Measures of Central Tendency Test

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Measures of Central Tendency Test - 26

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Question 1
1 / -0

The mean of the following distribution is
class
Frequency
0-5
4
5-10
5
10-15
7
15-20
12
20-25
7
25-30
5

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution

Mean $$=$$ $$\dfrac{sum \ of \ (Class mark \times frequency)}{Sum \ of \ frequency}$$

Mean $$=$$ $$\dfrac{2.5\times 4 + 7.5\times5 + 12.5\times 7+ 17.5\times12 + 22.5\times 7 + 27.5\times 5}{4 +5+7+12+7 +5}$$
Mean = $$\dfrac{640}{40} = 16$$
Question 2
1 / -0

The numbers $$ 4, 6\ and\ 8$$ are the frequencies of $$(x + 2), x \ and\ (x - 1) $$ respectively . If their arithmetic mean is $$8,$$ the value of $$x$$ is

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

$$8 =$$ $$\dfrac{4\times \left ( x+2 \right )+6x+8\times \left ( x-1 \right )} {4+6+8}$$
$$=\dfrac{4x+8+6x+8x-8}{18}$$
$$\therefore$$ $$x = 8$$

Question 3

1 / -0

The examination marks of $$250$$ pupils are recorded below. Find the mean.

Class interval	0-9	10-19	20-29	30-39	40-49
Frequency	0	2	6	24	36
Class interval	50-59	60-69	70-79	80-89	90-99
Frequency	47	55	40	27	13

$$50.12$$

$$90.31$$

$$60.82$$

$$75.12$$

Solution

$$Mean = \displaystyle \frac{Total (mid value \times frequency)}{Total frequency}$$

Class interval	Mid- value	Fequency.	Mid-value $$\times$$ frequency
0-9	4.5	0	0
10-19	14.5	2	29
20-29	24.5	6	147
30-39	34.5	24	828
40-49	44.5	36	1602
50-59	54.5	47	2561.5
60-69	64.5	55	3547.5
70-79	74.5	40	2980
80-89	84.5	27	2281.5
99-99	94.5	13	1228.5
		Total - 250	Total 15205

$$Mean = \displaystyle \frac{Total (mid value \times frequency)}{Total frequenct} = \frac{15205}{250}= 60.82$$

Question 4
1 / -0

A contractor employed $$18$$ labourers at Rs. $$12$$ per day, $$ 10$$ labourers at Rs.$$ 13.50 $$ per day,$$ 5$$ labourers at Rs.$$ 25 $$ per day and $$2$$ labourers at Rs.$$ 42$$ per day. The average wage of a labourer per day is

A
Rs. $$16$$

B
Rs. $$20$$

C
Rs.$$ 24$$

D
Rs. $$28$$

Solution

Mean = $$\dfrac{Total Sum}{Total Count}$$

Mean = $$\dfrac{18\times12 + 10\times13.50 + 5\times25 + 2\times42}{18+10+5+2}$$
= $$\dfrac{560}{35}$$
$$= 16$$
Question 5
1 / -0

In a class test in English $$10$$ students scored $$75$$ marks, $$12$$ students scored $$60$$ marks, $$8$$ scored $$40$$ marks and $$3$$ scored $$30$$ marks, the mode for their score is

A
$$75$$

B
$$30$$

C
$$60$$

D
$$25$$

Solution

$$\textbf{Step -1: Defining mode.}$$
$$\text{In a given data set, the mode is the value that appears most often among every other values.}$$
$$\text{Given data set says, }$$
$$\text{10 Students scored 75 marks.}$$
$$\text{12 Students scored 60 marks.}$$
$$\text{8 Students scored 40 marks.}$$
$$\text{3 Students scored 30 marks.}$$
$$\textbf{Step -2: Finding the mode of their score.}$$
$$\text{The mode of their score will be the marks occurring most often.}$$
$$\therefore \text{12 Students scored 60 marks.}$$
$$\text{Hence 60 is the mode of their score.}$$
$$\textbf{Hence,The mode of their score is 60. (Option C)}$$
Question 6
1 / -0

The marks of $$20$$ students in a test were as follows:
$$5, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20$$
The median is

