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

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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 Distributions

3 The Shape of Normal Distributions Normal distributions are bell shaped, and symmetrical around   Why symmetrical? Let  = 100. Suppose x = 110. Now suppose x =

4 Normal Probability Distributions The expected value (also called the mean) E(X) (or  )  can be any number n The standard deviation  can be any nonnegative number n The total area under every normal curve is 1 n There are infinitely many normal distributions

5 Total area =1; symmetric around µ

6 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)?

7 X  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

8 X X  µ = 3 and  = 1  µ = 6 and  = 1

9 X  X  µ = 6 and  = 2 µ = 6 and  = 1

10 X Probability = area under the density curve P(6 < X < 8) = area under the density curve between 6 and 8. ab b a P(6 < X < 8) µ = 6 and  = 2 0 X

11 X Probability = area under the density curve P(6 < X < 8) = area under the density curve between 6 and 8. ab b a P(6 < X < 8) µ = 6 and  = X

12 X Probability = area under the density curve P(6 < X < 8) = area under the density curve between 6 and 8. ab b a 68 68

13 ab Probabilities: area under graph of f(x) P(a < X < b) = area under the density curve between a and b. P(X=a) = 0 P(a < x < b) = P(a < x < b) f(x) P(a < X < b) X

14 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.

15 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.

16 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?

17 X µ = 6 and  = 2 Z µ = 0 and  = 1 (X-6)/2

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

19 Z = standard normal random variable  = 0 and  = 1 Z Standard Normal Distribution.5

20 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

21 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:

22 Areas Under the Z Curve: Using the Table P(0 < Z < 1) = = Z

23 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(- 

24 n Example – continued X~N(60, 8) In this example z 0 = = z P(z < 1.25)

25 Examples n P(0  z  1.27) = z Area= =.3980

26 P(Z .55) = A 1 = 1 - A 2 = = A2A2

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

28 P(z  -1.85) =.0322

29 Examples (cont.) A1A1 A2A z n P(-1.18  z  2.73)= A - A 1 n = n = A1A1 A

30 P(-1 ≤ Z ≤ 1) = =.6826 vi) P(-1≤ Z ≤ 1)

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

32 Examples (cont.) viii)

33 Examples (cont.) ix)

34 X~N(275, 43) find k so that P(x

35 P( Z < 2.16) = Z Area=

36 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

37 Solution

38 Solution (cont.)


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