Measures of Central Tendency Test

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Measures of Central Tendency Test - 30

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Weekly Quiz Competition

Question 1

1 / -0

The following table shows the frequency distribution of advertisements on T.V.

Duration (in sec.)	$$25 - 30$$	$$30 - 35$$	$$35 - 40$$	$$40 - 45$$	$$45 - 50$$	$$50 - 55$$
No. of advertisements	$$10$$	$$32$$	$$15$$	$$9$$	$$7$$	$$2$$

Obtain mean duration of an advertisement on T.V. by assumed mean method

$$31.97$$ seconds

$$32.97$$ seconds

$$34.97$$ seconds

$$35.97$$ seconds

Solution

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

$$x_i=$$ mid value of class interval

Let the assumed mean be $$a=42.5$$

Class interval	$$f_i$$	$$x_i$$	$$d_i=x_i-a$$	$$f_id_i$$
$$25-30$$	$$10$$	$$27.5$$	$$-15$$	$$-150$$
$$30-35$$	$$32$$	$$32.5$$	$$-10$$	$$-320$$
$$35-40$$	$$15$$	$$37.5$$	$$-5$$	$$-75$$
$$40-45$$	$$9$$	$$42.5$$	$$0$$	$$0$$
$$45-50$$	$$7$$	$$47.5$$	$$5$$	$$35$$
$$50-55$$	$$2$$	$$52.5$$	$$10$$	$$20$$

$$N=\Sigma f_i=75$$

$$\Sigma f_id_i=-490$$

Mean, $$\overline x=a +\cfrac {\Sigma f_id_i}{N}$$

$$\therefore \overline x=42.5 + \cfrac{-490}{75}=35.966667 $$

$$\implies x= 35.97$$

The mean duration of an advertisement on T.V is $$35.97$$ seconds

Hence, option $$D$$ is correct.

Question 2
1 / -0

The following data have been arranged in ascending order of magnitude.
$$63, 66, 69, x, x + 2, 76, 89$$ and $$103$$.
If the median of the given data is $$71$$, find the value of $$x$$.

A
$$60$$

B
$$69$$

C
$$70$$

D
None of these

Solution

median of the data: $$63,66,69,x,x+2,76,89,103$$ is $$71$$
The number of observations in the data are 8. Hence, the median will be the mean of $$3^{rd}$$ term and $$4^{th}$$ term,
Thus, $$\cfrac{x + x +2}{2} = 71$$
$$\Rightarrow 2x + 2 = 142$$
$$\Rightarrow x = 70$$
Question 3
1 / -0

The data is $$17, 26, 60, 45, 33, 32,29, 34$$ and $$56$$. If $$26$$ is replaced by $$62$$, then the new median is

A
$$17$$

B
$$29$$

C
$$33$$

D
$$34$$

Solution

Arranging data in ascending order- $$17,26,29,32,33,34,45,56,60$$
Count of observations $$=9 $$
Median for odd number of data is the middle most observation
So median $$= 5^{th}$$ observation $$= 33$$
If $$26$$ is replaced by $$62$$, then the observations in ascending order will be- $$17,29,32,33,34,45,56,60,62$$.
Again Median for odd number of data is the middle most observation
So new median $$= 5^{th}$$ observation $$= 34$$

Question 4

1 / -0

Number of calories (in' 00) consumed daily by a sample of 15 years old boys are given below.

Calories	1000 - 1500	1500 - 2000	2000 - 2500	2500 - 3000	3000 - 3500	3500 - 4000	4000 - 4500
No. of boys	5	13	16	18	27	10	4

Obtain mean calories consumed daily by a boy by step deviation method.

