Probability Density Distribution Function

 Tutorial is the teaching method. Through online tutors can teach the lesson or whatever doubts student have, tutors can clarify the doubts.  The equation used to describe a continuous probability distribution is a probability density function.The probability density of a continuous random variable is a function it can be integrated to obtain the probability of the random variable takes a value in a given interval. Probability density function is used to find the point of normal distribution curve.

 

Definition of Mean and Standard Deviation for Probability Density Distribution Function

 

Mean and standard deviation:

1)The mean of the normal distribution defines the location of the center of the graph.

2)The standard deviation of the normal distribution probability defines the height and width of the graph.

Standard normal distribution table

z 

Φ( z ) 

φ( z )

0.00

0.5000

0.3989

0.01

0.5040

0.3989

0.02

0.5080

0.3989

0.03

0.5120

0.3988

0.04

0.5160

0.3986

0.05

0.5199

0.3984

0.06

0.5239

0.3982

0.07

0.5279

0.3980

0.08

0.5319

0.3977

0.09

0.5359

0.3973

0.10

0.5398

0.3970

0.11

0.5438

0.3965

0.12

0.5478

0.3961

0.13

0.5517

0.3956

0.14

0.5557

0.3951

0.15

0.5596

0.3945

0.16

0.5636

0.3939

0.17

0.5675

0.3932

0.18

0.5714

0.3925

0.19

0.5753

0.3918

0.20

0.5793

0.3910

0.21

0.5832

0.3902

0.22

0.5871

0.3894

0.23

0.5910

0.3885

0.24

0.5948

0.3876

0.25

0.5987

0.3867

0.26

0.6026

0.3857

0.27

0.6064

0.3847

0.28

0.6103

0.3836

0.29

0.6141

0.3825

0.30

0.6179

0.3814

0.31

0.6217

0.3802

0.32

0.6255

0.3790

0.33

0.6293

0.3778

0.34

0.6331

0.3765

0.35

0.6368

0.3752

0.36

0.6406

0.3739

0.37

0.6443

0.3725

0.38

0.6480

0.3712

0.39

0.6517

0.3697

0.40

0.6554

0.3683

0.41

0.6591

0.3668

0.42

0.6628

0.3653

0.43

0.6664

0.3637

0.44

0.6700

0.3621

0.45

0.6736

0.3605

0.46

0.6772

0.3589

0.47

0.6808

0.3572

0.48

0.6844

0.3555

0.49

0.6879

0.3538

0.50

0.6915

0.3521


My forthcoming post is on Sample Size Formula, Confidence Interval Proportions will give you more understanding about mathematics. 

 

Example for Probability Density Distribution Function

 

Example 1

Change into normal distribution to probability distribution. Variable X is normal probability distribution with mean 2 and standard deviation 3. Find the mean and standard deviation..

1) P (0 ≤ X ≤ 4) 2) P (| X − 3 | < 4)

Solution:

          Given μ = 2, σ = 3

1) P (0 ≤ X ≤ 4)

          We know that Z = (X – μ) / σ

          When X = 0, Z = (0 – 2) / 3 = −2 / 3 =-0.67

          When X = 4, Z = (4 – 2) / 3 = 2/3 =0.67

            P (0 ≤ X ≤ 4) = P (−0.67 < Z < 0.67)

           = P (0< Z <0.67) + P (0 < Z < 0.67)

           = 2 P (0< Z <0.67)

           = 2* (0.3187)

           = 0.6374

2) P (| X − 3| < 4) = P (−4 < (X − 3) < 4) ⇒ P (−3 < X < 7)

Z = (X – μ) / σ

           When X = −3, Z = (−3 – 2)/3 = −5/3 = − 1.67

           When X = 7, Z = (7 – 2)/3 = 5/3 = 1.67

           P (−3< X < 7) = P (−1.67 < Z < 1.67)

           = P (0 < Z < 1.67) + P (0 < Z < 1.67)

           = = 2 P (0< Z <0.67)

           = 2* (0.3187)

           = 0.6374

Example 2 

Variable X is a normal random variable with mean 60 and standard deviation 30.To find the probability of P(X<60).

Solution:

Mean μ = 50

Standard deviation σ = 20

Normal probability distribution

Z = (X- μ)/ σ

Given value for X = 60

Z = (60-50)/20

   = 10/20

   = 0.5

Z = 0.5

Using the normal probability distribution table, the Z value 0.5

Z = 0.3989

If X is greater than μ then use below formula

X > μ = 0.5 + Z

60 > 50 = 0.5 + 0.3989

P(X) = 0.5 + 0.3989

         = 0.8989

P(X) = 0.8989

 

Example 3:  Variable X is a normal random variable with mean 20 and standard deviation 5.5.To find the probability of P(X< 35).

Solution:

Mean μ = 20

Standard deviation σ = 5.5

Using the formula

Z = (X- μ)/ σ

Given value for X = 35

Z = (35-20)/5.5

   = 15/5.5

   = 2.72

Z = 2.72

Using the normal probability distribution table, the Z value 2.72

Z = 0.4345

If X is greater than μ then use below formula

X > μ = 0.5 + Z

55 > 27 = 0.5 + 0.4345

P(X) = 0.5 + 0.4345

         = 0.9345

P(X) = 0.9345