1. 8 - 28 What does data from a normal distribution look like? The shape of histograms developed from small samples drawn from a normal population are somewhat unpredictable. Yet, the bell pattern is usually to some extent apparent. For large samples the representation of the bell curve is usually more visible. Small Sample n = 25 Large Sample n = 200

2. 8 - 29 The Normal Probability Density Function The two parameters are  (mean) and  2 (variance). The mean defines the location and the variance determines the dispersion.

3. 8 - 30 N(mu, sigma-squared) or x ~ N( ,  2 ) [1/{  (2  )}] exp[  (x  ) 2 / 2  2 ] Where exp(..) means e(.. ), where e is the transcendental constant. x ~ N( ,  2 ) means random variable x is distributed as Normal with the indicated mean and variance separated by comma in parentheses.

4. 8 - 31 Variance and Mean of Normal Distributions The figure above illustrates several normal distributions with identical variances. The only difference in the distributions is the central location, the mean. In the figure to the right there are two distributions with identical means, but with different variances.  = 1  = 2

5. 8 - 32 Using the Normal Probability Density Function Using the normal probability density function to determine the probability of some interval would be complicated. A special normal distribution, called the standard normal, can be used to determine probabilities for any normal random variable.

6. 8 - 33 The Standard Normal Distribution N(0,1) Sometimes called the Z - distribution

7. 8 - 34 Standard Normal Distribution: N(0,1) The standard normal distribution provides a basis for computing probabilities for all normal distributions. It has a mean of zero and a variance of one.

8. 8 - 35 Map from f(X) to f(Z) The techniques used to map or translate any normal random variable into a standard normal random variable is called a z-transform. Because the z-transform gives the standard normal unique status among normals, the standard normal is also referred to as the z-distribution.

9. 8 - 36 Reverse Map from f(Z) to f(X) The techniques used to map standard normal random variable to any normal random variable is reverse z- transform. z  =x , z  +  =x x is obtained by adding z standard deviations to the mean, x = z  +  When z= -4, x is  -4  z= 4, x is  +4  z= 0, x is .

10. 8 - 37 Standard Normal Distribution Table A table in the appendix (p. 521) contains probability calculations for various points in the distribution. The table provides the probability that a standard normal random variable will be between 0 and a specified value.

11. 8 - 38 Example 3 To compute the probability that a standard normal random variable will be between 0 and 1, look up the value 1.00 in the table. The table value of.3413 is the area under the curve between 0 and 1, which is also the probability that the random variable will assume a value in that interval..3413

12. 8 - 39 Example 4 Using the table, compute the probability that a standard normal random variable is between 0 and 2.34.

13. 8 - 40 Example 4 - Solution First, draw a picture. The table is constructed to give the probability between 0 and z. To determine the area under the standard normal curve use the table and find the area associated with z = 2.34. P(0 < z < 2.34) =.4904

14. 8 - 41 Example 5 Using the table, determine the probability that a standard normal random variable is between -1.29 and 0.

15. 8 - 42 Example 5 - Solution First, draw a picture. Unfortunately, z = -1.29 is not given in the table.

16. 8 - 43 Ex5 – Exploit symmetry Since the normal distribution is symmetric, P(-1.29 < z < 0) = P(0 < z < 1.29). P(0 < z < 1.29) =.4015. Thus, P(-1.29 < z < 0) =.4015.

17. 8 - 44 Area between two positive z values (Ex6) What is the probability that a standard normal random variable will be between 1.0 and 2.0? Problem: The table only contains probabilities from 0 to some value z.

18. 8 - 45 Ex6 First get two areas Determine the probability that z is between 0 and 2.0. P(0 < z < 2.0) =.4772 Determine the probability that z is between 0 and 1.0. P(0 < z < 1.0) =.3413

19. 8 - 46 Ex6 – Subtract the smaller # from Bigger # Subtract the unwanted probability. P(0 < z < 2) - P(0 < z < 1) = P(1 < z < 2).4772 -.3413 =.1359 what we got from the table the part we don’t want leaving the area of interest

20. 8 - 47 Z – transformations in computing probabilities

21. 8 - 48 Z - transformation The z-transformation can transform any normal random variable into a standard normal random variable. The z-transformation is denoted by z and given by the formula

22. 8 - 49 How does the transformation work? The numerator, x - , centers the z-distribution around zero. By subtracting the mean of the random variable from each data value, the mean of the resulting z random variable will be zero.

23. 8 - 50 Example of x before using the z transformation Find the probability that a normal random variable with a mean of 10 and a standard deviation of 20 will lie between 10 and 40.

24. 8 - 51 Applying z transform to each inequality term Applying the z-transformation to each term yields

25. 8 - 52 Map x~N( ,  2) on left to the z~N(0,1) on right P(10 < x < 40) = P(0 < z < 1.5)  = 10 mean  for z = 0  + 1.5  = 40  + 1.5  = 1.5

26. 8 - 53 Geometry before z transformation Find the probability that a normal random variable with a mean of 10 and standard deviation of 20 will be greater than 30.

27. 8 - 54 Algebra of z transform. Applying the z-transformation yields = P(z > 1). Then, write as: Part1 Minus Part2 P(z > 1) = P(0 < z < ) - P(0 < z < 1) =.5 -.3413 =.1587

28. 8 - 55 Map from x above to z below (Ex.8) Note that the X value of 30 transformed into the z-value of 1. In other words, 30 is one standard deviation away from the mean.

29. 8 - 56 Example 9, x~ N(12,1) A beer distributor believes that the amount of beer in a 12 ounce can of beer has a normal distribution with a mean of 12 ounces and a standard deviation of 1 ounce. If a 12 ounce beer can is randomly selected, find the following probabilities.

30. 8 - 57 Prob that beer makers are cheating. Example 9-A Find the probability that the 12 ounce can of beer will actually contain less than 11 ounces of beer. Let X = amount of beer in the 12 ounce can. X has a normal distribution with  = 12 and  = 1. x~ N(12,1) P(X < 11) = = P(z < -1) =.5 - P(-1 < z < 0) =.5 -.3413 =.1587

31. 8 - 58 Prob that they are cheating, x~ N(12,1).3413.1587

32. 8 - 59 Prob that beer makers are too generous: N(12,1) Find the probability that the 12 ounce can of beer will actually contain more than 12.5 ounces of beer. Let X = amount of beer in the 12 ounce can. X has a normal distribution with  = 12 and  = 1. P(X > 12.5) = = P(z >.5) =.5 - P(0 < z <.5) =.5 -.1915 =.3085

33. 8 - 60 Prob that beer makers are too generous.1915.3085

34. 8 - 61 Prob of slightly different kind of cheating N(12,1) Find the probability that the 12 ounce can of beer will actually contain between 10.5 and 11.5 ounces of beer. Let X = amount of beer in the 12 ounce can. X has a normal distribution with  = 12 and  = 1. P(10.5 < X < 11.5) = P(-1.5 < z < -.5) = P(-1.5 < z < 0) - P(-.5 < z < 0) =.4332 -.1915 =.2417

35. 8 - 62 Slightly different cheating prob from z.1915.2417