Measures of Central Tendency Test

Result

Measures of Central Tendency Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The distribution of income of $$60$$ teachers in a school are given below. Find the Arithmetic mean of the following data by Asumed mean method.

A
$$256$$

B
$$230$$

C
$$265$$

D
$$254$$

Solution

Let us first construct the table.
Let us take the assumed mean, $$a = 275$$
$$\bar { x } =a+\dfrac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } } $$
$$\bar { x } =275+\left[ \dfrac { -600 }{ 60 } \right] =275-10$$
$$\bar { x } =265$$

Question 2

1 / -0

Calculate the average income of the workers per week using step deviation method.

$$500$$

$$640$$

$$480$$

$$320$$

Solution

Weekly Income h=200	Mid-point	Number of workers	$${ d }_{ i }={ x }_{ i }-a$$	$${ u }_{ i }=\cfrac { { x }_{ i }-a }{ h } $$	$${ f }_{ i }{ u }_{ i }$$
200-400	300	4	-200	-1	-4
400-600	500(a)	6	0	0	0
600-800	700	2	200	1	2
800-1000	900	8	400	2	16
		$$\sum { { f }_{ i } } =20$$			14

Mean $$=a+\cfrac { \sum { { f }_{ i }u_{ i } } }{ \sum { { f }_{ i } } } \times h\\ =500+\cfrac { 14 }{ 20 } \times 200\\ =500+140\quad \\ =640Ans$$

Question 3

1 / -0

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

$$44.66$$

$$40.66$$

$$42.66$$

$$45.66$$

Solution

Answer:-

Class interval width $$(W)=21-11 = 10$$

Converting the given data into frequency distribution table:

1. Write the mid-point of each interval $$(X_i)$$.

2. Consider any value of the $$X_i$$ as $$A$$ or Assumed mean.

3. Find deviation $$(d_i)$$ of each value of $$X$$ from assumed mean as defined in the table.

4. Multiply each valye of $$d_i$$ with the corresponding frequency $$f_i$$.

Class interval	$$X_i$$	No. of employees $$f_i$	$$d_i=\cfrac{X-A}{W}$$	$$f_id_i$$
11-21	16	2	-2	-4
21-31	26	4	-1	-4
31-41	36=A	6	0	0
41-51	46	8	1	8
51-61	56	10	2	20
		$$\Sigma f = 30$$		$$\Sigma fd = 20$$

5. Calculate mean using the formula below:

Mean = $$A+\cfrac{\Sigma fd}{\Sigma f}=36+\cfrac{20}{30} \times 10 ={36+6.66}=42.66$$

Hence, option C is correct.

Question 4

1 / -0

Find the mean using step deviation method.

$$14$$

$$13$$

$$12$$

$$10$$

Solution

x	Mid-point	(frequency)	$${ d }_{ i }={ x }_{ i }-6$$	$${ u }_{ i }=\cfrac { { x }_{ i }-6 }{ 4 } $$	$${ f }_{ i }{ u }_{ i }$$
0-4	2	1	-4	-1	-1
4-8	6 (a)	2	0	0	0
8-12	10	3	4	1	3
12-16	14	4	8	2	8
		10			$$\sum { { f }_{ i }{ u }_{ i } } =10$$=10

Mean $$=a+\cfrac { \sum { { f }_{ i }u_{ i } } }{ \sum { { f }_{ i } } } \times h\\ =6+\cfrac { 10 }{ 10 } \times 4\\ =10\quad Ans$$

Question 5
1 / -0

Calculate the mean of the following data with the help of Assumed method.

A
$$7.58$$

B
$$5.78$$

C
$$6.87$$

D
$$8.67$$

Solution

Let us first construct the table.
Let us take the assumed mean, $$a = 5$$
$$\bar { x } =a+\dfrac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } } $$
$$\bar { x } =5+\left[ \dfrac { 85.12 }{ 33 } \right] =5+2.58$$
$$\bar { x } =7.58$$

Question 6

1 / -0

In a study on a certain population, the following data was given.

Population (X)	2000-2001	2001-2002	2002-2003	2003-2004	2004-2005	2005-2006	2006-2007
Number of people	10	20	30	40	50	60	70

Find the average number of population using step deviation method.

$$2002$$

$$2003$$

$$2004$$

$$2005$$

Solution

X	Midpoint (X)	Frequency (F)	i=class interval width	A=2003.5 Assumed mean	u=$$\dfrac{x-A}{i}$$	fu
2000-2001	2000.5	10	1	2003.5	-3	-30
2001-2002	2001.5	20	1	2003.5	-2	-40
2002-2003	2002.5	30	1	2003.5	-1	-30
2003-2004	2003.5=A	40	1	2003.5	0	0
2004-2005	2004.5	50	1	2003.5	1	50
2005-2006	2005.5	60	1	2003.5	2	120
2006-2007	2006.5	70	1	2003.5	3	210
		$$\Sigma f=280$$				$$\Sigma fu=280$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u = \dfrac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}$$ = arithmetic mean
$$\overline {X} =2003.5 +\dfrac{280}{280}\times 1$$
$$= 2003.5 + 1$$
$$= 2004.5$$ $$\approx$$ $$2005$$

Question 7

1 / -0

The frequency distribution of marks in English are given in the table:

Marks	50-60	60-70	70-80	80-90
Number of students	12	24	14	10

Find the mean by step deviation method.

