Measures of Central Tendency Test

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Measures of Central Tendency Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The numbers are arranged in the descending order : $$108,\,94,\,88,\,82,\,x+7,\,x-7,\,60,\,58,\,42,\,39$$. If the median is $$73$$, the value of $$x$$ is:

A
$$72$$

B
$$73$$

C
$$76$$

D
$$75$$

Solution

Numbers are $$108,94,88,82,x+7,x-7,60,58,42,39$$
Median $$=73$$
Median $$=\cfrac { { \left( \cfrac { n }{ 2 } \right) }+{ \left( \cfrac { n }{ 2 } +1 \right) } }{ 2 } $$
$$\left( \cfrac { n }{ 2 } \right) $$th obs $$=x+7$$
$$\left( \cfrac { n }{ 2 } +1 \right) $$th obs $$=x-7$$
$$73=\cfrac { (x+7)+(x-7) }{ 2 } \\ =>146=2x\\ =>x=73$$
Question 2
1 / -0

Numbers $$50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8$$ are written in descending order and their median is $$25$$ find $$x$$?

A
$$20$$

B
$$25$$

C
$$12$$

D
$$11$$

Solution

$$\displaystyle \because $$ Descending order of terms
$$= 50, 42, 35, (2x + 10,), (2x - 8), 12, 11, 8$$
$$\displaystyle \Rightarrow $$ Total number of terms $$= 8$$ (even)
$$\displaystyle \therefore $$ Median = $$\displaystyle \frac{1}{2}\left [ \left(\frac{8}{2}\right)^{th}\text {term}+\left ( \frac{8}{2}+1 \right )^{th} \text{term}\right ]$$
$$= \displaystyle \frac{1}{2}\left [ \left ( 2x+10 \right )+\left ( 2x-8 \right ) \right ]$$
$$= \displaystyle \frac{1}{2}\left [ 4x+2 \right ]=\left ( 2x+1 \right )$$
As per question--
$$\displaystyle \left ( 2x+1 \right )=25$$
$$\displaystyle \therefore x=\cfrac{25-1}{2}=12$$
Question 3
1 / -0

If the mean of x and $$\displaystyle \frac{1}{x}$$ is M, then the mean of x$$^3$$ and $$\displaystyle \frac{1}{x^3}$$ is

A
$$\displaystyle \frac{M (M^2 - 3)}{2}$$

B
$$M (4M^2 - 3)$$

C
$$M^3$$

D
$$M^3 + 3$$

Solution

Given,
$$M=\dfrac{x+\dfrac{1}{x}}{2}$$
$$2M=x+\dfrac{1}{x}$$ $$ ....... (1)$$

On cubing both sides, we get
$$(2M)^3=(x+\dfrac{1}{x})^3$$
$$8M^3=x^3+\dfrac{1}{x^3}+3(x+\dfrac{1}{x})$$
$$8M^3=x^3+\dfrac{1}{x^3}+3(2M)$$
$$8M^3=x^3+\dfrac{1}{x^3}+6M$$
$$x^3+\dfrac{1}{x^3}=8M^3-6M$$
$$\dfrac{x^3+\dfrac{1}{x^3}}{2}=M(4M^2-3)$$

Hence, this is the answer.
Question 4
1 / -0

Calculate the median for the following
Weight in kg. 20 22 24 27 28 30 31
No. of boys 8 10 11 9 7 5 3

A
$$23$$

B
$$26$$

C
$$23.5$$

D
$$24$$

Solution

Weight (x) No. of boys (f) cumulative frequency(c.f)
20 8 8
22 10 18
24 11 29
27 9 38
28 7 45
30 5 50
31 3 53
Total number of boys =53
$$\therefore Median=\left(\frac{n+1}{2}\right)^{th} term=\frac{53+1}{2}=\frac{54}{2}=27^{th} term$$
according to the table the weight of each child from 19th boy to 29th boy is 24 kg.
$$\therefore Median =24$$
Question 5
1 / -0

The mean of the following frequency distribution is
Class Interval Frequency
0-10 4
10-20 6
20-30 10
30-40 16
40-50 14

A
25

B
35

C
30

D
31

Solution

Class
interval Mid
point Freq. Diff. from
(A=25)(d) fd
0-10 5 4 -20 -80
10-20 15 6 -10 -60
20-30 25 10 0 0
30-40 35 16 10 160
40-50 45 14 20 280
Total $$\sum f = 50$$ $$\sum fd = 300$$
$$(\bar x) = A +\displaystyle \frac{\sum fd}{\sum f} = 25 + \frac{300}{50} = 31$$
Question 6
1 / -0

Calculate the mode
$$17, 13, 16, 17, 16, 21, 21, 9, 12, 17$$

A
$$17$$

B
$$16$$

C
$$21$$

D
$$13$$

Solution

Here the frequency of $$17$$ is highest.
So, the mode = $$17$$

Question 7

1 / -0

C.I.	0-10	10-20	20-30	30-40	40-50
Frequency	7	5	3	4	6

Find the mean of the following continuous distribution

$$23.8$$

$$22.8$$

$$25.8$$

$$24.8$$

Solution

C.I.	Frequency (f)	Class Mark (x)	fx
0-10	7	5	35
10-20	5	15	75
20-30	3	25	75
30-40	4	35	140
40-50 Mean $$ = \dfrac {\sum { fx} }{\sum { f }} = \dfrac { 595}{25} = 23.8 $$	6 $$ \sum { f} = 25 $$	45	270 $$\sum { fx } = 595 $$

