Measures of Central Tendency Test

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Measures of Central Tendency Test - 28

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Question 1

1 / -0

Below is given frequency distribution of marks (out of 100) obtained by the students.
Calculate mean marks scored by a student.

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100
No. of Students	3	5	7	10	12	15	12	6	2	8

$$48.75$$ marks

$$51.75$$ marks

$$54.75$$ marks

$$57.75$$ marks

Solution

$$ci$$	$$f_i$$	$$x_i$$	$$f_ix_i$$
0-10	3	5	15
10-20	5	15	75
20-30	7	25	175
30-40	10	35	350
40-50	12	45	540
50-60	15	55	825
60-70	12	65	780
70-80	6	75	450
80-90	2	85	170
90-100	8	95	760

Consider the following table, to calculate mean:

$$N=\Sigma f_i=80$$

$$\Sigma f_ix_i=4140$$

Mean $$\overline x=\dfrac {\Sigma f_ix_i}{N}$$

$$\therefore \overline x=\dfrac{4140}{80}=51.75$$

Mean marks scored by a student is $$51.75$$

Hence, option $$B$$ is correct.

Question 2
1 / -0

Find the mode of the following:
$$18,14,22,23,14,18,17,28,28,14,25,14$$

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

The numbers are: $$18,14,22,23,14,18,17,28,28,14,25,14$$
Arranging in ascending order: $$14, 14, 14, 14, 17, 18, 18, 22, 23,25, 28, 28$$
$$14$$ occurs $$4$$ times and maximum number of times.
Hence, the mode of the above data is $$14$$
Question 3
1 / -0

If $$25$$ is the arithmetic mean of $$ X$$ and $$46,$$ then find $$X.$$

A
$$2$$

B
$$4$$

C
$$8$$

D
$$16$$

Solution

Since $$25$$ is the arithmetic mean of $$X$$ and $$46$$,

$$\therefore 25=\displaystyle \frac{(X+46)}{2}$$
$$\therefore X+46=50$$
$$\therefore X=4$$
Question 4
1 / -0

Following table gives age groupwise distribution of people suffering from 'Asthama' due to air pollution in certain city. Find mean age of person suffering from 'Asthama' .
Age (in years) 7 - 11 11 - 15 15 - 19 19 - 23 23 - 27 27 - 31 31 - 35 35 - 39
No. of People 5 9 13 21 16 15 12 9

A
$$23.88$$years

B
$$24.88$$ years

C
$$25.88 $$years

D
$$26.88$$ years

Solution

Consider the following table, to calculate mean:
$$ci$$ $$ f_i$$ $$x_i$$ $$f_ix_i$$
7-11 5 9 45
11-15 9 13 117
15-19 13 17 221
19-23 21 21 441
23-27 16 25 400
27-31 15 29 435
31-35 12 33 396
35-39 9 37 333
$$N=\Sigma f_i=100$$
$$\Sigma f_ix_i=2388$$

Mean $$\overline x=\dfrac {\Sigma f_ix_i}{N}$$
$$\therefore \overline x=\dfrac{2388}{100}=23.88$$
Mean age of people suffering from Asthama is $$23.88$$years
Hence, option $$A$$ is correct.

Question 5

1 / -0

Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
Find average (mean) amount of milk given by a cow by 'shift of origin method'.

Milk (in litres)	24 - 30	30 - 36	36 - 42	42 - 48	48 - 54	54 - 60	60 - 66	66 - 72	72 - 78	78 - 84	84 - 90
No. of cows	1	3	8	5	5	5	8	4	6	2	3

$$51.12$$ litres

$$54.12$$ litres

$$57.12$$ litres

$$60.12$$ litres

Solution

Consider the following table, to calculate mean by "shift of origin method":

$$x_i$$=mid value of class interval

Assumed mean $$a=57$$

$$ci$$	$$f_i$$	$$x_i$$	$$d_i=x_i-a$$	$$f_id_i$$
24-30	1	27	27-57= -30	-30
30-36	3	33	33-57= -24	-72
36-42	8	39	39-57= -18	-144
42-48	5	45	45-57= -12	-60
48-54	5	51	51-57= -6	-30
54-60	5	57	57-57=0	0
60-66	8	63	63-57=6	48
66-72	4	69	69-57=12	48
72-78	6	75	75-57=18	108
78-84	2	81	81-57=24	48
84-90	3	87	87-57=30	90

$$N=\Sigma f_i=50$$

$$\Sigma f_id_i=6$$

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

$$\therefore \overline x=57 + \dfrac{6}{50}=57.12$$

Average amount of milk given by cow is $$ 57.12$$ litres

Hence, option $$C$$ is correct.

Question 6
1 / -0

The following observation are arranged in ascending order. If the median of the data is $$17$$, find $$x$$ if the observations are $$6, 8, 9, 15, x,x + 2, 21, 22, 25, 29.$$

A
$$x = 11$$

B
$$x = 13$$

C
$$x = 15$$

D
$$x = 16$$

Solution

The observations are: $$6, 8, 9, 15, x,x+2,21,22,25,29$$
The number of observations are $$10$$. Since, the observations are even, the median will be the mean of $$\cfrac{n}{2}$$ and $$\cfrac{n}{2}+1$$ observations
median $$=$$ mean of $$\cfrac{10}{2}$$ and $$\cfrac{10+2}{2}$$ observations
median $$=$$ mean of $$5^{th}$$ and $$6^{th}$$ observations
median = $$\cfrac{x + x+2}{2}$$
$$17 = x + 1$$
$$x = 16$$

Question 7

1 / -0

Below is given frequency distribution of dividend in percentage declared by 120 companies.