$$58$$

$$48$$

$$69$$

$$71$$

Solution

$$X$$	Mid point(x)	Frequency (F)	i=class interval width	A=65 Assumed mean	u=$$\dfrac{x-A}{i} $$	fu
$$50-60$$	$$55$$	$$12$$	$$10$$	$$65$$	$$-1$$	$$-12$$
$$60-70$$	$$65=A$$	$$24$$	$$10$$	$$65$$	$$0$$	$$0$$
$$70 -80$$	$$75$$	$$14$$	$$10$$	$$65$$	$$1$$	$$14$$
$$80-90 $$	$$85$$	$$10$$	$$10$$	$$65$$	$$2$$	$$20$$
		$$\Sigma f=60$$				$$\Sigma fu=22$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum f=$$ Summation of the frequencies given in the grouped data
$$\sum fu=$$ Summation of the frequencies and deviation of a given mean data
$$u= \dfrac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}=$$ arithmetic mean
$$\overline {X} =65 +\dfrac{22}{60}\times 10$$
$$= 65 + 3.666$$
$$= 68.666$$ $$\approx$$ $$69$$ marks

Question 8

1 / -0

Calculate the arithmetic mean for the following birthday gifts using direct method.

Children	0-12	12-24	24-36	36-48	48-60
Birthday gifts	3	5	7	9	11

$$36.85$$

$$23.56$$

$$24.87$$

$$12.24$$

Solution

Children	Birthday gifts$$({ f }_{ i }$$)	$$x_{ i }$$	$${ f }_{ i }x_{ i }$$
0-12	3	6	18
12-24	5	18	90
24-36	7	30	210
36-48	9	42	378
48-60	11	54	594
		$$\sum { { f }_{ i } } =35$$	$$\sum { { f }_{ i }x_{ i } } =1290$$

$$Mean=\cfrac { \sum { { f }_{ i }x_{ i } } }{ \sum { { f }_{ i } } } \\ =\cfrac { 1290 }{ 35 } =36.85$$

Question 9

1 / -0

Find the arithmetic mean using step deviation method for the following. The data shows distance covered by $$40$$ passengers to perform their work. (Round off your answer to the nearest whole number).

Distance (km)	1-5	5-9	9-13	13-17	17-21	21-25	25-29
Number of passengers	2	4	6	8	10	5	5

$$15$$

$$16$$

$$17$$

$$19$$

Solution

Distance (km)	Mid point $$x$$	Number of passengers $$f$$	$$i$$=class interval width	$$A$$= Assumed mean	$$u=\dfrac{x-A}{i}$$	$$fu$$
1-5	3	2	4	11	-2	-4
5-9	7	4	4	11	-1	-4
9-13	11	6	4	11	0	0
13-17	15	8	4	11	1	8
17-21	19	10	4	11	2	20
21-25	23	5	4	11	3	15
25-29	27	5	4	11	4	20
		$$\Sigma f=40$$				$$\Sigma f u=55$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {x} =A + \frac{\sum fu}{\sum f} \times i$$
A = Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u = \frac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {x}$$ = arithmetic mean
$$\overline {x} =11 +\frac{55}{40}\times 4$$
$$= 11 + 5.5$$
$$= 16.5$$ $$\approx$$ $$17$$

Question 10

1 / -0

Calculate the mean for the following data using step deviation method:

Class interval (X)	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	15	10	15	20	25	30	35

Round off your answer to the nearest whole number.

$$41$$

$$42$$

$$43$$

$$44$$

Solution

Consider the following table, to calculate mean by "step deviation method":

$$x_i$$=mid value of class interval

Assumed mean $$a=35$$

class interval $$c=10$$

$$ci$$	$$f_i$$	$$x_i$$	$$u_i=\dfrac{x_i-a}{c}$$	$$f_iu_i$$
$$0-10$$	$$15$$	$$5$$	$$-3$$	$$-45$$
$$10-20$$	$$10$$	$$15$$	$$-2$$	$$-20$$
$$20-30$$	$$15$$	$$25$$	$$-1$$	$$-15$$
$$30-40$$	$$20$$	$$35$$	$$0$$	$$0$$
$$40-50$$	$$25$$	$$45$$	$$1$$	$$25$$
$$50-60$$	$$30$$	$$55$$	$$2$$	$$60$$
$$60-70$$	$$35$$	$$65$$	$$3$$	$$105$$

$$N=\Sigma f_i=150$$

$$\Sigma f_iu_i=110$$

Mean $$\overline x=a +\frac {\Sigma f_iu_i}{N}\times c$$

$$\therefore \overline x=35 + \dfrac{110}{150} \times 10=42.3334 \approx 42$$

Hence, option $$B$$ is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 38

Result

Measures of Central Tendency Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

$$256$$

$$230$$

$$265$$

$$254$$

Solution

Question 2

$$500$$

$$640$$

$$480$$

$$320$$

Solution

Question 3

$$44.66$$

$$40.66$$

$$42.66$$

$$45.66$$

Solution

Question 4

$$14$$

$$13$$

$$12$$

$$10$$

Solution

Question 5

$$7.58$$

$$5.78$$

$$6.87$$

$$8.67$$

Solution

Question 6

$$2002$$

$$2003$$

$$2004$$

$$2005$$

Solution

Question 7

$$58$$

$$48$$

$$69$$

$$71$$

Solution

Question 8

$$36.85$$

$$23.56$$

$$24.87$$

$$12.24$$

Solution

Question 9

$$15$$

$$16$$

$$17$$

$$19$$

Solution

Question 10

$$41$$

$$42$$

$$43$$

$$44$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.