Math/Stat 352 Lecture 9. Section 4.5 Normal distribution

Transcription

1 Math/Stat 352 Lecture 9 Section 4.5 Normal distribution 1

2 Abraham de Moivre, Pierre-Simon Laplace ( ) A French mathematician, who introduced the Normal distribution in his book The doctrine of chances: or, a method for calculating the probabilities of events in play, first published in 1718 and considered the first textbook on Probability. A French mathematician and astronomer. Extended the Moivre s result in the book Analytical theory of probabilities, published in (So-called de Moivre-Laplace theorem) De Moivre-Laplace theorem suggests an approximation to the central part of Binomial distribution

3 Johann Carl Friedrich Gauss ( ) A German mathematician and scientist who contributed significantly to many fields, including number theory, statistics, analysis, differential geometry, geodesy, geophysics, astronomy and optics. Referred to mathematics as "the queen of sciences. Rigorously justified the method of least squares in 1809 using the Normal distribution for errors.

4 Charles Sanders Peirce, an American philosopher, logician, mathematician, and scientist Francis Galton, an English prolific scientist Wilhelm Lexis, an eminent German statistician, economist, social scientist, a founder of the interdisciplinary study of insurance. Coined the term Normal distribution around 1875

5 Karl Pearson (March 27, 1857 April 27, 1936) established the discipline of mathematical statistics Made the term Normal distribution popular

6 Binomial(100,0.9), np=90 dbinom(q, 100, 0.9) Concentrated around Number of succsses

7 Bin(100,0.1)np=1 0 Bin(100,0.9)np=90 dbinom(q, 100, 0.1) Only a small fraction of possible outcomes has not negligible probability (i.e. only small part can be seen in experiment) prob is very small (not 0!) here Number of succsses

8 Poisson? Binomial? Normal? Poisson(30) Binomial(1000,.03) N(30,30) dpois(q, 30) Poisson Binomial Normal Number of succsses

9 Poisson(1) Binomial(1000,.001] dpois(q, 1) dbinom(q, 1000, 1/1000) Number of succsses Number of succsses The distributions are not symmetric, np=λ=1 too small for normal apprxmtn.

10 Poisson? Binomial? Normal? Rule of thumb: If n is large (n > 100), p is small (p < 0.05), and both np and n(1-p) are not small (say >10) then B(n,p)~P(np)~N(np,np(1-p))

11 The Normal Distribution The normal distribution (also called the Gaussian distribution) is by far the most commonly used distribution in the sciences. It provides a good model for many, although not all, continuous populations. The normal distribution is continuous. The mean of a normal population may have any value, and the variance may have any positive value. Density Normal densities Mean= 0, Std=0.6 Mean= 7, Std=1 Mean= 0, Std=1 Mean= 0, Std=3 Mean= 5, Std=1 Properties of Normal pdf: Bell shaped curve, symmetric Centered at the mean Larger standard deviation gives flatter curve with longer tails Value of random variable 11

12 Normal R.V.: pdf, mean, and variance The pdf of a normal random variable with mean µ and variance σ 2, X ~ N(µ, σ 2 ), is s given by 1 ( ) 2 /2 2 x µ σ f( x) = e, < x< σ 2π If X ~ N(µ, σ 2 ), then the mean and variance of X are given by µ = µ σ X = σ 2 2 X Note: The normal pdf is symmetric, so the mean =median. 12

13 Normal distributions what are the most likely observations? % Rule X ~ N(µ, σ 2 ) pdf. About 68% of the observations are in the interval µ ± σ. About 95% of the observations are in the interval µ ± 2σ. About 99.7% of the observations are in the interval µ ± 3σ. The proportion of a normal observations that are within a given number of standard deviations of the mean is the same for any normal population. 13

14 Standard Normal Distribution The standard normal distribution is a normal distribution with mean 0 and variance 1, X ~ N(0, 1). Standard normal distribution is usually denoted by Z: Z ~ N(0, 1). We can convert the observations from any X ~ N(µ, σ 2 ), to the standard units : z = x µ σ This process is often called standardization. The value of X in standard units is often called z-score. Standard units tell how many standard deviations an observation is from the population mean. 14

15 Computing standard normal probabilities: Table A.2 P(Z < 0.47) =

16 Computing standard normal probabilities: Table A.2 P(Z 1.38) =P(Z < 1.38) = P(Z > 1.38)= Reminder: The total area under a pdf curve is 1.

17 Computing standard normal probabilities: Table A.2 P(0.71 < Z < 1.28) = P(Z < 1.28) P(Z < 0.71)

18 Computing standard normal probabilities: Table A.2 Symmetry in action P(Z < 0.67) = symmetry = P(Z > -0.67) P(Z < 0.67) = symmetry = 1- P(Z < -0.67)

19 Standard Normal Percentiles Given that P(Z < a)=0.95 find a. Here a is called 95 th percentile of Z. Inside the table I looked for Found and Used z-value corresponding to the midpoint (0.95) between the two available probabilities a= If an available probability is closer to the one we need, use the z-value corresponding to that probability. a =?

