Understanding Statistics
Standard Deviation of Data in a Frequency Table
Standard Deviation of Data in a Frequency Table
x̅̅̅̅ Review:
Review Std deviation
S=
𝑋𝑖−x̅̅̅̅ 2
𝑛−1
In order be able to apply this formula we need data for all the points in the sample
Frequency distribution:
- we can not use the S formula as we don’t know the grades
- with histogram we lose granularity. We know only that are in
range but not really where in the range
- So how do we do we find the Standard deviation in this
situation?
Class (Grade) Frequency
94-100 5
87-93 8
80-86 12
73-79 7
66-72 4
Standard Deviation of Data in a Frequency Table
It can be proven for frequency distribution that :
S=
[𝑛 𝑓∗𝑥2
− 𝑓∗𝑥 2
]
𝑛(𝑛−1)
n - sample size
f – frequency
X – midpoint of the class
n=36
One way to solve that equation (other than algorithms or specialised programing language like R) is to create additional rows
in the initial table
S=
36 253989 − 3009 2
36(36−1)
= 8.43
Class (Grade) Frequency Class midpoint (f*X) f*x2
94-100 5 97 485 47045
87-93 8 90 720 64800
80-86 12 83 996 82668
73-79 7 76 532 40432
66-72 4 69 276 19044
3009 253989
∑1 = 3009
∑2= 253989

Master statistics 1#09_ Standard Deviation of Data in a Frequency Table

  • 1.
    Understanding Statistics Standard Deviationof Data in a Frequency Table
  • 2.
    Standard Deviation ofData in a Frequency Table x̅̅̅̅ Review: Review Std deviation S= 𝑋𝑖−x̅̅̅̅ 2 𝑛−1 In order be able to apply this formula we need data for all the points in the sample Frequency distribution: - we can not use the S formula as we don’t know the grades - with histogram we lose granularity. We know only that are in range but not really where in the range - So how do we do we find the Standard deviation in this situation? Class (Grade) Frequency 94-100 5 87-93 8 80-86 12 73-79 7 66-72 4
  • 3.
    Standard Deviation ofData in a Frequency Table It can be proven for frequency distribution that : S= [𝑛 𝑓∗𝑥2 − 𝑓∗𝑥 2 ] 𝑛(𝑛−1) n - sample size f – frequency X – midpoint of the class n=36 One way to solve that equation (other than algorithms or specialised programing language like R) is to create additional rows in the initial table S= 36 253989 − 3009 2 36(36−1) = 8.43 Class (Grade) Frequency Class midpoint (f*X) f*x2 94-100 5 97 485 47045 87-93 8 90 720 64800 80-86 12 83 996 82668 73-79 7 76 532 40432 66-72 4 69 276 19044 3009 253989 ∑1 = 3009 ∑2= 253989