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4.3 NORMAL PROBABILITY DISTRIBUTIONS The Most Important Probability Distribution in Statistics Normal Distribution A random variable X with mean m and standard deviation s is normally distributed if its probability density function is given by x m (1/ 2) s e 2 1 f ( x) x s 2 where 3.14159 ... and e 2.71828 ... Normal Probability Distributions The expected value (also called the mean) mcan be any number The standard deviation s can be any nonnegative number There are infinitely many normal distributions Parameters μ and σ Normal pdfs have two parameters μ - expected value (mean “mu”) σ - standard deviation (sigma) μ controls location σ controls spread 7: Normal Probability 4 The effects of m and s How does the standard deviation affect the shape of f(x)? s= 2 s =3 s =4 How does the expected value affect the location of f(x)? m = 10 m = 11 m = 12 µ = 3 and s = 1 m 0 3 6 8 9 12 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 X µ = 3 and s = 1 m 0 3 6 3 12 µ = 6 and s = 1 m 0 9 X 6 9 12 X m 0 3 µ = 6 and s = 2 6 8 3 12 µ = 6 and s = 1 m 0 9 X 6 8 9 12 X 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? 36.0 34.7 34.4 33.2 35.1 38.3 37.2 36.2 33.4 35.7 36.1 35.2 33.6 39.3 34.8 37.4 37.9 35.2 34.4 39.8 36.0 38.2 39.3 35.6 36.7 37.0 34.6 31.5 34.0 33.0 36.0 37.2 38.4 37.7 36.9 36.8 36.0 34.8 35.4 36.9 35.1 33.5 35.7 35.7 36.8 34.0 37.0 35.0 35.7 38.9 68-95-99.7 Rule for Normal Distributions 68% of the AUC within ±1σ of μ 95% of the AUC within ±2σ of μ 99.7% of the AUC within ±3σ of μ 7: Normal Probability 14 Example: 68-95-99.7 Rule Wechsler adult intelligence scores: Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15) 68% of scores within μ±σ = 100 ± 15 = 85 to 115 95% of scores within μ ± 2σ = 100 ± (2)(15) = 70 to 130 99.7% of scores in μ ± 3σ = 100 ± (3)(15) = 55 to 145 7: Normal Probability 16 Property and Notation Property: normal density curves are symmetric around the population mean m, so the population mean = population median = population mode = m Notation: X ~ N(ms is written to denote that the random variable X has a normal distribution with mean m and standard deviation s. 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. 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. Standardizing (cont.) If X has mean m and stand. dev. s, standardizing a particular value of x tells how many standard deviations x is above or below the mean m. Exam 1: m=80, s=10; exam 1 score: 92 Exam 2: m=80, s=8; exam 2 score: 90 Which score is better? 92 80 12 z1 1.2 10 10 90 80 10 z2 1.25 8 8 90 on exam 2 is better than 92 on exam 1 µ = 6 and s = 2 0 3 6 8 9 12 X (X-6)/2 µ = 0 and s = 1 .5 -3 -2 -1 .5 0 1 2 3 Z Pdf of a standard normal rv A normal random variable x has the following pdf: f ( x) 1 s 2 e 12 ( xm2 ) s 2 , x Z ~ N (0,1) substitute0 for m and 1 for s pdf for the standard normal rv becomes ( z) 1 2 e 1 2 z 2 , z Standard Normal Distribution .5 -3 -2 -1 .5 0 1 2 3 Z = standard normal random variable m = 0 and s = 1 Z 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 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: Areas Under the Z Curve: Using the Table P(0 < Z < 1) = .8413 - .5 = .3413 .50 .3413 .1587 0 1 Z Standard normal probabilities have been calculated and are provided in table Z. P(- <Z<z0) The tabulated probabilities correspond to the area between Z= - and some z0 Z = z0 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 … 1 … … … … 1.2 … 0.01 … 0.00 … z … … … … Example – continued X~N(60, 8) X 60 70 60 P( X 70) P 8 8 P( z 1.25) 0.8944 0.8944 0.8944 0.8944 0.8944 0.8944 P(z < 1.25) = 0.8944 In this example z0 = 1.25 z 0.0 0.1 0.2 0.8438 0.8461 0.8849 0.8869 0.8888 0.05 0.5199 0.5596 0.5987 0.8485 0.8508 0.8531 0.8907 0.8925 0.8944 0.06 0.5239 0.5636 0.6026 0.07 0.5279 0.5675 0.6064 0.08 0.5319 0.5714 0.6103 0.09 0.5359 0.5753 0.6141 0.8554 0.8577 0.8599 0.8621 0.8962 0.8980 0.8997 0.9015 … … … 1.2 0.04 0.5160 0.5557 0.5948 … 0.8413 … 1 0.03 0.5120 0.5517 0.5910 … 0.02 0.5080 0.5478 0.5871 … 0.01 0.5040 0.5438 0.5832 … 0.00 0.5000 0.5398 0.5793 … … … … Examples Area=.3980 0 P(0 z 1.27) = 1.27 .8980-.5=.3980 z A2 0 .55 P(Z .55) = A1 = 1 - A2 = 1 - .7088 = .2912 Examples Area=.4875 Area=.0125 -2.24 0 P(-2.24 z 0) = .5 - .0125 = .4875 z P(z -1.85) = .0322 Examples (cont.) .9968 A1 A1 .1190 A2 A -1.18 0 P(-1.18 z 2.73) 2.73 z = A - A1 = .9968 - .1190 = .8778 vi) P(-1≤ Z ≤ 1) .6826 .1587 .8413 P(-1 ≤ Z ≤ 1) = .8413 - .1587 =.6826 6. P(z < k) = .2514 .2514 .5 .5 -.67 Is k positive or negative? Direction of inequality; magnitude of probability Look up .2514 in body of table; corresponding entry is -.67 Examples (cont.) viii) .7190 .2810 250 275 P( X 250) P( Z ) 43 25 P( Z ) P( Z .58) 1 .2810 .7190 43 Examples (cont.) ix) .8671 .1230 .9901 ix) P(225 x 375) P 22543 275 x 275 43 375 275 43 P(1.16 z 2.33) .9901 .1230 .8671 X~N(275, 43) find k so that P(x<k)=.9846 .9846 P ( x k ) P x 275 k 275 43 43 P z k 275 43 k 275 2.16 ( from standard normal table) 43 k 2.16 ( 43 )275 367 .88 P( Z < 2.16) = .9846 .9846 Area=.5 .4846 .1587 0 2.16 Z 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 Solution X = amount of dye discharged into can X ~ N( m, .4); determine m so that P( X 6) .01 x m 6 m 6 m .01 P( x 6) P .4 .4 P z .4 6 m .4 2.33 (from standard normal table) m = 6 - 2.33(.4) = 5.068