Question 8
1 / -0

The mean of the following distribution is

A
$$15$$

B
$$16$$

C
$$17$$

D
$$18$$

Solution

Class $$x_i$$ $$ f_i$$ $$f_i.x_i$$
0-5 2.5 4 10.0
5-10 7.5 5 37.5
10-15 12.5 7 287.5
15-20 17.5 12 210.5
20-25 22.5 7 157.5
25-30 27.5 5 137.5
Mean$$=\dfrac{\sum f_i.x_i}{\sum f_i}$$
$$\Rightarrow \dfrac{10+37.5+287.5+210.5+157.5+137.5}{4+5+7+12+7+5}$$
$$\Rightarrow \dfrac{640}{40}=16$$
Question 9
1 / -0

The mean of $$20$$ observations is $$15$$. One observation $$20$$ is deleted and two more observations are included to the data. If the mean of new set of observations is $$15$$, then find the sum of the two new observations included.

A
$$30$$

B
$$35$$

C
$$33$$

D
$$32$$

Solution

Mean $$ = \dfrac { \text{Sum of observations}}{\text{Total number of observations}} $$
Given, mean of $$ 20 $$ observations $$ = 15 $$.
Thus, sum of $$ 20 $$ observations $$ = 15 \times 20 = 300 $$
When $$ 20 $$ is deleted and two more observations are included in the data, sum of observation $$ = 300 - 20 + X + Y = 280 + X + Y $$.
And number of observations is $$ 20 - 1 + 2 = 21 $$
Given, Mean of wages of new set of observations $$ = 15 $$
$$ = \dfrac {280 + X + Y }{21} = 15 $$
$$ \Rightarrow 280 + X + Y= 15 \times 21 = 315 $$
$$ \Rightarrow X + Y = 315 - 280 = 35 $$
Question 10
1 / -0

The weighted arithmetic mean ofthe first $$n$$ natural numbers whose weights are equal to the corresponding numbers is given by,

A
$$\displaystyle \frac{1}{2}(n+1)$$

B
$$\displaystyle \frac{1}{2}(n+2)$$

C
$$\displaystyle \frac{1}{3}(2n+1)$$

D
$$\displaystyle \frac{1}{3}n (2n+1)$$

Solution

$$x_i$$ $$x$$ Weight $$(w)$$
  $$x_1$$   $$1$$   $$1$$
  $$x_2$$   $$2$$   $$2$$
  $$x_3$$
....
....   $$3$$   $$3$$
  $$x_n$$   $$n$$   $$n$$
First $$n$$ natural numbers are written in the table along with their respective weight.
Weighted arithmetic mean $$=\dfrac{\sum_{n}^{r-1}w_rx_r}{\sum_{n}^{r-1}w_r}$$
$$=\dfrac{1\times 1+2\times 2+3\times 3+...+n\times n}{1+2+3+4+...+n}$$

$$=\dfrac{1^2+2^2+3^2+...+n^2}{1+2+3+4+...+n}$$

Using formula of sum of series, we get
$$=\dfrac{n(n+1)(2n+1)}{6}\times \dfrac{2}{n(n+1)}$$

$$=\dfrac{(2n+1)}{3}$$
Hence, this is the required answer.

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Weight (x)	No. of boys (f)	cumulative frequency(c.f)
20	8	8
22	10	18
24	11	29
27	9	38
28	7	45
30	5	50
31	3	53

Class interval	Mid point	Freq.	Diff. from (A=25)(d)	fd
0-10	5	4	-20	-80
10-20	15	6	-10	-60
20-30	25	10	0	0
30-40	35	16	10	160
40-50	45	14	20	280
Total	$$\sum f = 50$$			$$\sum fd = 300$$

Class	$$x_i$$	$$ f_i$$	$$f_i.x_i$$
0-5	2.5	4	10.0
5-10	7.5	5	37.5
10-15	12.5	7	287.5
15-20	17.5	12	210.5
20-25	22.5	7	157.5
25-30	27.5	5	137.5

$$x_i$$	$$x$$	Weight $$(w)$$
$$x_1$$	$$1$$	$$1$$
$$x_2$$	$$2$$	$$2$$
$$x_3$$ .... ....	$$3$$	$$3$$
$$x_n$$	$$n$$	$$n$$

Measures of Central Tendency Test - 34

Result

Measures of Central Tendency Test - 34

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Question 1

$$72$$

$$73$$

$$76$$

$$75$$

Solution

Question 2

$$20$$

$$25$$

$$12$$

$$11$$

Solution

Question 3

$$\displaystyle \frac{M (M^2 - 3)}{2}$$

$$M (4M^2 - 3)$$

$$M^3$$

$$M^3 + 3$$

Solution

Question 4

$$23$$

$$26$$

$$23.5$$

$$24$$

Solution

Question 5

25

35

30

31

Solution

Question 6

$$17$$

$$16$$

$$21$$

$$13$$

Solution

Question 7

$$23.8$$

$$22.8$$

$$25.8$$

$$24.8$$

Solution

Question 8

$$15$$

$$16$$

$$17$$

$$18$$

Solution

Question 9

$$30$$

$$35$$

$$33$$

$$32$$

Solution

Question 10

$$\displaystyle \frac{1}{2}(n+1)$$

$$\displaystyle \frac{1}{2}(n+2)$$

$$\displaystyle \frac{1}{3}(2n+1)$$

$$\displaystyle \frac{1}{3}n (2n+1)$$

Solution

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