Measures of Central Tendency Test

Result

Measures of Central Tendency Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The median of the given data:
$$49$$, $$48$$, $$15$$, $$20$$, $$28$$, $$17$$, $$14$$ and $$10$$ is

A
$$18$$

B
$$20$$

C
$$22$$

D
$$24$$

Solution

$$10,14,15,17,20,28,48,49\\ n=8\\ \therefore median=\cfrac { n }{ 2 } +1\\ ={ (4+1) }^{ th }\quad obs\\ ={ (5) }^{ th }\quad obs\\ =20$$
Question 2
1 / -0

The following table shows the number of students and the time they utilized daily for their studies. Find the mean time spent by students for their studies by direct method.
Time (hrs.) $$0-2$$ $$2-4$$ $$4-6$$ $$6-8$$ $$8-10$$
No. of students $$7$$ $$18$$ $$12$$ $$10$$ $$3$$

A
$$4\ hrs$$

B
$$5\ hrs$$

C
$$4.36\ hrs$$

D
$$5.36\ hrs$$
Question 3
1 / -0

Median of $$8,4,6,18,5,12$$ and $$100$$ is __________.

A
$$8$$

B
$$6$$

C
$$12$$

D
$$100$$

Solution

Arrange in ascending order $$\Rightarrow$$
$$4, 5, 6,\underset{\uparrow}{8}, 12, 18, 100$$
$$7$$ observations $$\therefore$$ Median is $$4^{th}$$ observation
Median $$=8$$ Answer.
Question 4
1 / -0

The average annual income (in Rs.) of certain agricultural workers is $$S$$ and that of other workers is $$T$$. The number of agricultural workers is $$11$$ times that of other workers. Then the average monthly income (in Rs.) of all the workers is:

A
$$\cfrac{S+T}{2}$$

B
$$\cfrac{S+11T}{2}$$

C
$$\cfrac{1}{11S}+T$$

D
$$\cfrac{11S+T}{12}$$

Solution

Let the number of other workers $$ =x$$

Then number of agricultural worker $$=11x$$

Total number of worker $$=12x$$

Average monthly income $$=\dfrac{S\times11x+T\times x}{12x}$$

$$=\dfrac{11S+T }{12}$$
Question 5
1 / -0

The average age of the boys in a class is $$16$$ years and that of the girls is $$15$$ years. The average age for the whole class when there are equal number of girls and boys is:

A
$$15$$ years

B
$$15.5$$ years

C
$$16$$ years

D
Cannot be computed with the given information

Solution

Let No of boys = No of girls = x
Avg age of whole class = $$\dfrac{16\times x+15\times x}{x+x}$$
$$ = \dfrac{16+15}{2}$$
$$ = 15.5 $$ years
Question 6
1 / -0

The mean and median of proper divisors of $$360$$ are connected by

A
median> mean

B
median< mean

C
median = mean

D
median = 2(mean)
Question 7
1 / -0

The frequency distribution of the marks obtained by $$100$$ students in a test carrying $$50$$ marks then the mean is
Marks 0-9 10-19 20-29 30-39 40-49
No. of students 8 15 20 45 12

A
$$28.3$$

B
$$28$$

C
$$27.3$$

D
$$26.4$$

Solution

Here the mean of given data is given by $$\mu =\dfrac { \sum { { f }_{ i }{ x }_{ i } } }{ { f }_{ i } } $$,
Where $$x_i$$ are the mid values the mark range and $$f_i$$ are the corresponding frequencies or in this case the number of students.

$$\therefore \mu =\dfrac{4.5\times 8+14.5\times 15+24.5\times 20+34.5\times 45+44.5\times 12}{8+15+20+45+12}$$

$$=\dfrac{2830}{100}$$

$$=28.30$$
Hence, the answer is $$28.30.$$
Question 8
1 / -0

The sum of first $$n$$ natural numbers is given by the expression $$(2n^{2}+3n)$$. The mean of the given numbers is

A
$$\left(\dfrac{2}{n}+3\right)$$

B
$$\left(2+\dfrac{3}{n}\right)$$

C
$$(2+3n)$$

D
$$(2n+3)$$

Solution
Question 9
1 / -0

The median of $$\dfrac { x }{ 5 } ,x,\dfrac { x }{ 4 } ,\dfrac { x }{ 2 } ,\dfrac { x }{ 3 } $$ is $$8$$ then the value of $$x$$ is

A
24

B
18

C
27

D
51

Solution

Number of terms, n=5.
Given median = 8
$$\Rightarrow (\frac{n+1}{2})^{x} term=8$$
$$\Rightarrow (\frac{5+1}{2})^{x}term =8$$
$$\Rightarrow 3^{x}$$ term =8
$$\Rightarrow \frac{x}{4}=8$$
$$\Rightarrow x=32$$
Question 10
1 / -0

The mean of first ten prime numbers is

A
$$12$$

B
$$12.9$$

C
$$13$$

D
$$14$$

Solution

First 10 prime numbers are $$2,3,5,7,11,13,17,19,23,29$$
Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$

Apply the above formula, we get

Mean=$$\frac{2+3+5+7+11+13+17+19+23+29}{10}=12.9$$

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Time (hrs.)	$$0-2$$	$$2-4$$	$$4-6$$	$$6-8$$	$$8-10$$
No. of students	$$7$$	$$18$$	$$12$$	$$10$$	$$3$$

Marks	0-9	10-19	20-29	30-39	40-49
No. of students	8	15	20	45	12

Measures of Central Tendency Test - 49

Result

Measures of Central Tendency Test - 49

Score

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Rank

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SHARING IS CARING

Question 1

$$18$$

$$20$$

$$22$$

$$24$$

Solution

Question 2

$$4\ hrs$$

$$5\ hrs$$

$$4.36\ hrs$$

$$5.36\ hrs$$

Question 3

$$8$$

$$6$$

$$12$$

$$100$$

Solution

Question 4

$$\cfrac{S+T}{2}$$

$$\cfrac{S+11T}{2}$$

$$\cfrac{1}{11S}+T$$

$$\cfrac{11S+T}{12}$$

Solution

Question 5

$$15$$ years

$$15.5$$ years

$$16$$ years

Cannot be computed with the given information

Solution

Question 6

median> mean

median< mean

median = mean

median = 2(mean)

Question 7

$$28.3$$

$$28$$

$$27.3$$

$$26.4$$

Solution

Question 8

$$\left(\dfrac{2}{n}+3\right)$$

$$\left(2+\dfrac{3}{n}\right)$$

$$(2+3n)$$

$$(2n+3)$$

Solution

Question 9

24

18

27

51

Solution

Question 10

$$12$$

$$12.9$$

$$13$$

$$14$$

Solution

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