $$\cfrac{U.L.+L.L.}{2}$$	Students $$(F)$$	$$d=X-A$$	$$Fd$$
50-60	55	3	-20	-60
60-70	65	4	-10	-40
70-80	75 $$= A$$	8	0	0
80-90	85	8	10	80
90-100	95	2	20	40
		$$\Sigma F = 25$$		$$\Sigma Fd = 20$$

Mean by Shortcut Method $$=A+\cfrac{\Sigma Fd}{\Sigma F}$$

$$= 75+\cfrac{20}{25}$$

$$=75+\cfrac{4}{5}$$

$$= 75+0.8$$

$$= 75.8$$

Question 5

1 / -0

Find the mean of the above frequency distribution using the step deviation method.

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Class interval width $$i = 4-2 = 2$$

Class interval	$$X_i$$	Frequency $$F_i$$	$$u=\cfrac{X-A}{i}$$	$$F_iu_i$$
$$2-4$$	$$3$$	$$11$$	$$-1$$	$$-11$$
$$4-6$$	$$5=A$$	$$22$$	$$0$$	$$0$$
$$6-8$$	$$7$$	$$33$$	$$1$$	$$33$$
$$8-10$$	$$9$$	$$44$$	$$2$$	$$88$$
		$$\sum F_i = 110$$		$$\sum F_iu_i = 110$$

$$\overline x = A+\cfrac{\sum fu}{\sum f} \times i$$

$$=5+\left(\cfrac{110}{110} \times 2\right) $$

$$= 5+2$$

$$ = 7$$

Hence, option $$C$$ is correct.

Question 6

1 / -0

An Egg Seller distributes eggs to the shop in a city. The number of eggs he distributes for each shop has been recorded and the data obtained was grouped into a class shown in the table below. Find the mean using sdirect method.

$$91.5$$

$$83.07$$

$$87.34$$

$$89.76$$

Solution

Number of eggs	Mid-Value	$${ f }_{ i }$$	$${ f }_{ i }{ x }_{ i }$$
$$0-30$$	$$15$$	$$40$$	$$600$$
$$30-60$$	$$45$$	$$26$$	$$1170$$
$$60-90$$	$$75$$	$$38$$	$$2850$$
$$90-120$$	$$105$$	$$12$$	$$1260$$
$$120-150$	$$135$$	$$25$$	$$3375$$
$$150-180$$	$$165$$	$$30$$	$$4950$$
		$$\sum { { f }_{ i }=171 } $$	$$\sum { { f }_{ i }{ x }_{ i }=14205 } $$

$$Mean=\cfrac { \sum { { f }_{ i }x_{ i } } }{ \sum { { f }_{ i } } } \\ =\cfrac { 14205 }{ 171 } \\ =83.07$$

Question 7
1 / -0

Find the mean number of leaves per tree using step deviation method.

A
$$10.4$$

B
$$11$$

C
$$12$$

D
$$13$$

Solution

Answer:-
Class interval width (i) = 4
No. of leaves
X No. of trees
F $$u=\cfrac{X-A}{i}$$ Fu
0-4 2 10 -1 -10
4-8 6 = A 20 0 -0
8-12 10 30 1 30
12-16 14 40 2 80
$$\Sigma f = 100$$ $$\Sigma fu = 110$$
Mean = $$A+\cfrac{\Sigma fu}{\Sigma f}=6+\cfrac{110}{100} \times 4 ={6+4}=10.4$$
10.4 leaves
Question 8
1 / -0

The percentage of marks obtained by the students in a class of $$50$$ are given below. Find the mean for the following data.
Marks$$(\%)$$ $$40-50$$ $$50-60$$ $$60-70$$ $$70-80$$ $$80-90$$ $$90-100$$
Number of students $$6$$ $$12$$ $$14$$ $$8$$ $$6$$ $$4$$

