Measures of Central Tendency Test

Result

Measures of Central Tendency Test - 25

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 median of a set of $$9$$ distinct observations is $$20.5$$. If each of the largest $$4$$ observation of the set is increased by $$2$$, then the median of the new set

A
is increased by $$2$$

B
is decreased by $$2$$

C
is two times the original median

D
remains the same as that of the original set

Solution

$$Since$$ $$n= 9,$$ $$the$$ $$median$$ $$will$$ $$be$$ $$the$$ $$5th(\frac{9 + 1}{2}) term.$$
$$If$$ $$the$$ $$largest$$ $$4$$ $$terms$$ $$are$$ $$increased$$ $$by$$ $$2,$$ $$there$$ $$is$$ $$no$$ $$effect$$ $$on$$ $$the$$ $$5th$$ $$term,$$ $$implying$$ $$median$$ $$will$$ $$remain$$ $$same.$$
Question 2
1 / -0

In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class are 72, then the average marks of the girls :

A
$$73$$

B
$$65$$

C
$$68$$

D
$$74$$

Solution

Average of marks of $$70$$ boys =$$75$$

Sum of marks of $$70$$ boys = $$75\times70$$ = $$5250$$

Average of marks of whole class = $$72$$

Sum of marks of whole class = $$100\times72$$ = 7200

Sum of marks of $$30$$ girls =$$ 7200-5250 = 1950$$

Average of marks of girls = $$\dfrac{1950}{30}$$ =$$ 65$$
Question 3
1 / -0

If mode of the following data is $$7$$, then the value of $$k$$ in the $$2, 4, 6, 7, 5, 6, 10, 6, 7, 2k + 1, 9, 7, 13$$ is:

A
$$3$$

B
$$7$$

C
$$4$$

D
$$2$$

Solution

Mode is the value which occurs most often in the data set of values.
Given data values are $$2,4,6,7,5,6,10,6,7,2k+1,9,7,13$$
In the above data set, values $$6,7$$ have occured more times i.e., 3 times

But given that mode is $$7$$. So, should occur more times than $$6$$. Hence the variable $$2k+1$$ must be 7

$$\implies 2k+1=7$$

$$\implies 2k=6 \implies k=\dfrac 62$$

$$\implies k=3$$
Question 4
1 / -0

The numbers $$3, 5, 7$$ and $$9$$ have their respective frequencies $$ x-2, x+2, x-3 $$ and $$ x+3$$. If the arithmetic mean is $$6.5$$, then the value of $$ x$$ is

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

Mean $$= 6.5 = \dfrac{3 \times (x-2) + 5 \times (x + 2) + 7 \times (x - 3) + 9 \times (x + 3)}{x-2+x+2+x-3+x+3}$$
$$\therefore$$ $$6.5 = \dfrac{3x - 6 + 5x + 10 + 7x - 21 + 9x + 27}{4x}$$
$$\therefore$$ $$6.5 \times 4x = 24x + 10$$
$$\therefore$$ $$26x - 24x = 10$$
$$\therefore$$ $$x = 5$$
Question 5
1 / -0

Mode of the data
$$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$$ is

A
$$14$$

B
$$15$$

C
$$16$$

D
$$17$$

Solution

Given Data:
$$15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15, 17, 15$$

Arranging it in ascending order: $$14,14,14,14,15$$,$$15,15,15,15,16,17,18,19,19,20$$

$$15$$ occurs $$5$$ times and most number of time.

Hence, it is the mode of the data.
Question 6
1 / -0

Median of the following numbers:
$$4, 4, 5, 7, 6, 7, 7, 12, 3$$ is

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

Solution

Terms are: $$4, 4, 5, 7, 6, 7, 7, 12, 3$$.
Arranging the terms in ascending order: $$3,4,4,5,6,7,7,7,12$$.

Since the number of terms is odd the median will be the middle term i.e. $$5^{th}$$ term which is $$6$$.

