1. CHAPTER 6 6-1:Normal Distribution Instructor: Alaa saud Note: This PowerPoint is only a summary and your main source should be the book.

2. Outline: Introduction 6-1 Normal Distribution. 6-2 Application of the Normal Distribution. 6-3 The central Limit Theorem. Note: This PowerPoint is only a summary and your main source should be the book.

3. random variable  A random variable is a variable whose values are determined by chance. Discrete R.VContinuous R.V have a finite No. of possible value or an infinite No. of values can be counted. Variables that can assume all values in the interval between any two given values. CH.(5) CH.(6) Note: This PowerPoint is only a summary and your main source should be the book.

4. Many continuous variables, have distributions that are bell-shaped, and these are called approximately normally distributed variables Introduction

6. Mean' Median Mode. (b) Negatively skewed (c) Positively skewed (a) Normal

7. Normal Distribution The equation for a normal disribution Position parameters Shape parameter

8. (b) Different means but same standard deviations (a) Same means but different standard deviations (e) Different means and different standard deviations

9. A normal distribution is a continuous,symmetric, bell-shaped distribution of a variable. 1.A normal distribution curve is bell shaped. 2.The mean, median and mode are equaled and located at the center. 3.A normal distribution curve is unimodal. 4.The curve is symmetric about the mean. 5.The curve is continuous. 6.The curve never touches the x axis. 7.The total area under a normal distribution curve is equal to 1 or 100%. 8.Embirical Rule.. Properties of Normal Distribution :

10. Normal Distribution Empirical Rule: Normal Distribution 68% 95% 99.7 %

11. Standard Normal Distribution : The equation for a normal disribution

12. Standard Normal Distribution Empirical Rule: Standard Normal Distribution

13.  Procedure To Finding the area under Standard Normal Distribution: 1. To the left of any Z value 2.To the right of any Z value 3.Between any two Z values P(Z<a) P(Z>a)=1-P(Z<a) P(a<Z<b)=P(Z<b)-P(Z<a)

14. Example 6-1: Find the area to the left of z=1.99 P(Z<1.99)=0.9767

15. Example 6-2: Find the area to the right of z=-1.16 P(Z>-1.16)=1-P(Z<-1.16) =1-0.1230 = 0.8770

16. Find the area between z=1.68 and z=-1.37 Example 6-3: P(-1.37<Z<1.68)=P(Z<1.68)-P(Z<-1.37) =0.9535-0.0853 =0.8682

17. Example 6-4: Find probability for each a.P(0<z<2.32) b.P(z<1.65) c.P(z>1.91) P(0<Z<2.32)=P(Z<2.32)-P(Z<0) =0.9898-0.5000 =0.4898

18. P(Z>1.91)=1-P(Z<1.91) =1-0.9719 =0.0281 P(Z<1.65)=0.9505

19. Example 6-5: Find the z value such that the area under the standard normal distribution curve between 0 and the z value is 0.2123 0.5000+0.2123=0.7123 Z=0.56

20. Q(41): find the Z value Q(46):Find the Z value to the right of the mean so that (b)69.85% of the area under the distribution curve lies to the left of it ANC. Z=0.52 ANC. Z=-1.39

21. Q(4 ):Find the Z value to the left of the mean so that : (a) 98.87% of the area under the distribution curve lies to the right of it ? 1-0.9887=0.113 Z=-0.28