1. 4.3 NORMAL PROBABILITY DISTRIBUTIONS The Most Important Probability Distribution in Statistics

2. A random variable X with mean  and standard deviation  is normally distributed if its probability density function is given by Normal Distribution

3. Normal Probability Distributions The expected value (also called the mean)  can be any number The standard deviation   can be any nonnegative number n There are infinitely many normal distributions

4. 7: Normal Probability Distributions 4 Parameters μ and σ n Normal pdfs have two parameters μ - expected value (mean “mu”) σ - standard deviation (sigma) σ controls spreadμ controls location

5. The effects of  and  How does the standard deviation affect the shape of f(x)?  = 2  =3  =4  = 10  = 11  = 12 How does the expected value affect the location of f(x)?

6. X 8 369120  A family of bell-shaped curves that differ only in their means and standard deviations. µ = the mean of the distribution  = the standard deviation µ = 3 and  = 1

7. X 369120 X 369 0  µ = 3 and  = 1  µ = 6 and  = 1

8. X 8 369120  X 8 369 0  µ = 6 and  = 2 µ = 6 and  = 1

9. EXAMPLE A Normal Random Variable The following data represent the heights (in inches) of a random sample of 50 two-year old males. (a) Create a relative frequency distribution with the lower class limit of the first class equal to 31.5 and a class width of 1. (b) Draw a histogram of the data. (c ) Do you think that the variable “height of 2-year old males” is normally distributed?

10. 36.036.234.836.034.638.435.436.8 34.733.437.438.231.537.736.934.0 34.435.737.939.334.036.935.137.0 33.236.135.235.633.036.833.535.0 35.135.234.436.736.036.035.735.7 38.333.639.837.037.234.835.738.9 37.239.3

14. 7: Normal Probability Distributions 14 68-95-99.7 Rule for Normal Distributions n 68% of the AUC within ±1σ of μ n 95% of the AUC within ±2σ of μ n 99.7% of the AUC within ±3σ of μ

16. 7: Normal Probability Distributions 16 Example: 68-95-99.7 Rule Wechsler adult intelligence scores: Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15) n 68% of scores within μ ± σ = 100 ± 15 = 85 to 115 n 95% of scores within μ ± 2σ = 100 ± (2)(15) = 70 to 130 n 99.7% of scores in μ ± 3σ = 100 ± (3)(15) = 55 to 145

17. Property and Notation Property: normal density curves are symmetric around the population mean , so the population mean = population median = population mode =  Notation: X ~ N(  is written to denote that the random variable X has a normal distribution with mean  and standard deviation .

18. Standardizing Suppose X~N(  Form a new random variable by subtracting the mean  from X and dividing by the standard deviation  : (X  n This process is called standardizing the random variable X.

19. Standardizing (cont.) (X  is also a normal random variable; we will denote it by Z: Z = (X   has mean 0 and standard deviation 1: E(Z) =  = 0; SD(Z) =   n The probability distribution of Z is called the standard normal distribution.

20. Standardizing (cont.) n If X has mean  and stand. dev. , standardizing a particular value of x tells how many standard deviations x is above or below the mean . n Exam 1:  =80,  =10; exam 1 score: 92 Exam 2:  =80,  =8; exam 2 score: 90 Which score is better?

21. X 8 369120 µ = 6 and  = 2 Z 0123-2-3.5 µ = 0 and  = 1 (X-6)/2

22. Pdf of a standard normal rv n A normal random variable x has the following pdf:

23. Z = standard normal random variable  = 0 and  = 1 Z 0123-2-3.5 Standard Normal Distribution.5

24. Important Properties of Z #1. The standard normal curve is symmetric around the mean 0 #2.The total area under the curve is 1; so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2

25. Finding Normal Percentiles by Hand (cont.) n Table Z is the standard Normal table. We have to convert our data to z-scores before using the table. n The figure shows us how to find the area to the left when we have a z-score of 1.80:

26. Areas Under the Z Curve: Using the Table P(0 < Z < 1) =.8413 -.5 =.3413 0 1 Z.1587.3413.50

27. Standard normal probabilities have been calculated and are provided in table Z. The tabulated probabilities correspond to the area between Z= -  and some z 0 Z = z 0 P(-  <Z<z 0 ) z0.000.010.020.030.040.050.060.070.080.09 0.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 10.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015

28. n Example – continued X~N(60, 8) In this example z 0 = 1.25 0.8944 = 0.8944 z0.000.010.020.030.040.050.060.070.080.09 0.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359 0.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.5753 0.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141 10.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621 1.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015 P(z < 1.25)

29. Examples n P(0  z  1.27) = 1.270 z Area=.3980.8980-.5=.3980

30. P(Z .55) = A 1 = 1 - A 2 = 1 -.7088 =.2912 0.55 A2A2

31. Examples n P(-2.24  z  0) = Area=.4875.5 -.0125 =.4875 z -2.24 0 Area=.0125

32. P(z  -1.85) =.0322

33. Examples (cont.) A1A1 A2A2 02.73 z -1.18 n P(-1.18  z  2.73)= A - A 1 n =.9968 -.1190 n =.8778.1190.9968 A1A1 A

34. P(-1 ≤ Z ≤ 1) =.8413 -.1587 =.6826 vi) P(-1≤ Z ≤ 1).8413.1587.6826

35. Is k positive or negative? Direction of inequality; magnitude of probability Look up.2514 in body of table; corresponding entry is -.67 6. P(z < k) =.2514.5.2514 -.67

36. Examples (cont.) viii).2810.7190

37. Examples (cont.) ix).9901.1230.8671

38. X~N(275, 43) find k so that P(x<k)=.9846

39. P( Z < 2.16) =.9846 0 2.16 Z.1587.4846 Area=.5.9846

40. Example Regulate blue dye for mixing paint; machine can be set to discharge an average of  ml./can of paint. Amount discharged: N( ,.4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable. Determine the setting  so that only 1% of the cans of paint will be unacceptable

41. Solution