Measures of Dispersion Test

Result

Measures of Dispersion Test - 7

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Weekly Quiz Competition

Question 1

1 / -0

Calculate the standard deviation from the following data.14, 22, 9, 15, 20, 17, 12, 11

4.18

4.15

2.15

3.18 Solution

X	(X-x̅)	(X-x̅)²
14	-1	1
22	7	49
9	-6	36
15	0	0
20	5	25
17	2	4
12	-3	9
11	-4	16
ΣX= 120	Σ(X-x̅)	Σ(X-x̅)²=140

x̅= ΣX/n= 120/8= 15

σ= √Σ(X-x̅)²\n= √140/8= 4.18(approx)

Question 2
1 / -0

Find the range of the group of numbers -10, -8, 1, 11, 19.

A
25

B
29

C
22

D
24

Solution

R= L-S, Where L is largest item and S is smallest item

R=19-(-10)= 19+10= 29
Question 3
1 / -0

If mean and coefficient of variation of a set of data is 10 and 5, then S.D. is

A
20

B
15

C
10

D
50

Solution

coefficient of variation= σ/x̅, where σ= standard deviation and x̅ is mean

5= σ/10

σ=10*5=50
Question 4
1 / -0

Calculate mean deviation from median and its coefficient from the following data : 100, 150, 80, 90, 160, 200, 140

A
32.14

B
34.28,0.24

C
35.74

D
33.65

Solution

Arrange all items in ascending order,

X |X-M|

80 60

90 50

100 40

140 0

150 10

160
20

200 60

ΣX=920 Σ|X-M|= 240

Calculate median:-

Median= size of( n+1/2 )the item = (7+1)/2= 4th item =140

MD(about median)= Σ|X-Med|/n=240/7=34.28

Coefficient of MD about mean = MD about median/median =34.28/140=0.24

Question 5

1 / -0

Calculate the standard deviation for the following data - 5, 8, 7, 11, 14

3.14

3.15

3.16

3.12 Solution

x(items)	x-x̅	(x-x̅)²
5	-4	16
8	-1	1
7	-2	4
11	2	4
14	5	25
Σx= 45	Σ(x-x̅)=0	Σ(x-x̅)²= 50

x̅= Σx/n= 45/5= 9

σ = √Σ(x-x̅)²/n=√ 50/5= 3.16

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Re-Attempt Weekly Quiz Competition

X	\|X-M\|
80	60
90	50
100	40
140	0
150	10
160	20
200	60
ΣX=920	Σ\|X-M\|= 240

Measures of Dispersion Test - 7

Result

Measures of Dispersion Test - 7

Score

-

Rank

-

SHARING IS CARING

Question 1

4.18

4.15

2.15

3.18

Solution

Question 2

25

29

22

24

Solution

Question 3

20

15

10

50

Solution

Question 4

32.14

34.28,0.24

35.74

33.65

Solution

Question 5

3.14

3.15

3.16

3.12

Solution

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