Question 7

1 / -0

Directions For Questions

The following table gives the weekly wages of workers in a factory :

Weekly wages	Mid-value $$(x_i)$$	No. of $$(f_i)$$ workers	$$f_ix_i$$	Cumulativefrequency
50-55	52.5	5	262.5	5
55-60	57.5	20	1150.0	25
60-65	62.5	10	625.0	35
65-70	67.5	10	675.0	45
70-75	72.5	9	652.0	54
75-80	77.5	6	465.0	60
80-85	82.5	12	990.0	72
85-90	87.5	8	700.0	80
		$$\sum f_i=80$$	$$\sum f_ix_i =5520$$

...view full instructions

The mean is

$$70$$

$$68$$

$$71$$

$$69$$

Solution

Mean of a frequency distribution =$$\displaystyle \dfrac{\sum f_i x_i}{\sum f_i}$$
Mean $$= \dfrac {5520}{80}$$
Mean $$= 69$$

Question 8
1 / -0

If three sets of data had means of $$15, 22.5$$ and $$24$$ based on $$6, 4,$$ and $$5$$ observations respectively, then the mean of these three sets combined is

A
$$20.0$$

B
$$20.5$$

C
$$22.5$$

D
$$24.0$$

Solution

Mean of combined three sets $$=\dfrac{Total \space Sum}{Total \space Count}$$
$$=\dfrac{15\times6 + 22.5\times4 + 24\times5}{6+4+5}$$
$$=\dfrac{300}{15}$$
$$= 20$$
Question 9
1 / -0

The mean weight of $$20$$ students is $$25$$ kg and the mean weight of another $$10$$ students is $$40$$ kg. Find the mean weight of the $$30$$ students.

A
$$20$$

B
$$30$$

C
$$40$$

D
$$50$$

Solution

$$
\text{Given Mean weight of 20 students}=25\\ \Rightarrow \dfrac { \text{Sum of weight of 20 students} }{ 20 } =25\\
\Rightarrow \text{Sum of weight of 20 students}=25\times 20=500kg\\
\text{Given Mean weight of 10 students}=40\\ \Rightarrow \dfrac { \text{Sum of weight of 10 students} }{ 10 } =40\\
\Rightarrow \text{Sum of weight of 10 students}=40\times 10=400kg$$
Mean weight of all 30 students
$$\Rightarrow \dfrac { \text{Sum of weight of 30 students} }{ 30 } \\
\Rightarrow \dfrac { \text{Sum of weight of 20 students+Sum of weight of 10 students} }{ 30 } \\
=\dfrac { 500+400 }{ 30 } =\dfrac { 900 }{ 30 } =30kg
$$
Question 10
1 / -0

The number of observations in a group is $$40$$ . If the average of first $$10$$ is $$4.5$$ and that of the remaining $$30$$ is $$3.5$$, then the average of the whole group is

A
$$\displaystyle \frac{1}{5}$$

B
$$\displaystyle \frac{15}{4}$$

C
$$4$$

D
$$8$$

Solution

$$\textbf{Step -1: Finding the sum of first 10 observations.}$$

$$\text{Average of first 10 observations}=4.5$$

$$\mathbf{\because\text{Average}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$$

$$\therefore4.5=\dfrac{\text{Sum of first 10 observations}}{10}$$

$$\Rightarrow\text{Sum of first 10 observations}=4.5\times10=45$$

$$\textbf{Step -2: Finding the sum of remaining 30 observations.}$$

$$\text{Average of remaining 30 observations}=3.5$$

$$\therefore3.5=\dfrac{\text{Sum of remaining 30 observations}}{30}$$

$$\Rightarrow\text{Sum of remaining 30 observations}=3.5\times30=105$$

$$\textbf{Step -3: Finding the average of the whole group.}$$

$$\text{Sum of total 40 students}=45+105=150$$

$$\text{Number of total observations}=40$$

$$\text{Average of the whole group}=\dfrac{150}{40}$$

$$=\dfrac{15}{4}$$

$$\textbf{ Hence, option B is correct.}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 25

Result

Measures of Central Tendency Test - 25

Score

-

Rank

-

SHARING IS CARING

Question 1

is increased by $$2$$

is decreased by $$2$$

is two times the original median

remains the same as that of the original set

Solution

Question 2

$$73$$

$$65$$

$$68$$

$$74$$

Solution

Question 3

$$3$$

$$7$$

$$4$$

$$2$$

Solution

Question 4

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 5

$$14$$

$$15$$

$$16$$

$$17$$

Solution

Question 6

$$4$$

$$5$$

$$6$$

$$7$$

Solution

Question 7

$$70$$

$$68$$

$$71$$

$$69$$

Solution

Question 8

$$20.0$$

$$20.5$$

$$22.5$$

$$24.0$$

Solution

Question 9

$$20$$

$$30$$

$$40$$

$$50$$

Solution

Question 10

$$\displaystyle \frac{1}{5}$$

$$\displaystyle \frac{15}{4}$$

$$4$$

$$8$$

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.