A
$$13.5$$

B
$$12.5$$

C
$$14.5$$

D
$$15.5$$

Solution

Here, $$n$$ $$=$$ $$20$$ which is an even number.
$$\therefore$$ Median $$= \displaystyle \cfrac{\left ( \dfrac{n}{2} \right )^{th} \text{term} + \left ( \dfrac{n}{2}+1\right )^{th} \text{term}}{2}$$
$$\therefore$$ Median $$= \displaystyle \cfrac{\left ( \dfrac{20}{2} \right )^{th} \text{term} + \left ( \dfrac{20}{2}+1\right )^{th} \text{term}}{2}$$
$$\displaystyle =\cfrac{10^{th}\: \text{term} + 11^{th}\: \text{term}}{2}$$
$$\displaystyle =\cfrac{13+14}{2}=13.5$$
Question 7
1 / -0

The median of variables 5, 8, 10, 13, 17, 21, 27, 30, 33, 42 is

A
21

B
17

C
38

D
19

Solution
Question 8
1 / -0

Arithmetic mean for the given data is
Marks
No. of students
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
50 - 60
5
10
25
30
20
10

A
$$55$$

B
$$35$$

C
$$2$$

D
$$33$$

Solution

Mean $$=$$ $$\dfrac{\text {Sum of} (\text {class mark} \times \text {frequency})}{\text {Sum of frequency}}$$

Mean $$=$$ $$\dfrac{5\times 5 + 15\times10 + 25\times 25+ 35\times30 + 45\times 20 + 55\times 10}{5 +10+25+30+20 +10}$$

Mean $$=$$ $$\dfrac{3300}{100}$$ = $$33$$
Question 9
1 / -0

The average weight of $$10$$ men is decreased by $$ 3 \space kg$$ when one of them whose weight is $$80 kg $$ is replaced by a new person. The weight of the new person is

A
$$70$$

B
$$60$$

C
$$50$$

D
$$73$$

Solution

Let the mean of $$10$$ persons =$$ M$$
Sum of weight of the persons $$ = 10M$$

After replacing a person weighing 80 Kg with another weighing say$$ x$$ Kg, $$sum = 10M - 80 +x$$
New Mean$$ = M-3$$
hence,$$ 10(M-3) = 10M - 80 +x $$
$$10M- 30= 10M -80 +x$$
$$x = 50 Kg$$
Question 10
1 / -0

In a family, there are $$8$$ men, $$7$$ women and $$5$$ children whose mean ages separately are respectively $$24,20$$ and $$6$$ years. The mean age of the family is

A
$$17.1$$ years

B
$$18.1$$ years

C
$$19.1$$ years

D
none of these

Solution

Here we have three collections for which $${ A }_{ 1 }=24,{ n }_{ 1 }=8;{ A }_{ 2 }=20,{ n }_{ 2 }=7$$ and $${ A }_{ 3 }=6,{ n }_{ 3 }=5$$.
Their combined mean is the required mean.
By the formula $$\displaystyle A=\frac { { n }_{ 1 }{ A }_{ 1 }+{ n }_{ 2 }{ A }_{ 2 }+{ n }_{ 3 }{ A }_{ 3 } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 } } =\frac { 8\times 24+7\times 20+5\times 6 }{ 8+7+5 } $$
$$\displaystyle =\frac { 192+140+30 }{ 20 } =\frac { 362 }{ 20 } =18.1$$
$$\therefore$$ The mean age of the family $$=18.1$$ years

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Marks	No. of students
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60	5 10 25 30 20 10

class	Frequency
0-5	4
5-10	5
10-15	7
15-20	12
20-25	7
25-30	5

Measures of Central Tendency Test - 26

Result

Measures of Central Tendency Test - 26

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SHARING IS CARING

Question 1

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 2

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 3

$$50.12$$

$$90.31$$

$$60.82$$

$$75.12$$

Solution

Question 4

Rs. $$16$$

Rs. $$20$$

Rs.$$ 24$$

Rs. $$28$$

Solution

Question 5

$$75$$

$$30$$

$$60$$

$$25$$

Solution

Question 6

$$13.5$$

$$12.5$$

$$14.5$$

$$15.5$$

Solution

Question 7

21

17

38

19

Solution

Question 8

$$55$$

$$35$$

$$2$$

$$33$$

Solution

Question 9

$$70$$

$$60$$

$$50$$

$$73$$

Solution

Question 10

$$17.1$$ years

$$18.1$$ years

$$19.1$$ years

none of these

Solution

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