Measures of Central Tendency Test

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Measures of Central Tendency Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

A group of $$10$$ items has arithmetic mean $$6$$. If the arithmetic mean of $$4$$ of these items is $$7.5$$, then the mean of the remaining items is

A
$$6.5$$

B
$$5.5$$

C
$$4.5$$

D
$$5.0$$

Solution

We have, $$\bar{X}=6,\bar{X_1} = 7.5, \bar{X_2}=?, w_1=4, w_2=6$$
$$\therefore \displaystyle \bar{X} = \frac{\bar{X_1}w_1+
\bar{X_2}w_1
}{w_1+w_2}=\frac{7.5\times 4+\bar{X_2}\times 6}{10}=6\Rightarrow \bar{X_2} = 5$$
Question 2
1 / -0

Marks obtained by 60 students of a class are shown as
Marks 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Number
of students
8 7 $$f_1$$ 15 5 $$f_2$$ 7
the mean marks are 64, then $$f_1$$ : $$f_2$$ is

A
$$7 : 4$$

B
$$4 : 5$$

C
$$5 : 4$$

D
None of these

Solution

Total number of student is 28. $$\Rightarrow f_1+f_2+42=60\Rightarrow f_1+f_2 = 18$$ (i)

And mean $$ =\cfrac{35\times 8+45\times 7+55f_1+65\times 15+75\times 5+85f_2+95\times 7}{60} = 64$$ (given)

$$\Rightarrow 11f_1+17f_2 = 246$$ (ii)

solving these two equation we get, $$f_1 = 10, f_2 = 8$$
Question 3
1 / -0

The daily wages (in Rs.) of ten workers are $$20,25,17,18,8,15,22,11,9,14$$. The median of wages is

A
$$15$$

B
$$16$$

C
$$17$$

D
$$14$$

Solution

Arranging the wages in ascending order of magnitude, we have
$$\quad 8,9,11,14,15,17,18,20,22,25$$.
Since there are $$10$$ observations. Therefore, Median is arithmetic mean of $$\left(\displaystyle\frac{10}{2}\right)^{th}$$ and $$\left(\displaystyle\frac{10}{2} + 1\right)^{th}$$ observations.
So, the Median = $$\displaystyle\frac{15+17}{2} = 16$$
Question 4
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
$$\displaystyle\frac{15}{4}$$

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

C
$$8$$

D
$$4$$

Solution

$$\textbf{Step -1: Finding the sum of first 10 observations.}$$
$$\text{Average of first 10 observations}=4.5$$
$$\mathbf{\because\text{Average}=\dfrac{\text{Sum of observations}}{\text{Number of observations}}.}$$
$$\therefore4.5=\dfrac{\text{Sum of first 10 observations}}{10}$$
$$\Rightarrow\text{Sum of first 10 observations}=4.5\times10=45$$
$$\textbf{Step -2: Finding the sum of remaining 30 observations.}$$
$$\text{Average of remaining 30 observations}=3.5$$
$$\therefore3.5=\dfrac{\text{Sum of remaining 30 observations}}{30}$$
$$\Rightarrow\text{Sum of remaining 30 observations}=3.5\times30=105$$
$$\textbf{Step -3: Finding the average of the whole group.}$$
$$\text{Sum of total 40 students}=45+105=150$$
$$\text{Number of total observations}=40$$
$$\text{Average of the whole group}=\dfrac{150}{40}$$
$$=\dfrac{15}{4}$$
$$\textbf{ Hence, option A is correct.}$$
Question 5
1 / -0

Obtain the median for the following frequency distribution
x:
1
2
3
4
5
6
7
8
9
f :
8
10
11
16
20
25
15
9
6

A
$$6$$

B
$$4$$

C
$$3$$

D
$$5$$

Solution

Calculate of Median
Here N$$= 120 \Rightarrow \displaystyle \frac{N}{2}=60$$
we find that the cumulative frequency just greater than N/2, is 65 and the value of x corresponding to 65 is 5. Therefore, median is 5.

x
f
c.f
1
8
8
2
10
18
3
11
29
4
16
45
5
20
65
6
25
90
7
15
105
8
9
114
9
6
120

N=120
Question 6
1 / -0

If $$x_i$$'s are the mid points of the class intervals of grouped data, $$f_1$$'s are the corresponding frequencies and $$\bar x$$ is the mean, then $$\sum (f_i x_i - \bar x)$$ is equal to

A
$$0$$

B
$$-1$$

C
$$1$$

D
$$2$$

Solution

Given that, $$x_i$$ are the mid points of the class intervals.
$$f_i$$ are the corresponding frequencies.
The mean $$\bar x$$ for the grouped data is $$\bar x=\dfrac{\sum f_ix_i}{n}$$

$$\implies \bar xn=\sum f_ix_i$$ ------(1)

$$\sum (f_ix_i-\bar x)=\sum f_ix_i-\sum \bar x$$

$$\implies \sum (f_ix_i-\bar x)=\bar xn-\sum \bar x$$ (from (1))

$$\implies \sum (f_ix_i-\bar x)=\bar xn-\bar xn=0$$
Question 7
1 / -0

If $$u_i \displaystyle = \frac{x_i - 25}{10} , \sum f_i u_i=20$$ and $$\sum f_i =100$$, then $$\overline {x}$$ is equal to

