This presentation contains details of binomial, exponential and normal distribution

1. BRAINWARE UNIVERSITY
Name: SAMPAD KAR
Roll No: BWU/BTA/22/225
Section: D
Group: D2
Course Name: Probability & Statistics
Course Code: BSCM202

2. Contents
• Random Variables
• Probability Distribution of Discrete random variables
• Probability Distribution of Continuous random variables
• Mean and Variance of Binomial Distribution
• Mean and Variance of Exponential Distribution
• Mean and Variance of Normal Distribution

3. Random Variables
A random variable is a variable which represents the outcome of a trial, an
experiment or an event. It is a number which is different each time the trial or event
is repeated.
For Example,
A coin is tossed 3 times simultaneously. Find the Probability of getting 3 heads?
S = {(H,H,,H),(H,H,T),(H,T,T),(T,H,T),(H,T,H),(T,H,H),(T,T,H),(T,T,T)}
P(getting at least 3 heads) =
1
8

4. Probability Distribution of Discrete random variables
A discrete variable is a variable that can "only" take-on certain numbers on the number line.
The Probability Mass Function (PMF) is also called a probability function or frequency function which
characterizes the distribution of a discrete random variable. Let X be a discrete random variable of a
function, then the probability mass function of a random variable X is given by
Px (x) = P( X=x ), For all x belongs to the range of X
It is noted that the probability function should fall on the condition :
•Px (x) ≥ 0 and
•∑xϵRange(x) Px (x) = 1
Example
When we roll a single dice, the possible outcomes are:
1, 2, 3, 4, 5, 6
The probability of each of these outcomes is 16.
If we define the discrete variable X as:
X: the number obtained when rolling a dice.
Then this is a discrete random variable since the sum of the probabilities of each of these possible
outcomes is equal to 1, indeed:
1
6
+
1
6
+
1
6
+
1
6
+
1
6
+
1
6
=1

5. A continuous random variable is a random variable that has only continuous values. Continuous values are
uncountable and are related to real numbers.
The probability density function (pdf) is used to describe the probabilities associated with a continuous random
variable.
Probability Distribution of Continuous random variables

6. Mean and Variance of Binomial Distribution
For binomial distribution,
pmf = P(X=x) = 𝑥
𝑛
𝐶𝑃𝑥
(1 − 𝑃)𝑛−𝑥
Mean = E(x) = 𝑥=1
𝑛
𝑥. 𝑥
𝑛
𝐶𝑃𝑥
(1 − 𝑝)𝑛−𝑥
= 𝑥=1
𝑛
𝑥.
𝑛!
𝑛−𝑥 ! 𝑥!
𝑃𝑥
(1 − 𝑃)𝑛−𝑥
= 𝑥=1
𝑛
𝑥.
𝑛!
𝑛−𝑥 ! 𝑥(𝑥−1)!
𝑃𝑥(1 − 𝑃)𝑛−𝑥
= 𝑥=1
𝑛 𝑛.(𝑛−1)!
(𝑛−1)−(𝑥−1) ! (𝑥−1)!
𝑃. 𝑃𝑥−1
(1 − 𝑃)(𝑛−1)−(𝑥−1)
= nP 𝑥=1
𝑛 (𝑛−1)!
(𝑛−1)−(𝑥−1) ! (𝑥−1)!
𝑃𝑥−1
(1 − 𝑃)(𝑛−1)−(𝑥−1)
= nP 𝑥=1
𝑛
𝑥−1
𝑛−1
𝐶 𝑃𝑥−1
(1 − 𝑃)(𝑛−1)−(𝑥−1)
= 𝑛𝑃(P + (1 − 𝑃))(𝑛−1)
= nP

7. E(x(x-1)) = 𝑥=1
𝑛
𝑥(𝑥 − 1). 𝑥
𝑛𝐶𝑃𝑥(1 − 𝑝)𝑛−𝑥
= 𝑥=1
𝑛
𝑥(𝑥 − 1).
𝑛!
𝑛−𝑥 ! 𝑥!
𝑃𝑥(1 − 𝑃)𝑛−𝑥
= 𝑥=1
𝑛
𝑥(𝑥 − 1).
𝑛!
𝑛−𝑥 ! 𝑥(𝑥−1)(𝑥−2)!
𝑃𝑥(1 − 𝑃)𝑛−𝑥
= 𝑥=1
𝑛 𝑛.(𝑛−1)(𝑛−2)!
(𝑛−1)−(𝑥−1) ! (𝑥−2)!
𝑃2
. 𝑃𝑥−2
(1 − 𝑃)(𝑛−2)−(𝑥−2)
= (𝑛2−𝑛)𝑃2
𝑥=1
𝑛 (𝑛−2)!
(𝑛−2)−(𝑥−2) ! (𝑥−2)!
𝑃𝑥−2(1 − 𝑃)(𝑛−2)−(𝑥−2)
= (𝑛2
−𝑛)𝑃2
𝑥=1
𝑛
𝑥−2
𝑛−2
𝐶 𝑃𝑥−2
(1 − 𝑃)(𝑛−2)−(𝑥−2)
= (𝑛2−𝑛)𝑃2 (P + (1 − 𝑃))(𝑛−2)
= (𝑛2−𝑛)𝑃2
E(𝑥2
) = E(x(x-1)) + E(x)
= (𝑛2
−𝑛)𝑃2
+ nP
Variance = E(𝑥2) – {E(x)}2
= (𝑛2
−𝑛)𝑃2
+ nP – 𝑛2
𝑃2
= nP−𝑛𝑃2
= nP(1-P)

8. Mean and Variance of Exponential Distribution
For exponential distribution,
f(x) =
Mean = E(x) = −∞
0
0𝑑𝑥 + 0
∞
𝑥.λ 𝑒−λ𝑥
𝑑𝑥
= 0 + λ 0
∞
𝑥𝑒−λ𝑥𝑑𝑥
= λ.
𝛤(1)
λ
2
=
1
λ
Now,
E(𝑥2
) = λ 0
∞
𝑥2
𝑒−λ𝑥
𝑑𝑥
= λ.
𝛤(3)
λ
3 =
2
λ
2
{λ 𝑒−λ𝑥
, x > 0
0 , elsewhere
Variance = E(𝑥2) – {E(x)}2
=
2
λ
2 -
1
λ
2
=
1
λ
2

9. Mean and Variance of Normal Distribution
For Normal Distribution,
σ = Standard Deviation
µ = Mean
σ2= Variance
Standard Normal Distribution, N(0,1):
σ = 1 , μ = 0
f(x) =
1
2Π
𝑒−
𝑧2
2 z =
𝑥−𝜇
𝜎

10. THANK YOU