Lesson 100: The Normal Distribution. HL Math - Santowski

Transcription

1 Lesson 100: The Normal Distribution HL Math - Santowski

2 Objectives Introduce the Normal Distribution Properties of the Standard Normal Distribution Introduce the Central Limit Theorem

3 Normal Distributions A random variable X with mean m and standard deviation s is normally distributed if its probability density function is given by f(x) where (1/ ) 1 e s and xm s e x

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

5 Total area =1; symmetric around µ

6 The effects of m and s How does the standard deviation affect the shape of f(x)? s= s =3 s =4 How does the expected value affect the location of f(x)? m = 10 m = 11 m = 1

7 m µ = 3 and s = X A family of bell-shaped curves that differ only in their means and standard deviations. µ = the mean of the distribution s = the standard deviation

8 m µ = 3 and s = X m µ = 6 and s = 1 0 X

9 m µ = 6 and s = X m µ = 6 and s = X

10 P(6 < X < 8) µ = 6 and s = 0 X Probability = area under the density curve P(6 a < X < 8) b = area under the density curve between 6 a and b8. X

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

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

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

14 The Normal Distribution: as mathematical function (pdf) f 1 xm 1 ( ( x) e s s ) Note constants: = e=.7188 This is a bell shaped curve with different centers and spreads depending on m and s

15 The Normal PDF It s a probability function, so no matter what the values of m and s, must integrate to 1! s 1 e 1 ( xm ) s dx 1

16 Normal distribution is defined by its mean and standard dev. E(X)=m = x s 1 xm 1 ( ) s e dx Var(X)=s = ( x 1 e s 1 xm ( ) s dx) m Standard Deviation(X)=s

17 **The beauty of the normal curve: No matter what m and s are, the area between m-s and m+s is about 68%; the area between m-s and m+s is about 95%; and the area between m-3s and m+3s is about 99.7%. Almost all values fall within 3 standard deviations.

18 Rule 68% of the data 95% of the data 99.7% of the data

19 Rule in Math terms ) ( 1 ) ( 1 ) ( 1 s m s m s m s m s m s m s m s m s m s s s dx e dx e dx e x x x

20 Standardizing Suppose X~N(ms Form a new random variable by subtracting the mean m from X and dividing by the standard deviation s: (Xms This process is called standardizing the random variable X.

21 Standardizing (cont.) (Xms is also a normal random variable; we will denote it by Z: Z = (Xms has mean 0 and standard deviation 1: E(Z) = m = 0; SD(Z) = s 1. 1 The probability distribution of Z is called the standard normal distribution.

22 µ = 6 and s = X (X-6)/ µ = 0 and s = Z

23 Pdf of a standard normal rv A normal random variable x has the following pdf: f ( x) Z pdf ( z) 1 s ~ N(0,1) e 1 e 1 ( xm ) s substitute z, 0, x z for m and 1 fors for thestandard normal rv becomes 1

24 Standard Normal Distribution Z Z = standard normal random variable m = 0 and s = 1

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

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

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

28 Standard normal probabilities have been calculated and are provided in table Z. P(- <Z<z 0 ) The tabulated probabilities correspond to the area between Z= - and some z 0 z Z = z

29 Example continued X~N(60, 8) X P( X 70) P 8 8 Pz ( 1.5) P(z < 1.5) = In this example z 0 = 1.5 z

30 Examples Area= z P(0 z 1.7) = =.3980

31 Examples A 0.55 P(Z.55) = A 1 = 1 - A = =.91

32 Examples Area=.4875 Area= P(-.4 z 0) = =.4875 z

33 Examples P(z -1.85) =.03

34 Examples (cont.) A A 1 A A 0.73 z P(-1.18 z.73) = A - A 1 = =.8778

35 vi) P(-1 Z 1) P(-1 Z 1) = =.686

36 6. P(z < k) = Is k positive or negative? Direction of inequality; magnitude of probability Look up.514 in body of table; corresponding entry is -.67

37 Examples (cont.) viii) P( X 50) P( Z ) 43 5 P( Z ) P( Z.58)

38 Examples (cont.) ix) ix) P(5 x 375) 575 x P P( 1.16 z.33)

39 X~N(75, 43) find k so that P(x<k)= P( x k) P P z k x k k (from standard normal table) 43 k.16(43)

40 P( Z <.16) = Area= Z

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

42 Solution X X =amount of dye discharged into can ~N( m,.4); determine m so that PX ( 6).01

43 X Solution (cont.) =amount of dye discharged into can X ~N( m,.4); determine 6m.4 so that PX ( 6).01 xm 6m 6m P( x 6) P P z m m.33(from standard normal table) = 6-.33(.4) = 5.068