A
$$27$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

$$\text{Mean} = \displaystyle a + \frac{\sum f_i u_i}{\sum f_i} \times h , u_i = \frac{x_i -a}{h}$$
$$a = 25 , h = 10, \sum f_i u_i = 20, \sum f_i = 100$$
Mean $$= 25 + \displaystyle \frac{20}{100} \times 10 = 27$$
Question 8
1 / -0

A firm of readymade garments make both men's and women's shirt. Its profit average is $$6$$% of sales. Its profits in men's shirt average $$8$$% of sales and women's shirts comprise $$60$$% of output. The average profit per sales rupee in women's shirts is :

A
$$0.0466$$

B
$$0.0166$$

C
$$0.0666$$

D
None of these

Solution

Here $$\overline { x } =6,\overline { { x }_{ 1 } } =8,{ n }_{ 1 }=40,{ n }_{ 2 }=60$$.
Assuming that the total output is $$100,$$ we are required to find out $$\overline { { x }_{ 2 } } $$, we know that
$$\displaystyle \overline { x } =\frac { { n }_{ 1 }\overline { { x }_{ 1 } } +{ n }_{ 2 }\overline { { x }_{ 2 } } }{ { n }_{ 1 }+{ n }_{ 2 } } =\frac { 40\times 8+60\overline { { x }_{ 2 } } }{ 40+60 } $$
$$\displaystyle \Rightarrow 6=\frac { 320+60\overline { { x }_{ 2 } } }{ 100 } \Rightarrow \overline { { x }_{ 2 } } =\frac { 600-320 }{ 60 } =\frac { 280 }{ 60 } =4.66$$
Thus, the average profit in womens shirt is $$4.66$$% of sales or $$Rs \space 0.0466$$ per sale rupee.

Question 9

1 / -0

Following table gives frequency distribution of trees planted by different housing societies in a particular locality.
Find mean number of trees planted by housing society by using 'step deviation method'.

No. of tress	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
No. of Societies	2	7	9	8	6	4

$$25.42$$ trees

$$27.42$$ trees

$$29.42$$ trees

$$31.42$$ trees

Solution

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

$$x_i$$=mid value of class interval

Assumed mean $$a=27.5$$

class interval $$c=5$$

$$ci$$	$$f_i$$	$$x_i$$	$$u_i=\dfrac{x_i-a}{c}$$	$$f_iu_i$$
10-15	2	12.5	$$\dfrac{12.5-27.5}{5}=$$-3	-6
15-20	7	17.5	-2	-14
20-25	9	22.5	-1	-9
25-30	8	27.5	0	0
30-35	6	32.5	1	6
35-40	4	37.5	2	8

$$N=\Sigma f_i=36$$

$$\Sigma f_iu_i=-15$$

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

$$\therefore \overline x=27.5 + \dfrac{-15}{36} \times 5=25.42$$

Hence, option $$A$$ is corect.

Question 10
1 / -0

The results of two colleges are as follows: then
M.A M.Sc. B.A. B.Sc. Total
A Appeared

100 60 120 200 480
passed 90 45 75 150 360
B Appeared

240 200 160 200 800
Passed 200 160 100 140 600

A
A is better than B

B
None is better than the other

C
A is inferior to B

D
can not say

Solution

Total number of students of College A appeared $$=480$$
Total number of students of College A passed$$=360$$
$$\therefore$$ passing percentage $$=\dfrac{360}{480}\times 100=75$$ %

Total number of students of College B appeared $$=800$$
Total number of students of College A passed$$=600$$
$$\therefore$$ passing percentage $$=\dfrac{600}{800}\times 100=75%$$ %

Since Both colleges have passing percentage as same,none is better than other.
Hence, option $$B$$ is correct.

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Marks	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Number of students	8	7	$$f_1$$	15	5	$$f_2$$	7

		M.A	M.Sc.	B.A.	B.Sc.	Total
A	Appeared	100	60	120	200	480
	passed	90	45	75	150	360
B	Appeared	240	200	160	200	800
	Passed	200	160	100	140	600

x	f	c.f
1	8	8
2	10	18
3	11	29
4	16	45
5	20	65
6	25	90
7	15	105
8	9	114
9	6	120
		N=120

Measures of Central Tendency Test - 27

Result

Measures of Central Tendency Test - 27

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SHARING IS CARING

Question 1

$$6.5$$

$$5.5$$

$$4.5$$

$$5.0$$

Solution

Question 2

$$7 : 4$$

$$4 : 5$$

$$5 : 4$$

None of these

Solution

Question 3

$$15$$

$$16$$

$$17$$

$$14$$

Solution

Question 4

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

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

$$8$$

$$4$$

Solution

Question 5

$$6$$

$$4$$

$$3$$

$$5$$

Solution

Question 6

$$0$$

$$-1$$

$$1$$

$$2$$

Solution

Question 7

$$27$$

$$25$$

$$30$$

$$35$$

Solution

Question 8

$$0.0466$$

$$0.0166$$

$$0.0666$$

None of these

Solution

Question 9

$$25.42$$ trees

$$27.42$$ trees

$$29.42$$ trees

$$31.42$$ trees

Solution

Question 10

A is better than B

None is better than the other

A is inferior to B

can not say

Solution

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