Measures of Central Tendency Test

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Measures of Central Tendency Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The median of the set of the observation 1,3,5,7,11,13,17 is

A
$$1$$

B
$$7$$

C
$$9$$

D
$$17$$

Solution

The set of the observation $$=1,3,5,7,11,13,17$$

$$Median=\dfrac{n+1}{2}\ term$$
$$Median=\dfrac{7+1}{2}\ term$$
$$Median=\dfrac{8}{2}\ term$$
$$Median=4\ term$$
$$Median=7$$

Hence, this is the answer.
Question 2
1 / -0

The median of the observations $$3,18,6,16,12$$ and $$10$$ is

A
$$11$$

B
$$10$$

C
$$12$$

D
$$20$$

Solution

Arranging given number in ascending order,
$$3,6,10,12,16,18$$
Number of observations $$=6$$
$$Median=\dfrac{1}{2}\left[\left(\dfrac{6}{2}\right)^{th}term+\left(\dfrac{6}{2}+1\right)^{th}term\right]$$

$$Median=\dfrac{1}{2}[3^{rd}term+4^{th}term]$$
$$3^{rd}term=10$$ and $$4^{th}term=12$$

$$\therefore$$ $$Median=\dfrac{1}{2}[10+12]$$

$$=\dfrac{1}{2}\times 22$$
$$=11$$
Question 3
1 / -0

In a test in Mathematics, $$7$$ student scored $$19$$ marks, $$11$$ student scored $$15$$ marks, $$16$$ students $$13$$ marks and $$12$$ student scored $$10\ marks$$. The mode of the data is

A
$$19\ marks$$

B
$$15\ marks$$

C
$$13\ marks$$

D
$$10\ marks$$

Solution

As $$13$$ marks is obtained by $$16$$ studends.
(maximum number of students )
Hence,
Mode of data is $$13$$ marks.
Option $$C$$ is correct answer.
Question 4
1 / -0

The mean height of 25 male workers in a factory is 61 cm, and the mean height of 35 female workers in the same factory is 58 cm. The combined mean height of 60 workers in the factory is

A
59.25

B
59.5

C
59.75

D
58.75

Solution

The mean height of $$25$$ male workers in factory is $$61\,cm$$
$$\therefore$$ Total height of $$25$$ male worker $$=25\times 61=1525\,cm$$
The mean height of $$35$$ female workers in factory is $$58\,cm$$
$$\therefore$$ Total height of $$25$$ female worker $$=35\times 58=2030\,cm$$
$$\Rightarrow$$ Total height of $$60$$ workers $$=1525+2030=3555\,cm$$
$$\Rightarrow$$ Required mean $$=\dfrac{3555}{60}=59.25\,cm$$
Question 5
1 / -0

The mean weight of 9 students is 25 kg. If one more student is joined in the group the mean is unaltered, then the weight of the 10th student is

A
25 kg

B
24 kg

C
26 kg

D
23 kg

Solution

Formula used:
$$Arithmetic\ mean= \dfrac{Sum\ of\ given\ numbers}{Total\ numbers}$$

Apply the above formula, we get the sum of weights of the 9 students = $$25\times 9=225$$ kg
If one more student is joined in the group, then total number of students is 10, and the mean isn 25.
The sum of weights of the 10 students is $$25\times 10=250$$ kg
The weight of the tenth student is $$250-225=25$$ kg
Question 6
1 / -0

The sum of $$15$$ observations is $$434+x.$$ If the mean of the data is $$x,$$ then find $$x.$$

A
$$25$$

B
$$27$$

C
$$31$$

D
$$33$$

Solution

Given, sum of all observations $$=434+x$$
Number of observations$$=15$$

and
$$Mean=x$$

$$\Rightarrow \dfrac{434+x}{15}=x$$

$$\Rightarrow 434+x=15x$$

$$\Rightarrow x-15x=-434$$

$$\Rightarrow -14x=-434$$

$$\Rightarrow x=\dfrac{-434}{-14}$$

$$\Rightarrow x=31$$

Hence, the value of $$x$$ is $$31$$ .
Question 7
1 / -0

The mode of the unimodular data $$7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 12$$ and $$x$$ is $$10$$. The value of $$x$$ is

A
$$10$$

B
$$9$$

C
$$8$$

D
$$11$$

Solution

Given data-
$$7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 12, x$$
Mode of the data $$= 10$$
Frequency of $$9 = 3$$
Frequency of $$10 = 3$$
Since frequency of $$9$$ is same as frequency of $$10$$ but mode of data is $$10$$. Thus the value of $$x$$ is $$10$$.
Question 8
1 / -0

Out of 5 brands of chocolates in a shop, a boy has to purchase the brand which is most liked by children . What measure of central tendency would be most appropriate if the data is provided to him?

A
Mean

B
Mode

C
Median

D
Any of the three

Solution

As we know mode is the measure of central tendency used to find that what occurs most often.
Hence, Mode would be most appropriate for him.
Question 9
1 / -0

The mean of first ten odd natural number is

A
5

B
10

C
20

D
19

Solution

Since, first ten odd natural numbers are$$1, 3, 5, 7, 9, 11, 13, 15, 17, 19$$

We know that
$$\text{Mean}=\dfrac{\text{sum of observation}}{\text{number of observation}}$$
$$\therefore$$ Mean $$=\dfrac{(1 + 19 + 3 + 17 + 5 + 15 + 7 + 13 + 9 + 11) }{ 10} = \dfrac{100}{10} = 10$$
Question 10
1 / -0

From a frequency distribution table if $$N = 100, h = 10 ;c.f. = 38 ;f = 18 ;L = 50$$, then find the median for the distribution. Choose the correct alternative.

A
$$56.67 $$

B
$$55.76 $$

C
$$56.76 $$

D
$$ 55.87 $$

Solution

Medium $$= I + \left(\dfrac{\dfrac{N}{2}- c.f}{f}\right)\times h$$
$$= 50 + \dfrac{\left(\dfrac{100}{2}-38\right)}{18}\times 10$$
$$50+\dfrac{(50-38)}{18} \times 10 = 50 + \dfrac{12}{18} \times 10 = 56.67$$

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Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 23

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Measures of Central Tendency Test - 23

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Question 1

$$1$$

$$7$$

$$9$$

$$17$$

Solution

Question 2

$$11$$

$$10$$

$$12$$

$$20$$

Solution

Question 3

$$19\ marks$$

$$15\ marks$$

$$13\ marks$$

$$10\ marks$$

Solution

Question 4

59.25

59.5

59.75

58.75

Solution

Question 5

25 kg

24 kg

26 kg

23 kg

Solution

Question 6

$$25$$

$$27$$

$$31$$

$$33$$

Solution

Question 7

$$10$$

$$9$$

$$8$$

$$11$$

Solution

Question 8

Mean

Mode

Median

Any of the three

Solution

Question 9

5

10

20

19

Solution

Question 10

$$56.67 $$

$$55.76 $$

$$56.76 $$

$$ 55.87 $$

Solution

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