Difference between revisions of "Activity- statistics problems"

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Line 204: Line 204:
 
A=assumed average.<br>
 
A=assumed average.<br>
 
c=4<br>
 
c=4<br>
d=<math>\frac{x-A}{c}</math><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>
 
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=[<math>\2.125-0.180]16</math> <br>
  
 
Standard deviation σ=<math>\sqrt{\frac {\sum {fx^2}}{n}-({\frac{\sum fx}{n})^2}}</math> <br>
 
Standard deviation σ=<math>\sqrt{\frac {\sum {fx^2}}{n}-({\frac{\sum fx}{n})^2}}</math> <br>

Revision as of 12:03, 14 August 2014

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 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 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=[Failed to parse (syntax error): {\displaystyle \2.125-0.180]16}

Standard deviation σ=
σ=
σ=
σ=
σ=6.3