Difference between revisions of "Activity- statistics problems"

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=Problem 1=
 
=Problem 1=
−
If n =10, =  12 and = 1530 find coefficient of variation
+
If n =10, <math>\bar x  </math>  =  12 and<br>
 +
<math>\sum{x^2}= 1530 find the standard deviation </math> <br>
 +
 
 
=INTERPRETATION OF PROBLEM=
 
=INTERPRETATION OF PROBLEM=
  
 
=Previous knowledge=
 
=Previous knowledge=
  
−
=Solution=
+
=Solution:=
  
 
<math>\frac{\sum x}{n}=\bar x  </math> <br>
 
<math>\frac{\sum x}{n}=\bar x  </math> <br>
Line 15: Line 17:
 
<math>\sum {x} =120 </math> <br>
 
<math>\sum {x} =120 </math> <br>
 
σ=<math>\sqrt{\frac {\sum {x^2}}{n}-({\frac{\sum x}{n})^2}}</math> <br>
 
σ=<math>\sqrt{\frac {\sum {x^2}}{n}-({\frac{\sum x}{n})^2}}</math> <br>
 +
σ=<math>\sqrt{\frac{1530}{10}-{144}}</math> <br>
 
σ=<math>\sqrt{\frac{1530}{10}-{144}}</math> <br>
 
σ=<math>\sqrt{\frac{1530}{10}-{144}}</math> <br>
 
σ=<math>\sqrt{153-144}</math> <br>
 
σ=<math>\sqrt{153-144}</math> <br>
 
σ=<math>\sqrt{9}</math> <br>
 
σ=<math>\sqrt{9}</math> <br>
 
σ=3
 
σ=3
 +
 +
=Problem 2=
 +
Problem No.1 of excercise No.6.1 <br>
 +
Find the Standard deviation for the following data.
 +
{|class="wikitable"
 +
|-
 +
|x
 +
|03
 +
|08
 +
|13
 +
|18
 +
|23
 +
|-
 +
|f
 +
|07
 +
|10
 +
|15
 +
|10
 +
|08
 +
|}
 +
 +
=INTERPRETATION OF PROBLEM=
 +
 +
=Previous knowledge=
 +
 +
=Solution:=
 +
{|class="wikitable"
 +
|-
 +
|x
 +
|f
 +
|fx
 +
|x²
 +
|fx²
 +
|-
 +
|03
 +
|07
 +
|021
 +
|009
 +
|0063
 +
|-
 +
|08
 +
|10
 +
|080
 +
|064
 +
|0640
 +
|-
 +
|13
 +
|15
 +
|195
 +
|169
 +
|2535
 +
|-
 +
|18
 +
|10
 +
|180
 +
|324
 +
|3240
 +
|-
 +
|23
 +
|08
 +
|184
 +
|529
 +
|4232
 +
|-
 +
|
 +
|n=50
 +
|Σfx=660
 +
|
 +
|Σfx²=10710
 +
|}
 +
 +
 +
 +
Standard deviation σ=<math>\sqrt{\frac {\sum {fx^2}}{n}-({\frac{\sum fx}{n})^2}}</math> <br>
 +
σ=<math>\sqrt{\frac{10700}{50}-({\frac{660}{50})^2}}</math> <br>
 +
σ=<math>\sqrt{214-174.24}</math> <br>
 +
σ=<math>\sqrt{39.96}</math> <br>
 +
σ=6.3
 +
 +
=Problem 3=
 +
Problem No.5 of excercise No.6.3 <br>
 +
Find the varience and  Standard deviation for the following data.
 +
{|class="wikitable"
 +
|-
 +
|class intervals(CI)
 +
|30-34
 +
|34-38
 +
|38-42
 +
|42-46
 +
|46-50
 +
|50-54
 +
|-
 +
|freequency(f)
 +
|04
 +
|07
 +
|09
 +
|11
 +
|06
 +
|03
 +
|}
 +
 +
=INTERPRETATION OF PROBLEM=
 +
 +
=Previous knowledge=
 +
 +
=Solution:=
 +
{|class="wikitable"
 +
|-
 +
|C.I.
 +
|f
 +
|x
 +
|fx
 +
|d=<math>\frac{x-A}{c}</math>
 +
|fd
 +
|d²
 +
|fd²
 +
|-
 +
|30-34
 +
|4
 +
|32
 +
|128
 +
| -2
 +
| -8
 +
|4
 +
|16
 +
|-
 +
|34-38
 +
|7
 +
|36
 +
|252
 +
| -1
 +
| -7
 +
|1
 +
|7
 +
|-
 +
|38-42
 +
|9
 +
|40
 +
|360
 +
|0
 +
|0
 +
|0
 +
|0
 +
|-
 +
|42-46
 +
|11
 +
|44
 +
|484
 +
|1
 +
|11
 +
|1
 +
|11
 +
|-
 +
|46-50
 +
|6
 +
|48
 +
|288
 +
|2
 +
|12
 +
|4
 +
|24
 +
|-
 +
|50-54
 +
|3
 +
|52
 +
|156
 +
|3
 +
|9
 +
|9
 +
|27
 +
|-
 +
|
 +
|n=40
 +
|
 +
|Σfx=1668
 +
|
 +
|Σfd=17
 +
|
 +
|Σfd²=85
 +
|}
 +
A=assumed average.<br>
 +
c=4<br>
 +
d=<math>\frac{x-A}{c}</math>=<math>\frac{32-40}{4}</math>=<math>\frac{-8}{4}=-2</math><br>
 +
 +
assumed mean A=<math>\frac{\sum fx}{n}</math>=<math>\frac{1668}{40}=41.7</math><br>
 +
 +
Varience σ²=[<math>\frac{\sum {fd^2}}{n}-({\frac{\sum fx}{n})^2}]c^2</math> <br>
 +
 +
σ²=[<math>\frac{85}{40}-({\frac{17}{40})^2}]4^2</math> <br>
 +
 +
σ²=[2.125-0.180]16<br>
 +
 +
σ²=[1.945]16<br>
 +
 +
σ²=31.12<br>
 +
standard deviation, σ=<math>\sqrt{varience}</math> <br>
 +
σ=<math>\sqrt{31.12}</math> <br>
 +
σ=5.58
 +
 +
[[Category:Statistics]]