Dividend (in %)	10 - 19	20 - 29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79
No. of Companies	5	15	28	42	15	12	3

Obtain mean dividend declared by a company by step deviation method.

$$40.42$$ %

$$42.42$$ %

$$46.42$$ %

$$48.42$$ %

Solution

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

$$x_i$$=mid value of class interval

Assumed mean $$a=44.5$$

class interval $$c=10$$

$$ci$$	$$f_i$$	$$x_i$$	$$u_i=\dfrac{x_i-a}{c}$$	$$f_iu_i$$
$$9.5-19.5$$	$$5$$	$$14.5$$	$$\dfrac{14.5-44.5}{10}=-3$$	$$-15$$
$$19.5-29.5$$	$$15$$	$$24.5$$	$$-2$$	$$-30$$
$$29.5-39.5$$	$$28$$	$$34.5$$	$$-1$$	$$-28$$
$$39.5-49.5$$	$$42$$	$$44.5$$	$$0$$	$$0$$
$$49.5-59.5$$	$$15$$	$$54.5$$	$$1$$	$$15$$
$$59.5-69.5$$	$$12$$	$$64.5$$	$$2$$	$$24$$
$$69.5-79.5$$	$$3$$	$$74.5$$	$$3$$	$$9$$

$$N=\Sigma f_i=120$$

$$\Sigma f_iu_i=-25$$

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

$$\therefore \overline x=44.5 + \dfrac{-25}{120} \times 10=42.4166667 \Rightarrow 42.42$$

Hence, option $$B$$ is correct.

Question 8
1 / -0

Find the median of:
$$25,16,26,16,32,31,19,28$$ and $$35$$

A
$$16$$

B
$$28$$

C
$$26$$

D
$$19$$

Solution

$$\textbf{Step -1: Arrange the given data in ascending order.}$$

$$\text{Arranging data in ascending order:}$$

$$16,16,19,25,26,28,31,32,35$$

$$\textbf{Step -2: Find median of the given data.}$$

$$\text{Here, number of observations, }n=9(\text{odd number})$$

$$\mathbf{\because\text{Median}=\left(\dfrac{n+1}{2}\right)^\text{th}observation.}$$

$$\therefore \text{Median}=\left(\dfrac{9+1}{2}\right)^\text{th}\text{observation}$$
$$=\bigg(\dfrac{10}{2}\bigg)^\text{th}\text{observation}$$

$$=5^\text{th}\text{ observation}$$

$$=26$$

$$\textbf{Hence, the correct option is C.}$$
Question 9
1 / -0

Find the median of the following data:
$$18,19, 20, 23, 22, 20, 17, 19, 25$$ and $$21$$

A
$$20$$

B
$$21$$

C
$$22$$

D
$$23$$

Solution

The given data series is: $$18,19, 20, 23, 22, 20, 17, 19, 25. 21$$
Arranging it in increasing order: $$17, 18, 19, 19, 20, 20, 21, 22, 23, 25$$
The number of terms in the series are $$10$$, which is even. Hence, the median will be the mean of the two middle numbers.
Thus, median $$=$$ mean of $$5^{th}$$ term and $$6^{th}$$ term
Median $$= \dfrac{20 + 20}{2}$$
Median $$= 20$$
Question 10
1 / -0

What is the mean of first 8 whole numbers?

A
$$3$$

B
$$3.5$$

C
$$4$$

D
$$4.5$$

Solution

First $$8$$ whole numbers are $$ 0,1,2,3,4,5,6,7$$

We know that, $$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}$$

$$\therefore \ \text{Mean}=\dfrac {0+1+2+3+4+5+6+7}{8}$$
$$\Rightarrow \dfrac {28}{8}=3.5$$
Hence, the mean of first $$8$$ whole numbers is $$3.5$$.

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Age (in years)	7 - 11	11 - 15	15 - 19	19 - 23	23 - 27	27 - 31	31 - 35	35 - 39
No. of People	5	9	13	21	16	15	12	9

$$ci$$	$$ f_i$$	$$x_i$$	$$f_ix_i$$
7-11	5	9	45
11-15	9	13	117
15-19	13	17	221
19-23	21	21	441
23-27	16	25	400
27-31	15	29	435
31-35	12	33	396
35-39	9	37	333

Measures of Central Tendency Test - 28

Result

Measures of Central Tendency Test - 28

Score

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Rank

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SHARING IS CARING

Question 1

$$48.75$$ marks

$$51.75$$ marks

$$54.75$$ marks

$$57.75$$ marks

Solution

Question 2

$$12$$

$$13$$

$$14$$

$$15$$

Solution

Question 3

$$2$$

$$4$$

$$8$$

$$16$$

Solution

Question 4

$$23.88$$years

$$24.88$$ years

$$25.88 $$years

$$26.88$$ years

Solution

Question 5

$$51.12$$ litres

$$54.12$$ litres

$$57.12$$ litres

$$60.12$$ litres

Solution

Question 6

$$x = 11$$

$$x = 13$$

$$x = 15$$

$$x = 16$$

Solution

Question 7

$$40.42$$ %

$$42.42$$ %

$$46.42$$ %

$$48.42$$ %

Solution

Question 8

$$16$$

$$28$$

$$26$$

$$19$$

Solution

Question 9

$$20$$

$$21$$

$$22$$

$$23$$

Solution

Question 10

$$3$$

$$3.5$$

$$4$$

$$4.5$$

Solution

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