Difference between revisions of "Activity- statistics problems"
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|Σfd²=85 | |Σfd²=85 | ||
|} | |} | ||
− | + | A=assumed average.<br> | |
− | + | c=4<br> | |
+ | d=<math>\frac{x-A}{c}</math><br> | ||
+ | assumed mean A=<math>\frac{\sum fx}{n}</math>=<math>\frac{1668}{40}=41.7</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 06:06, 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 | 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==
Standard deviation σ=
σ=
σ=
σ=
σ=6.3