Uploaded by Md. Shofiul Bashar Rafi

Assignment-510-1

```Assignment- 01
Course Title:
Applied Categorical Data Analysis
Course Code:
AST-509
Submitted to:
M.H.M Imrul Kabir
Senior Lecturer
Department of Mathematical and Physical Sciences
Submitted by:
MD. Shofiul Bashar Rafi
ID-2019-3-82-002
Section# 01
Date: 05-07-2020
MODE for Binomial Distribution:
Let, m= Mode, and p is probability of success and q is the probability of failure , also note that, p+q=1
According to concept of mood for binomial distribution, we can write:
𝑷𝒓𝒐𝒃(𝒙 = 𝒎 + 𝟏) ≤ 𝑷𝒓𝒐𝒃(𝒙 = 𝒎) ≥ 𝑷𝒓𝒐𝒃(𝒙 = 𝒎 − 𝟏)
Now,
𝑃𝑟𝑜𝑏(𝑥 = 𝑚 + 1) ≤ 𝑃𝑟𝑜𝑏(𝑥 = 𝑚)
⟹ nCm+1 𝑝𝑚+1 𝑞 𝑛−𝑚−1 ≤ nCm 𝑝𝑚 𝑞 𝑛−𝑚
𝑛!
⟹ (𝑚+1)!(𝑛−𝑚−1)! 𝑝𝑚 𝑝 𝑞 𝑛−𝑚 𝑞 −1 ≤
𝑛!
𝑚!(𝑛−𝑚)!
(𝑛−𝑚)! 𝑞
𝑚!
⟹ (𝑚+1)! (𝑛−𝑚−1)! ≤ 1
𝑝
⟹
(𝑛−𝑚) 𝑝
(𝑚+1) 𝑞
≤1
⟹ 𝑝𝑛 − 𝑝𝑚 ≤ 𝑞𝑚 + 𝑞
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑞𝑚 + 𝑝𝑚
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑚(𝑝 + 𝑞)
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑚(𝑝 + 𝑞)
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑚 &times; 1[ 𝑠𝑖𝑛𝑐𝑒 𝑝 + 𝑞 = 1]
⟹ 𝒎 ≥ 𝒑𝒏 − 𝒒
Again,
𝑃𝑟𝑜𝑏(𝑥 = 𝑚) ≥ 𝑃𝑟𝑜𝑏(𝑥 = 𝑚 − 1)
⟹ nCm 𝑝𝑚 𝑞 𝑛−𝑚 ≥ nCm-1 𝑝𝑚−1 𝑞 𝑛−𝑚+1
⟹
⟹
⟹
𝑛!
𝑛!
𝑚!(𝑛−𝑚)!
𝑝𝑚 𝑞 𝑛−𝑚 ≥ (𝑚−1)!(𝑛−𝑚+1)! 𝑝𝑚 𝑝 −1 𝑞 𝑛−𝑚 𝑞
(𝑚−1)!(𝑛−𝑚+1)! 𝑝
𝑚!(𝑛−𝑚)!
(𝑛−𝑚+1) 𝑝
𝑚
𝑞
𝑞
≥1
≥1
⟹ 𝑝𝑛 − 𝑝𝑚 + 𝑝 ≥ 𝑞𝑚
⟹ 𝑝(𝑛 + 1) + 𝑝 ≥ 𝑚(𝑝 + 𝑞)
⟹ 𝒎 ≤ 𝒑(𝒏 + 𝟏)
So, 𝑃𝑟𝑜𝑏(𝑥 = 𝑚 + 1) ≤ 𝑃𝑟𝑜𝑏(𝑥 = 𝑚) ≥ 𝑃𝑟𝑜𝑏(𝑥 = 𝑚 − 1)
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑚 ≤ 𝑝(𝑛 + 1)
⟹ 𝑝𝑛 − 𝑞 ≤ 𝑚 ≤ 𝑝(𝑛 + 1)
⟹ 𝑝(𝑛 + 1)-1 ≤ 𝑚 ≤ 𝑝(𝑛 + 1) ,
This is the formula of Mode of binomial distribution
𝑝𝑚 𝑞 𝑛−𝑚
 Median of Binomial Distribution:
There is no exact formula of Median for Binomial distribution. In many journals, researchers
describe their own developed formula based on following characteristics:



If, p, and n is are integer, then Median≥ mean of binomial(np)
When, p=.5, and n is an odd number then median lies within the interval,
0.5(n-1)≤0.5(n+1)
And when n is even number, then
𝑛
Median= 2
𝑛
∑𝑚
0 𝑝𝑟𝑜𝑏(𝑥 = 𝑚𝑒𝑑𝑖𝑎𝑛) ≥ 0.5 𝑜𝑟 ∑𝑚 𝑝𝑟𝑜𝑏(𝑥 = 𝑚𝑒𝑑𝑖𝑎𝑛) ≥ 0.5
```