Normal Probabilty Distribution and its Problems
- 2. Normal Probability Distribution And its Problems
- 3. Group Members WALEED AHMED 14-ARID-1458 ALI RAZA 14-ARID-1402 NOMAN ALI 14-ARID-1449
- 5. What is Normal Distribution? • A probability distribution that plots all of its values in a symmetrical fashion and most of the results are situated around the probability's mean. • A normal distribution is a continuous, symmetric, bell-shaped distribution of a variable.
- 6. History • In 1733, the French Mathematician , Abraham DeMoivre. • About 100 years later • Pierre Laplace in France • Carl Gauss in Germany (“Gaussian Distribution”) • In 1924, Karl Pearson found that DeMoivre was Correct
- 7. The Normal Curve • Represented by the Bell-shaped Curve. • Symmetric Curve • Continuous Curve.
- 8. The Normal Equation The Mathematical Equation for Normal Distribution is Where x = Normal Random Variable μ = Mean “mu” Standard Deviation “sigma” 2 2 2 1 2 x f x e x
- 10. Effect of Mean • Mean Effects On The Position Of The Curve.
- 11. Effect of Standard Deviation • Standard Deviation Effects On the Disperse of the Curve.
- 12. The Standard Normal Distribution • The Standard Normal Distribution is a Normal Distribution With a Mean of 0 and a Standard Deviation Of 1. • Formula Becomes: 2 2 2 z e f x
- 13. Representation of Standard curve • Total area under the curve is 1.
- 14. Properties of a Normal Curve 1. The mean, median, and mode are equal and are located at the center of the distribution (Highest Point = μ). 2. A normal distribution curve is continuous, unimodal and symmetric about the mean. 3. The total area under every normal curve is 1. 4. It is completely determined by its Mean and S.D.
- 15. Properties • The Area that Lies Within 1 Standard Deviation Of the Mean is Approximately 0.68, Or 68% • Within 2 S.Ds, About 0.95, Or 95% • And Within 3 S.Ds, About 0.997, Or 99.7%.
- 16. Properties
- 17. Standardizing the Variables • All Normally Distributed variables can be transformed into Standard Normally distributed variables by the given formula. x z value - mean standard deviation z
- 19. Questions Find Probability for each. a) P(0< z <2.32) b) P(z < 1.65) c) P(z > 1.91)
- 20. Questions Sol.a) P(0 < z < 2.32) 1. Look up the Area Corresponding to 2.32. It is 0.9898. 2. Then lookup the Area Corresponding to 0 in z-table. It is 0.500 3. Subtract the Two Areas: 0.9898 - 0.5000 = 0.4898. 4. Hence the probability is 0.4898, or 48.98%.
- 21. Questions Sol.b) P(z < 1.65) 1. Look up the area corresponding to z < 1.65 in Table Z. It is 0.9505. 2. Hence, P(z < 1.65) = 0.9505, or 95.05%.
- 22. Questions Sol.c) P(z > 1.91) 1. Look up the area that corresponds to z > 1.91. It is 0.9719. 2. Then subtract this area from 1.0000. 3. P(z > 1.91) = 1.0000 - 0.9719 =0.0281, or 2.81%.