Latest revision as of 09:56, 30 October 2019

Problem 1

If n =10, = 12 and

INTERPRETATION OF PROBLEM

Previous knowledge

Solution:





σ=
σ=
σ=
σ=
σ=
σ=3

Problem 2

Problem No.1 of excercise No.6.1
Find the Standard deviation for the following data.

x 03 08 13 18 23
f 07 10 15 10 08

INTERPRETATION OF PROBLEM

Previous knowledge

Solution:

x f fx x² fx²
03 07 021 009 0063
08 10 080 064 0640
13 15 195 169 2535
18 10 180 324 3240
23 08 184 529 4232
n=50 Σfx=660 Σfx²=10710


Standard deviation σ=
σ=
σ=
σ=
σ=6.3

Problem 3

Problem No.5 of excercise No.6.3
Find the varience and Standard deviation for the following data.

class intervals(CI) 30-34 34-38 38-42 42-46 46-50 50-54
freequency(f) 04 07 09 11 06 03

INTERPRETATION OF PROBLEM

Previous knowledge

Solution:

C.I. f x fx d= fd d² fd²
30-34 4 32 128 -2 -8 4 16
34-38 7 36 252 -1 -7 1 7
38-42 9 40 360 0 0 0 0
42-46 11 44 484 1 11 1 11
46-50 6 48 288 2 12 4 24
50-54 3 52 156 3 9 9 27
n=40 Σfx=1668 Σfd=17 Σfd²=85

A=assumed average.
c=4
d===

assumed mean A==

Varience σ²=[

σ²=[

σ²=[2.125-0.180]16

σ²=[1.945]16

σ²=31.12
standard deviation, σ=
σ=
σ=5.58

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