A
$$64.6\%$$

B
$$65.6\%$$

C
$$66.6\%$$

D
$$67.6\%$$

Solution

Marks $$f_i$$ $$x_i$$   $$f_ix_i$$
$$40-50$$ $$6$$ $$45$$   $$270$$
$$50-60$$ $$12$$ $$55$$ $$660$$
$$60-70$$ $$14$$   $$65$$ $$910$$
$$70-80$$ $$8$$ $$75$$   $$600$$
$$80-90$$ $$6$$   $$85$$   $$510$$
$$90-100$$ $$4$$   $$95$$ $$380$$

$$\Sigma f_i=50$$ $$\Sigma f_ix_i=3330$$

We know that, the arithmetic mean using direct method is
$$\overline{x}=\displaystyle\frac{\Sigma f_i x_i}{\Sigma f_i}$$

$$\overline{x}=\dfrac{3330}{50}$$

$$\Rightarrow 66.6$$

Therefore, the mean for the given data is $$66.6\%$$.

Question 9

1 / -0

Calculate the mean for the following table using step deviation method.

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Answer:- By shortcut Method

Class interval width (i) = $$5-2 = 3$$

Age	No. of children F	Mid Value (X)	$$u=\cfrac{X-A}{i}$$	Fu
2-5	1	3.5	-1	-1
5-8	2	6.5 = A	0	0
8-11	3	9.5	1	3
11-14	4	12.5	2	8
14-17	5	15.5	3	15
	$$\Sigma F = 15$$			$$\Sigma Fu = 25$$

By step deviation method

Mean = $$A+\cfrac{\Sigma fu}{\Sigma f} \times i=6.5+\cfrac{25}{15} \times 3 ={6.5+5}=11.5 \simeq 12$$

B) 12

Question 10
1 / -0

Find the Arithmetic mean of the following data by Assumed mean method.

A
$$21$$

B
$$22$$

C
$$30$$

D
$$33$$

Solution

The above table represents the given data.
Let us take the assumed mean, $$a = 25$$
We know that, by assumed mean method, the mean of the distribution is given by:

$$\bar { x } =a+\dfrac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } } $$

$$\therefore \ \bar { x } =25+\left[ \dfrac { 520 }{ 65 } \right] =25+8$$

$$ \bar { x } =33$$

Hence, the mean of the given distribution, by assumed mean method, is $$33$$.

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Marks	X	Students (F)	$$d=\cfrac{X-A}{i}$$	Fd
100-120	110	4	-2	-8
120-140	130	5	-1	-5
140-160	150 = A	11	0	0
160-180	170	20	1	20
		$$\Sigma F = 40$$		$$\Sigma Fd = 7$$

No. of leaves	X	No. of trees F	$$u=\cfrac{X-A}{i}$$	Fu
0-4	2	10	-1	-10
4-8	6 = A	20	0	-0
8-12	10	30	1	30
12-16	14	40	2	80
		$$\Sigma f = 100$$		$$\Sigma fu = 110$$

Marks$$(\%)$$	$$40-50$$	$$50-60$$	$$60-70$$	$$70-80$$	$$80-90$$	$$90-100$$
Number of students	$$6$$	$$12$$	$$14$$	$$8$$	$$6$$	$$4$$

Measures of Central Tendency Test - 37

Result

Measures of Central Tendency Test - 37

Score

-

Rank

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SHARING IS CARING

Question 1

$$50$$

$$49.14$$

$$51.83$$

$$61.37$$

Solution

Question 2

$$153.5$$

$$154.5$$

$$150.2$$

$$156.5$$

Solution

Question 3

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Question 4

$$73.1$$

$$75.8$$

$$77.4$$

$$76.9$$

Solution

Question 5

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 6

$$91.5$$

$$83.07$$

$$87.34$$

$$89.76$$

Solution

Question 7

$$10.4$$

$$11$$

$$12$$

$$13$$

Solution

Question 8

$$64.6\%$$

$$65.6\%$$

$$66.6\%$$

$$67.6\%$$

Solution

Question 9

$$11$$

$$12$$

$$13$$

$$14$$

Solution

Question 10

$$21$$

$$22$$

$$30$$

$$33$$

Solution

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