4.3 NORMAL PROBABILITY DISTRIBUTIONS The Most Important Probability Distribution in Statistics

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

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

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

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)?

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

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

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

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?

7: Normal Probability Distributions 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

7: Normal Probability Distributions 16 Example: 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

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 .

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.

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.

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?

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

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

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

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.) 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:

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

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 ) z

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

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

P(Z .55) = A 1 = 1 - A 2 = = A2A2

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

P(z  -1.85) =.0322

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

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

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

Examples (cont.) viii)

Examples (cont.) ix)

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

P( Z < 2.16) = Z Area=

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

Solution