$$2222$$ calories

$$2548$$ calories

$$2563$$ calories

$$2761$$ calories

Solution

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

$$x_i$$=mid value of class interval

Assumed mean $$a=2750$$

class interval $$c=500$$

$$ci$$	$$f_i$$	$$x_i$$	$$u_i=\dfrac{x_i-a}{c}$$	$$f_iu_i$$
$$1000-1500$$	$$5$$	$$1250$$	$$-3$$	$$-15$$
$$1500-2000$$	$$13$$	$$1750$$	$$-2$$	$$-26$$
$$2000-2500$$	$$16$$	$$2250$$	$$-1$$	$$-16$$
$$2500-3000$$	$$18$$	$$2750$$	$$0$$	$$0$$
$$3000-3500$$	$$27$$	$$3250$$	$$1$$	$$27$$
$$3500-4000$$	$$10$$	$$3750$$	$$2$$	$$20$$
$$4000-4500$$	$$4$$	$$4250$$	$$3$$	$$12$$

$$N=\Sigma f_i=93$$

$$\Sigma f_iu_i=2$$

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

$$\therefore \overline x=2750 + \dfrac{2}{93} \times 500=2760.75 \approx 2761$$

mean calories consumed daily by a boy is $$2761$$ calories.

Hence, option $$D$$ is correct.

Question 5
1 / -0

The A.M. of the first ten odd numbers is

A
$$10$$

B
$$100$$

C
$$1000$$

D
$$1$$

Solution

First ten odd numbers are $$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$$ respectively.
$$A.M. (\overline {x}) = \dfrac{\text {sum of all the terms}}{\text {total number of terms}}$$

$$A.M. \left(\overline { x } \right) = \displaystyle\frac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$$

$$= \displaystyle\frac{100}{10}$$

$$=10$$

$$\therefore$$ the $$A.M$$ of first ten odd numbers is $$10$$
Question 6
1 / -0

Directions For Questions

Find the mean of the following data:

...view full instructions

Number of plants
0-2
2-4
4-6
6-8
8-10
10-12
12-14
Number of houses
1
2
1
5
6
2
3
mean is $$8.4$$
Is it true or false?

A
True

B
False

C
Ambiguous

D
Data insuffcient

Solution

For grouped data when Class Interval is given, we find the Class Mark.
Class Mark $$=$$ mid point of an interval $$= \dfrac { \text{Lower limit + Upper limit} }{ 2 }$$
Class Mark is taken as $${ x }_{ i }$$ for each class interval.
Find $$ { x }_{ i }{ f }_{ i }$$
From the table, we have $$\Sigma { x }_{ i }{ f }_{ i } = 162 $$ & $$\Sigma { f }_{ i } = 20$$
Mean $$=\dfrac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } }$$
$$= \dfrac { 162 }{ 20 } = 8.1$$
Question 7
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Class Interval
Frequency
100-12
12
120-140
14
140-160
8
160-180
6
180-200
10
mean is 145.20
If true then enter $$1$$ and if false then enter $$0$$

A
non of thease

B
0

C
1

D
can not det

Solution

$$
For\quad grouped\quad data\quad when\quad Class\quad Interval\quad \quad is\quad given,\quad we\quad find\quad the\quad Class\quad Mark.\\ Class\quad Mark\quad =\quad mid\quad point\quad of\quad an\quad interval\quad =\quad \frac { \left( Lower\quad limit\quad +\quad Upper\quad limit \right) }{ 2 } \\ Class\quad Mark\quad is\quad taken\quad as\quad { x }_{ i }for\quad each\quad class\quad interval.\quad Find\quad { x }_{ i }{ f }_{ i }.\quad \\ From\quad the\quad table,\quad we\quad have\quad \Sigma { x }_{ i }{ f }_{ i }\quad =\quad 7260\quad \& \quad \quad \Sigma { f }_{ i }\quad =\quad 50\\ Mean\quad =\quad \frac { \Sigma { x }_{ i }{ f }_{ i } }{ \Sigma { f }_{ i } } \quad =\quad \frac { 7260 }{ 50 } \quad =\quad 145.20
$$
Question 8
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Find the mean marks from the following data:
Marks
No. of students
Below $$10$$
$$4$$
Below $$20$$
$$10$$
Below $$30$$
$$18$$
Below $$40$$
$$28$$
Below $$50$$
$$40$$
Below $$60$$
$$70$$

