Difference between revisions of "Activity- statistics problems"
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=Problem 1= | =Problem 1= | ||
− | If n =10, = 12 and = 1530 find | + | If n =10, <math>\bar x </math> = 12 and<br> |
− | ==INTERPRETATION OF PROBLEM== | + | <math>\sum{x^2}= 1530 find the standard deviation </math> <br> |
− | ==Previous knowledge== | + | |
− | == | + | =INTERPRETATION OF PROBLEM= |
− | <math> | + | |
+ | =Previous knowledge= | ||
+ | |||
+ | =Solution:= | ||
+ | |||
+ | <math>\frac{\sum x}{n}=\bar x </math> <br> | ||
+ | |||
+ | <math>\sum {x} ={\bar x \times {n}} </math> <br> | ||
+ | |||
+ | <math>\sum {x} ={12 \times {10}} </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{1530}{10}-{144}}</math> <br> | ||
+ | σ=<math>\sqrt{\frac{1530}{10}-{144}}</math> <br> | ||
+ | σ=<math>\sqrt{153-144}</math> <br> | ||
+ | σ=<math>\sqrt{9}</math> <br> | ||
+ | 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