Types of Data Distribution

Introduction to types of data distribution:
The data distribution is also said to be a theoretical distribution. There are three types of data distribution given below. Random variables are distributed regarding to probability law corresponding to probability distribution is called theoretical distribution. Here we are going to discuss about types of data distribution with suitable example problem.

Types of Data Distribution:

There are three types data distribution, such as

Binomial distribution
Normal distribution
Poisson distribution

Example Problems – Types of Data Distribution:

Example problem 1 – types of data distribution:

Obtain the given Poisson distribution where, e = 2.718, x = 10  and `lambda` = 7, x = 10

Solution:

Given: `lambda` = 7; x = 10

Formula to find Poisson distribution = `((e^(-lambda))(lambda^(x)))/(x!)`

First we find the value of `e^(-lambda)` .

e-3 = (2.718)-7

= 0.000911

Then you will find` lambda^(x)`

Given: `lambda` = 7; x = 10

Therefore, `lambda^(x) = ` `7^(10)`

= 282475249

Then you will substitute the values in given formula

`((e^(-lambda))(lambda^(x)))/(x!)` ` = (0.000911(282475249))/(10!)`

= `257334.951839/3628800`

= 0.07091

Answer: Poisson distribution = 0.07091

Is this topic Arithmetic Means hard for you? Watch out for my coming posts.

Example problem 2 – Types of data distribution:

Obtain the value of p. In a Binomial distribution if n = 4 and P(r = 3) = 2P(r = 2).

Solution:

P(X = r) = nCr pr qn−r

P(X = 3) = 4C3 p3q1 and P(X = 2) = 4C2 p2q2

∴ 4C3 p3q1 = 2 (4C2 ) p2q2

4 p3q1 = 2 * 6 * p2q2

∴ 4 p = 12q

p = 3 q

p = 3 (1 − p) ⇒ 4p = 3; p =`3/4.`

Answer: The value of p = `3/4`

Example problem 3 – types of data distribution

Find the normal distribution of P( X `lt` 200) if the normal random variable has mean(m) = 150, standard deviation(`sigma` )  = 30.

Solutions:

Step 1: Given:

Mean (m) = 150

Standard deviation (`sigma` ) = 30

Step 2: Formula:

X `lt`  mean = 0.5 - Z

X `gt` mean = 0.5 + Z

X = mean = 0.5.

Z = `(X-m)/(sigma)`

Step 3: Solve:

Z = `(X-m)/(sigma)`

= `(200 - 150) / 30`

= `50/30`

= 1.67

Step 3: Find Z = 1.67, Refer Z table,

Z = 1.67 = 0.4525

Step 4: Solving:

X value greater than mean

P(X) = 0.5 + 0.4525

= 0.9525

Result: Normal Distribution P( X `lt` 200) = 0.9525