A
$$40\dfrac {5}{7} marks$$

B
$$40\dfrac {4}{7} marks$$

C
$$30\dfrac {5}{7} marks$$

D
$$20\dfrac {4}{7} marks$$

Solution

For the grouped data, when class interval is given we find class marks by using the formula
Class mark$$(x_i)=\dfrac { \left( Lower\ limit\ + Upper\ limit \right) }{ 2 }$$
Find $$x_{i} f_{i}$$ and then

Mean $$=\dfrac { \sum { x }_{ i }{ f }_{ i } }{ \sum { f }_{ i } } =\dfrac {2850}{70} =\dfrac {285}{7} =40\dfrac {5}{7}$$ marks

Question 9

1 / -0

The following data shows that the age distribution of patients of malaria in a village during a particular month. Find the average age of the patients.

Age (in years)	No. of cases
5-14	6
15-24	11
25-34	21
35-44	23
45-54	14
55-64	5
65-74	3

$$36.12$$ years

$$36.13$$ years

$$13.36$$ years

$$23.36$$ years

Solution

Age (in years)	No. of Cases (Fi)	Mid Point (Xi)	$$FiXi$$
5-14	6	9.5	57
15-24	11	19.5	214.5
25-34	21	29.5	619.5
35-44	23	39.5	908.5
45-54	14	49.5	693
55-64	5	59.5	2975.
65-74	3	69.54	208.5
	$$\ Sigma Fi=83$$		$$\Sigma FiXi =2998.5$$

For grouped data when class interval is given, we find the class mark.

Class mark $$=$$ midpoint of an interval $$=\dfrac {(Lower \quad limit + Upper \quad limit)}{2}$$

Class mark is taken as $$x_i$$ for each class interval, find $$x_if_i$$.

Now mean age $$=\dfrac {\Sigma x_if_i}{\Sigma f_i}=\dfrac {2998.5}{83}=36.1265=36.13$$

Question 10
1 / -0

The mean of first five prime numbers is

A
$$5.6$$

B
$$3.6$$

C
$$6.83$$

D
$$5.2$$

Solution

First five prime numbers are: $$2,3,5,7,11$$
Mean $$= \cfrac{\text{Sum}}{\text{Count of Numbers}}$$
Mean $$= \cfrac{2 + 3 + 5 + 7 + 11}{5}$$
Mean $$= \cfrac{28}{5}$$
Mean $$= 5.6$$

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Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

Class Interval	Frequency
100-12	12
120-140	14
140-160	8
160-180	6
180-200	10

Marks	No. of students
Below $$10$$	$$4$$
Below $$20$$	$$10$$
Below $$30$$	$$18$$
Below $$40$$	$$28$$
Below $$50$$	$$40$$
Below $$60$$	$$70$$

Measures of Central Tendency Test - 30

Result

Measures of Central Tendency Test - 30

Score

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Rank

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SHARING IS CARING

Question 1

$$31.97$$ seconds

$$32.97$$ seconds

$$34.97$$ seconds

$$35.97$$ seconds

Solution

Question 2

$$60$$

$$69$$

$$70$$

None of these

Solution

Question 3

$$17$$

$$29$$

$$33$$

$$34$$

Solution

Question 4

$$2222$$ calories

$$2548$$ calories

$$2563$$ calories

$$2761$$ calories

Solution

Question 5

$$10$$

$$100$$

$$1000$$

$$1$$

Solution

Question 6

True

False

Ambiguous

Data insuffcient

Solution

Question 7

non of thease

0

1

can not det

Solution

Question 8

$$40\dfrac {5}{7} marks$$

$$40\dfrac {4}{7} marks$$

$$30\dfrac {5}{7} marks$$

$$20\dfrac {4}{7} marks$$

Solution

Question 9

$$36.12$$ years

$$36.13$$ years

$$13.36$$ years

$$23.36$$ years

Solution

Question 10

$$5.6$$

$$3.6$$

$$6.83$$

$$5.2$$

Solution

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