20 Finding Probabilities for any Normal Variable Problem: Let X ~ N(µ, σ 2 ), find P(a< X < b). The probability that X lies within any interval is given by the integral of the pdf of X over that interval: where P a < X < b = f x dd, 1 ( ) 2 /2 2 x µ σ f( x) = e, < x< σ 2π The integral that provides probability does not have a closed form solution. How to proceed? Solution: Convert X to standard normal ( standardize ) and use z-table. a b 20

21 Computing normal probabilities: any X ~ N(µ, σ 2 ) Let X ~ N(50, 25). Find P( 42 < X < 52) P(42 < X < 52)= P( 5 < Z < ) = P( 1.6 < Z < 0.4) = =

22 Computing Normal Percentiles: any Normal distribution Let X ~ N(50, 25). Find the 40 th percentile of X. Find 40 th percentile of standard normal: z = Destandardize : = a 55, thus a=

23 Example. A process manufactures ball bearings whose diameters are normally distributed with mean cm and standard deviation of cm. Specifications call for the diameter to be in the interval 2.5±0.01 cm. What proportion of the ball bearings will meet the specifications? Soln: Let X = diameter of a ball bearing. X ~ N(2.505, ). P( 2.49 < X < 2.51)= standardize= P( < Z < 0.63)= =

24 Example: ball bearings contd. Suppose the machine was recalibrated, so that the mean diameter is now 2.5 cm. To what value must the standard deviation be lowered, so that 95% of the diameters will meet the specifications. Soln. X = diameter of a ball bearing, X ~ N(2.5, σ 2 ). Want σ s.th. P( 2.49 < X < 2.51) = Standardize: P( σ < Z < ) = P( 0.00 σ σ < Z < 0.00 σ ) =0.95. Need z-values that satisfy this equation: = -0.01/σ and 1.96 = 0.01/σ. Thus σ=0.01/1.96=0.0051cm.

25 Linear Functions of Normal Random Variables Let X ~ N(µ, σ 2 ) and let a 0 and b be constants. Then ax +b ~ N(aµ+b, a 2 σ 2 ) Let X 1, X 2,, X n be independent and normally distributed with means µ 1, µ 2,, µ n and variances σ 1 2, σ 2 2,, σ n 2. Let c 1, c 2,, c n be constants, and c 1 X 1 + c 2 X c n X n be a linear combination. Then c 1 X 1 + c 2 X c n X n ~ N(c 1 μ 1 + c 2 μ c n μ n, c 2 1 σ c 2 2 σ c 2 n σ n 2 ) 25

26 Example A chemist measures the temperature of a solution in o C. The measurement is denoted C, and is normally distributed with mean 40 o C and standard deviation 1 o C. The measurement is converted to o F by the equation F = 1.8C What is the distribution of F? Soln: let X=temperature in o C, X ~ N(40, 1). Let Y= temperature in o F. Then Y = 1.8X +32. Since Y is a linear function of a normal random variable, Y has a normal distribution. Mean of Y= 1.8(40)+32= 104 o F. Variance of Y = (1.8) 2 (1)= So, Y ~ N(104, 3.24). 26

27 Distributions of Functions of Normals Let X 1, X 2,, X n be independent and identically normally distributed with mean µ and variance σ 2. Then 2 X ~ N μ σ,. n Let X and Y be independent, with X ~ N(µ X, 2 2 ) and Y ~ N(µ Y, σ ). Then Y σ X X + Y N μ + μ σ + σ 2 2 ~ ( X Y, X Y) X Y N μ μ σ + σ 2 2 ~ ( X Y, X Y) 27

28 SAMPLING DISTRIBUTION OF THE SAMPLE MEAN EXAMPLE: Students in an university have a weight distribution that is known to be N(150, 20). Let X1, X2,, X16 represent the weights of 16 randomly selected students from this university. If X is the average weight for this sample, find P( > 160). Solution: Since the sample came from a normal distribution, the sample mean has a normal distribution as well. X X ~N(μ, σ 2 /n )=N(150, 20 2 /16)=N(150, 25). Thus, X PX ( > 160) = P( > ) = PZ ( > 2) = 1 PZ ( 2) = =

29 EXAMPLE, CONTD. An elevator at this university has a capacity of 1500 pounds. What is the probability that 9 students who enter the elevator will have a safe ride, i.e. their total weight is less than 1,500 lb? Solution: The sample mean has a normal distribution: X ~ N(μ, σ 2 /n )= N(150, 20 2 /9)=N(150, 44.44). Also, X P( Total weight < 1500)=P( <1500/9)=P( <166.67). So, X PX ( < ) = P( > ) = PZ ( < 2.5) = X

30 Estimating the Parameters If X 1,, X n are a random sample from a N(µ,σ 2 ) distribution, µ is estimated with the sample mean X and σ 2 is estimated with the sample variance deviation s 2 = 1 n i=1 (X i X ) 2. n 1 As with any sample mean, the uncertainty (standard deviation) in which we replace with estimator of µ. s/ n X is σ /, if σ is unknown. The mean is an unbiased n 30