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Probability Density Function or not Example
In this class, We discuss Probability Density Function or not Example.
The reader should have prior knowledge of continuous probability distribution. Click Here.
Example 1:
A function f(x) is defined as follows.
f(x) = 0 when x lt2
= 1/18(2x+ 3) when 2 lt= x lt= 4
= 0 when x gt 4
1) Show that f(x) is a probability density function?
2) Find the probability P(2 lt= x lt= 3)?
Solution:
The below equation shows the area under the function f(x) = 1. To accept f(x) as a probability density function.
∫_(-∞)^∞▒f(x)dx
= ∫_(-∞)^2▒0dx + ∫_2^4▒〖1/18(2x+3)dx〗 + ∫_4^∞▒〖0 dx〗
= 1/18 ∫_2^4▒〖2x+3〗
= 1/18[x2 + 3x]42
= 1/18[28-10]
= 1


∫_2^3▒f(x)dx = 1/18[x2 + 3x]32
= 1/18[18 – 10]
= 8/18
= 4/9

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